#data-science-and-ml

1 messages · Page 31 of 1

mighty patio
pastel warren
#

can some one help me out with this

#

perform a new Esemble Learning method called "Bagging" based on Voting on 19 decision tree classifiers

feral cave
#

cheers, I made it to separate the name with subjects, but how to split those subjects' column is still tricky for me🤦‍

alpine spruce
steady crypt
#

.

mint palm
#

whats that notation??

#

, i mean

boreal gale
# mint palm

just in case you haven't figured it out, cos is not plain cosine here, it's cosine similarity (it even said so above the equation)

mint palm
#

actually idk how but math notation that i know seems quite different

steady basalt
# mint palm

Doesn’t that mean that both satisfy the equation ?

#

@wooden sail your time to shine!

merry ridge
#

I'm confused why a comma is the part that you find confusing. The operator cos in this case is a function of two variables.

#

This is no different from writing f(x,y)

latent swan
#

not really data science related as much, but dont know where else ask this... i am using the mip library for a constraints programming problem. and when i optimize i get back a solution with status "OPTIMAL" but some constraints have been violated. doesnt it mean that it should be infeasible and only feasible solutions can be optimal??? i am so confused pls someone help :((

image.png image.png
bold timber
#

Hello guys, can anyone enlighten me on the purpose of save_best_only as true or false? I don't really understand this for making a decision on when to use save_best_only= True or save_best_only=False

hasty mountain
#

@desert oar using data from 30 pages of a book instead of simply 6 sentences also provided a better output, with a higher variety of words.

To be honest, I think that assigning arbitrary numbers to each word would work for a dataset of 6 sentences, as my vector embedding model assigned numbers that were, like... within range [6,7] for the most common characters and numbers within range ]0, 1] to the ones that were more rare, which made my model outputs being translated as almost always being the word corresponding to 1 and 6(the most common outputs were within range [1,6])
But, when I used 30 pages of a PDF book to make a word predictor model, this vector embedding indeed worked, generating a higher diversity of words, generating values that were way more closer to the values in the encoded data.
Thanks again for your patience. Now I think I'll try to adapt this for my RL algorithm.

#

However, I should say that, though my data had vectors like 2.4253 and my model almost correctly predicted 2.4162, it still couldn't predict correctly any of the words because of how close each vector value is from each other in my dictionary

#

I suppose this could be helped if I made a better model...or used more data...or even optimized my embedding model(which had a loss of around 3.618, while the actual word predictor had 0.496)
Also, I didn't arrange my data into sequences, so it could be directly fitted into the FCC layers. Maybe using attention layers instead of simple matrix multiplication could also help.

serene scaffold
bold timber
#

This image is the result of both which is on the left when I used save_best_only = False and on the right when I used save_best_only = True what's different between both? @serene scaffold

image.png
tacit basin
bold timber
tacit basin
#

The screen shot looks random to me 😜. I mean in relation to the question

warm jungle
#

pnumpy (https://github.com/Quansight/pnumpy) seems to speed up numpy.argsort quite a bit for large arrays, which is useful in my application. But it looks a bit like it might be abandonware (no commits for 2 years). Is there a fork or something similar that's actively maintained?

GitHub
GitHub - Quansight/pnumpy: Parallel NumPy seamlessly speeds up NumP...

Parallel NumPy seamlessly speeds up NumPy for large arrays (64K+ elements) with no change required to existing code. - GitHub - Quansight/pnumpy: Parallel NumPy seamlessly speeds up NumPy for large...

rugged comet
bold timber
#

But can you elaborate on the sentence "save the model at the end of every epoch"?

Is the difference between both (True/False) only referred to in epoch? @tacit basin

tacit basin
#

as opposed to false, where all model checkpoints will be saved, after each epoch regardless of the model performance

bold timber
bold timber
tacit basin
bold timber
tacit basin
#

filepath: string or PathLike, path to save the model file. e.g. filepath = os.path.join(working_dir, 'ckpt', file_name). filepath can contain named formatting options, which will be filled the value of epoch and keys in logs (passed in on_epoch_end). For example: if filepath is weights.{epoch:02d}-{val_loss:.2f}.hdf5, then the model checkpoints will be saved with the epoch number and the validation loss in the filename. The directory of the filepath should not be reused by any other callbacks to avoid conflicts.

bold timber
tacit basin
#

hope docs are correct in this case 🙂

young granite
#

someone knows a better way to insert new values to a df instead of creating a new df reindexing and then concat?

array_1 = df.index[df["X"] >= -140]
first_index = array_1[0]
array_2 = df.index[df["X"] <= 170]
last_index = array_2[len(array_2)-1]
new_df = df.loc[first_index:last_index]
new_df.at[first_index, "X"] = -140.00
new_df.at[last_index, "X"] = 170.00
new_df = new_df[['X', "X"]]
x = new_df.iloc[:, 1]
y = new_df.iloc[:, 0]

f = interp1d(x, y, kind='cubic')
x_int = np.linspace(start=x.min(), stop=x.max(), num=621)
y_int = f(x_int)
int_df = pd.DataFrame({"X": x_int, 'Y': y_int})

x_range = np.linspace(start=-160, stop=-140.5, num=40)
range_df = pd.DataFrame({"X":x_range, 'Y':np.NaN})
concat_df = pd.concat([range_df, int_df])
concat_df = concat_df.reset_index(drop=True)
concat_df.at[concat_df.index[0], 'Y'] = concat_df['Y'][concat_df.index[-1]]
concat_df = concat_df.interpolate(method='linear', axis=0)```
bold timber
tacit basin
bold timber
# tacit basin A little bit

I've run the EfficientNetB0 from tf.hub. In there, I used the data with scaling first and I get a score is 87%, but when I try to re-run again without scaling the data first I get a worse score of around 10%. Why did it happen?

As I know, EfficientNet has a compound scaling which is no need to scale the data first.

vestal spruce
#

does anyone knows an article or journal that tackles with fixing the tilted object with rotation so that it is parallel to the x and/or y axis?

bright pasture
#

ValueError: zero-size array to reduction operation maximum which has no identity

#

That's the error I'm getting while trying to preprocess audio files.

wooden sail
#

you're calling some flavor of np.max on an empty numpy array

#

!e

import numpy as np
x = np.array([])
np.amax(x)
arctic wedgeBOT
#

@wooden sail :x: Your 3.11 eval job has completed with return code 1.

001 | Traceback (most recent call last):
002 |   File "<string>", line 3, in <module>
003 |   File "<__array_function__ internals>", line 180, in amax
004 |   File "/snekbox/user_base/lib/python3.11/site-packages/numpy/core/fromnumeric.py", line 2793, in amax
005 |     return _wrapreduction(a, np.maximum, 'max', axis, None, out,
006 |            ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
007 |   File "/snekbox/user_base/lib/python3.11/site-packages/numpy/core/fromnumeric.py", line 86, in _wrapreduction
008 |     return ufunc.reduce(obj, axis, dtype, out, **passkwargs)
009 |            ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
010 | ValueError: zero-size array to reduction operation maximum which has no identity
bright pasture
wooden sail
#

i know you're not doing it like that, i'm just exemplifying. the idea is that you somehow loaded the audio file incorrectly, and so your array is empty

bright pasture
#

It also constantly keeps saying, "Torch not compiled with CUDA enabled"

bright pasture
#

I'm using conda, btw.

wooden sail
#

seems like you installed pytorch without cuda

bright pasture
#

But I can prove conda is on here.

wooden sail
#

you have to point pytorch to the correct cuda version when you install it though

#

if you install it without any flags, it doesn't bring gpu support

#

try this. do conda remove pytorch

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and then do conda install -c pytorch cudatoolkit=12.0 pytorch

bright pasture
#

U H

wooden sail
#

i guess you're remoting into a cluster or smth?

bright pasture
#

What on earth does that mean?

wooden sail
#

you're running this on your own computer?

bright pasture
#

Yes.

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I am

wooden sail
#

ok. sadly idk about that SSL error, never seen it before

#

does conda update --all work?

bright pasture
#

Same error.

wooden sail
#

ok this appears to be a moderately recent issue

mint palm
#

what exactly is attention weight? is it softmax_number*(key*query)??

bright pasture
#

But now...

bright pasture
wooden sail
#

seems conda doesn't have this version of cuda yet?

bright pasture
#

So... what do I do?

#

In your opinion, of course?

wooden sail
#

not use anaconda, if this is the only gpu you have, or ignore the warning and have stuff run on cpu (this'll be slow)

bright pasture
#

u h

#

Problem is, that error happens on python itself too.

wooden sail
#

wdym "on python itself"

#

conda is just a package manager

#

you could use pip to install the libs you want

bright pasture
#

I just checked.

#

Would I have to downgrade cuda somehow?

#

How would I do that?

#

And if so, would this mean that the latest version of cuda is incompatible?

bright pasture
wooden sail
#

yes, you'd have to downgrade cuda. the problem is that cuda is backwards compatible, but not forwards. i don't know if your gpu can work with an older cudatoolkit

bright pasture
#

For reference, I have a 3090.

coral nimbus
#

Hi, I have some questions about the data input for object recognition

#

I used roboflow for this project, which is to recognise plastic bottles

#

I labelled about 1000 plastic bottle images but the result turned out to be useless as it confuses other things with plastic bottles, and I have to hold the camera at a certain angle for it to recognise plastic bottles

#

What can I do to improve the quality of data input?

tacit basin
steady basalt
#

Mooooods

mighty patio
#

<@&831776746206265384>

silk garden
neat crescent
#

👍

mint palm
#

when we use transformer for video, while feeding for say 8 frames a time, and there are n patch from each frames, do we feed array of (8*n, something)?

fast rivet
#

error: "No values given for wildcard 'i'"
code:

configfile: "/home/comp/user/projects/python/emoClassifier/config.yaml"
n=[i for i in range(50)]
output = ["{work_dir}/emoClassifier/output/models/weighted/pt_save_pretrained_{n}/train_args/checkpoint-{i}/hamming_loss.png"]


rule all:
    input:
        expand(output,
                work_dir=config['workdir'],
                n=n,
                )


checkpoint fine_tuner:
    input:
        script="{work_dir}/emoClassifier/emo_classifier.py",
        dataset="{work_dir}/emoClassifier/output/bootstrapped_datasets/dataset_{n}"
    output:
        directory("{work_dir}/emoClassifier/output/models/weighted/pt_save_pretrained_{n}")
    conda:
        "emoClassifier_new.yaml"
    shell:
        "python3 {input.script} -w -ds {input.dataset} -sd {output}"


def loss_plotter_input(wildcards):
    checkpoint_output = checkpoints.fine_tuner.get(**wildcards).output[0]
    return expand("{work_dir}/emoClassifier/output/models/weighted/pt_save_pretrained_{n}/train_args/checkpoint-{i}/trainer_state.json",
        work_dir=wildcards.workdir,
        n=wildcards.n,
        i=glob_wildcards(os.path.join(checkpoint_output,"/train_args/checkpoint-{i}")).i,
    )


rule loss_plotter:
    input:
        script="{work_dir}/emoClassifier/loss_plotter.py",
        json=loss_plotter_input
    output:
        "{work_dir}/emoClassifier/output/models/weighted/pt_save_pretrained_{n}/train_args/checkpoint-{i}/hamming_loss.png"
    conda:
        "loss_plotter.yaml"
    shell:
        "python3 {input.script} {input.json} {output}"

question: How can I run this when wildcard i doesn't exist yet?

#

I'm using snakemake btw

candid garnet
# 
cbar.set_label("Reflectivity")```

If I'm generating plots (colormesh plots) for an experiment at a range of different angles using a for loop, how should I include a color bar in a way that it isn't constantly adding on a new bar each time?

Code:
```def plot_function(angle, params):
    rotation = np.degrees(rotation)

    params = {'mathtext.default': 'regular' }          
    plt.rcParams.update(params)
    plt.pcolormesh(x_axis, y_axis, reflectivity)
    cbar = plt.colorbar() # This is causing duplication
    cbar.set_label("Reflectivity")
    plt.savefig("path_xyz")

def main_heatmap_all_rotations():
    rotation_angles = np.linspace(0, np.pi/4., 90)
    
    for rotation_angle in rotation_angles:
        #Processing stuff removed
        plot_function(rotation_angle, other_params)```
#

for example 0 degrees vs 17 degrees

image.png image.png
mint palm
#

WHY decode side has 2 attentions but one in encoder?

silk garden
#

Hi! I wrote an answer but seemed to be self-promotion. So I would be really interested to continue the discussion with you. Here's my email : taha@edenai.co (or add me on Discord)

mighty patio
# candid garnet ```cbar = plt.colorbar() cbar.set_label("Reflectivity")``` If I'm generating pl...

Easiest way is probably to close and remake the figure every time:

import matplotlib.pyplot as plt
params = {'mathtext.default': 'regular' }          
plt.rcParams.update(params)

def plot_function(angle):
    rotation = np.degrees(rotation)
    fig, ax = plt.subplots()
    cs = ax.pcolormesh(x_axis, y_axis, reflectivity)
    cbar = fig.colorbar(cs, ax = ax)
    cbar.set_label("Reflectivity")
    fig.savefig("path_xyz")
    plt.close(fig)

I also removed params as a parameter, as you did not actually use the provided params in the function, and moved plt.rcParams.update() outside
I also changed to the object-oriented interface (fig.-, ax.-) instead of the pyplot functional interface (plt.-)

deft socket
#

Does anyone know how I can add multiple exogenous variables to the statsmodels ARIMA

#

I cant find anything about this in the documentation, and for now can only add one

bold timber
mint palm
#

can someone explain how GFLOPS is known?

silk garden
#

Hi @desert oar: I would like to continue this discussion with you without spamming the others or being accused of self-promotion. Happy to talk on Discord or by email: taha@edenai.co

gilded oasis
#

Hi, how can I extract only number of "O" in this dataframe

sly wadi
sly wadi
# gilded oasis

You should be able to fix it with mapping the formula to this function.

def fix_dict(s):
s = s.replace("'", '"')
return json.loads(s)

df["formula"] = df["formula"].map(fix_dict)

gilded oasis
dusty valve
#

Yellow line is real Walmart stock, blue line is what my model predicted

20221117_115049.jpg
#

Not half bad

deft socket
dusty valve
#

My own

#

Tensorflow

deft socket
#

Ahh an ML model

dusty valve
#

Yes

deft socket
#

I thought you were using an econometrics model like ARIMA

tidal bough
deft socket
#

The fit seems pretty good

dusty valve
#

There are a few in accuracies such as at the 120 - 150 days

bright pasture
#

When trying to train on my 3090 GPU, I'm getting this error.
OSError: [WinError 1455] The paging file is too small for this operation to complete. Error loading "C:\Users\phill\AppData\Local\Programs\Python\Python310\lib\site-packages\torch\lib\cufft64_10.dll" or one of its dependencies.

dusty valve
bright pasture
#

But for which drive? And for which initial and maximum size?

dusty valve
#

Type in performance in search bar

#

And open performance settings

#

I'm not on PC rn

#

So, there should be an advanced option

#

Look around and there should be paging size options

#

Set to manual and crank it up

#

Maybe 2 -3 gb

crisp jackal
#

Has anyone already implemented Item to Item collaborative filtering using pyspark?

bright pasture
#

How's this?

dusty valve
#

Depends

#

You put how much you need

#

I dunno how much you'll need

bright pasture
#

It's not like the paging file error tells me how much I need...

dusty valve
#

2-3 gb

deft socket
#

my out-of-sample fit for tesla stock realised variance

dusty valve
#

Ngl kinda looks over fitting

#

I did mine for 2 epochs

#

But other than that solid

deft socket
bright pasture
dusty valve
bright pasture
#

But it's 8 gb already.

dusty valve
#

Try C drive then

bright pasture
#

Why would the maximum be smaller than the initial?

#

No no, I mean that the initial is already set to 8 GB.

unknown.png
#

At least I think it is?

dusty valve
#

Then

bright pasture
#

How's this?

dusty valve
#

Alr

#

Try it urself

#

I got period 4 now

deft socket
#

like when you have a turning point

#

im facing that issue atm

dusty valve
#

If it's linear model

#

I used ELU for my activation which seemed to reduce that

deft socket
#

whats ELU?

dusty valve
#

Like relu but goes below zero

deft socket
#

As you can see, the lags here are pretty bad

dusty valve
#

Yes

#

What do your model layers look like

#

I used lstm with elu and dropouts

deft socket
dusty valve
#

K

bright pasture
# 
  File "D:\DiffSVC\diff-svc-main\run.py", line 15, in <module>
    run_task()
  File "D:\DiffSVC\diff-svc-main\run.py", line 11, in run_task
    task_cls.start()
  File "D:\DiffSVC\diff-svc-main\training\task\base_task.py", line 234, in start
    trainer.fit(task)
  File "D:\DiffSVC\diff-svc-main\utils\pl_utils.py", line 495, in fit
    self.run_pretrain_routine(model)
  File "D:\DiffSVC\diff-svc-main\utils\pl_utils.py", line 588, in run_pretrain_routine
    self.train()
  File "D:\DiffSVC\diff-svc-main\utils\pl_utils.py", line 1364, in train
    self.run_training_epoch()
  File "D:\DiffSVC\diff-svc-main\utils\pl_utils.py", line 1385, in run_training_epoch
    for batch_idx, batch in enumerate(self.get_train_dataloader()):
  File "C:\Users\phill\AppData\Local\Programs\Python\Python310\lib\site-packages\torch\utils\data\dataloader.py", line 681, in __next__
    data = self._next_data()
  File "C:\Users\phill\AppData\Local\Programs\Python\Python310\lib\site-packages\torch\utils\data\dataloader.py", line 1359, in _next_data
    idx, data = self._get_data()
  File "C:\Users\phill\AppData\Local\Programs\Python\Python310\lib\site-packages\torch\utils\data\dataloader.py", line 1325, in _get_data
    success, data = self._try_get_data()
  File "C:\Users\phill\AppData\Local\Programs\Python\Python310\lib\site-packages\torch\utils\data\dataloader.py", line 1176, in _try_get_data
    raise RuntimeError('DataLoader worker (pid(s) {}) exited unexpectedly'.format(pids_str)) from e
RuntimeError: DataLoader worker (pid(s) 18644, 17332) exited unexpectedly
Epoch 1: : 0batch [00:10, ?batch/s]```
#

I just got this error.

deft socket
#

I intend on trying some ML models, then combining them to get a combined forecast

dusty valve
#

Never heard of that

deft socket
#

this is when I add in the exogenous regressor, it tracks the direction better , but the magnitude is dampened for some reason

deft socket
#

to show that their model is better than the baseline

dusty valve
#

Whats the regressor look like

deft socket
#

*ARIMA(1,1)

deft socket
dusty valve
#

I gtg now, send all your stuff and ping me and I'll get back after

bright pasture
#

Can someone help?

#

I know HRL has to go.

deft socket
deft socket
#

ARIMA(p,q) is a combo of AR(p) and MA(q)

#

AR is a regression model with p lags of Y

#

MA is a regression model with q lags of error terms

#

(I is for integrated, but thats not important here)

agile cobalt
plush jungle
#

what's the difference between

np.sum(weights * x)```
and
```py
np.dot(weights, x)```
sly wadi
plush jungle
sly wadi
plush jungle
#

!e

import numpy as np

mat1 = np.array([1,2,3,4])
mat2 = np.array([1,2,3,4])

print(np.dot(mat1, mat2))
print(np.sum(mat1 * mat2))```
arctic wedgeBOT
#

@plush jungle :white_check_mark: Your 3.11 eval job has completed with return code 0.

001 | 30
002 | 30
plush jungle
#

these are the same so long as the inputs are equal length vectors right?

#

so this distinction only matters when multiplying vector * scalar or vector * matrix etc

sly wadi
#

I think they are equal as long as both arrays are 1-d arrays. (But I'm no expert)

plush jungle
#

ok thanks

gilded oasis
sly wadi
gilded oasis
gilded oasis
sly wadi
gilded oasis
#

I copied the code wrong, sorry.

sly wadi
gilded oasis
#

Its good thx !!!

3.png
sly wadi
iron basalt
# plush jungle so this distinction only matters when multiplying vector * scalar or vector * ma...

Yes, for two 1-d arrays: https://en.wikipedia.org/wiki/Dot_product#Algebraic_definition

Dot product

In mathematics, the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. In Euclidean geometry, the dot product of the Cartesian coordinates of two vectors is widely used. It is often called the inner product (or rarely projection prod...

#

Not sure why this wiki page decided to use colors.

tacit basin
#

what are the most hot / best paid 🙂 keywords in AI these days 🙂 ?

plush jungle
#

can someone help me understand backpropagation? If the forward function is like this:

    def forward(self,x):
        return sigmoid(np.sum(self.weights*x) + self.bias)```
then derivative of Error with respect to weights for the last layer what?
#

in my lecture notes it's formulated like this:

#

where z is the weighted sum

#

I only really understand the first term, which is just the error

#

but how do I get

δsigmoid/δz```
#

I've got this

def derivative_of_sigmoid(z):
    return np.exp(z)/((np.exp(z) + 1)**2)```
#

but what do I pass it?

#

and how do I know what

δz/δwi```
is in code?
inland eagle
#

does anyone know how to convert a column in a df from a float to int values using pandas?

serene scaffold
iron basalt
plush jungle
iron basalt
#

This function was chosen in part due to this property.

plush jungle
iron basalt
plush jungle
#

or just the last one?

iron basalt
#

Whenever simgoid(z) is needed.

plush jungle
iron basalt
thorn sable
#

s

#

alright thats very cool

#

ever since

vestal siren
#

Can someone help me with the negative log likelihood?

#

I don't understand what I did wrong?

#

Maybe the N?

delicate apex
# vestal siren

first glance, N is supposed to be the count of objects, and you have n = target_pred, which makes it an ndarray. I believe the mismatch means you might be getting unintended matrix multiplication instead of scalar-matrix

tidal bough
#

debugging without any sort of error whatsoever is a bad idea, but I notice something more obvious: surely np.divide is a two-argument function, not one-argument? 😛

delicate apex
#

!d numpy.divide

arctic wedgeBOT
#
numpy.divide 

numpy.divide(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'divide'>```
Divide arguments element-wise.
vestal siren
vestal siren
delicate apex
vestal siren
#

That worked, thank you!

runic igloo
vestal siren
# 
# Implement the following function to return all rows in X and Y such that the left 
# child gets all examples that are less than the split value and vice versa.
def w1_tree_split_data_left(X, Y, feature_index, split_value):
    """Split the data `X` and `Y`, at the feature indexed by `feature_index`.
    If the value is less than `split_value` then return it as part of the left group.
    
    # Arguments
        X: np.array of size `(n_objects, n_in)`
        Y: np.array of size `(n_objects, 1)`
        feature_index: index of the feature to split at 
        split_value: value to split between
    # Output
        (XY_left): np.array of size `(n_objects_left, n_in + 1)`
    """
    X_left, Y_left = None, None
    #################
    ### YOUR CODE ###
    #################
    return XY_left
#

Can someone help me with this problem? I received a hint that I have to append the output (Y) to the X vector of features, so that XY will have n_in + 1 columns. But I don't quite understand this hint?

umbral charm
#

But yea

copper willow
#

Hi guys, anyone have some advanced experiencie with google sheets that can assist me with something really quick? I dont think it'll take more than 10 mins

rugged comet
serene plume
# 
pairwise_similarities = 1 - pdist(self.encode(sentences), metric="cosine").astype(np.float16)
pairwise_similarities[pairwise_similarities < sim_threshold] = 0

Much cleaner and way less redundant work. Thanks for the pointer!

bright pasture
rugged comet
deft spire
#

Hey guys, I am absolute beginner in machine learning and neural networks and am willing to learn them, I don't have knowledge of higher maths (I'm at like full school level, knowing functions, trigonometry, partially exponent, logarithm), can you recommend any good books/free courses for such beginners?

rugged comet
# deft spire Hey guys, I am absolute beginner in machine learning and neural networks and am ...

Books
https://www.oreilly.com/library/view/ai-and-machine/9781492078180/
https://www.manning.com/books/deep-learning-with-python-second-edition

Courses
https://www.coursera.org/learn/introduction-tensorflow
https://www.udacity.com/course/intro-to-tensorflow-for-deep-learning--ud187

O’Reilly Online Learning
AI and Machine Learning for Coders
Manning Publications
Deep Learning with Python, Second Edition

In this extensively revised new edition of the bestselling original, Keras creator offers insights for both novice and experienced machine learning practitioners.

Coursera
Introduction to TensorFlow for Artificial Intelligence, Machine Lea...

Offered by DeepLearning.AI. If you are a software developer who wants to build scalable AI-powered algorithms, you need to understand how to ... Enroll for free.

Introduction to Virtual Reality | Deep Learning with TensorFlow | U...

Developed by Google and Udacity, the Intro to TensorFlow free online course teaches a practical approach to virtual reality for software developers. Learn online with Udacity.

deft spire
#

@rugged comet thanks a lot

deft spire
strange elbowBOT
bold timber
#

Anyone can enlighten me why I get so different results when I uses load_weights with the cloned model?

Please give me an insight about this because I've been stuck in there for 4 days 🙏

image.png
#

this is my checkpoint path

image.png
grand quarry
bold timber
grand quarry
bold timber
grand quarry
#

The variable test_data should be the same when you call evaluate()

#

The first time you evaluate and second time after you load the model

bold timber
bold timber
grand quarry
bold timber
grand quarry
#

Do

from copy import deepcopy

copied_test_data = deepcopy(test_data)

And do both evaluations on copied_test_data. Maybe that will work

bold timber
#

I get an error like this

image.png
#

I know another way by using .set_weights(model.get_weights()) but I think it just take from the last epoch during training not the best epoch

prime knot
#

push that

#

<_>

bold timber
prime knot
bold timber
prime knot
#

@bold timber can u send me ALL your code

grand quarry
vestal siren
# 
#At each node in the tree, the data is split according to a split criterion and each split is passed
#onto the left/right child respectively. Implement the following function to return all rows in X and Y 
#such that the left child gets all examples that are less than the split value and vice versa.
def w1_tree_split_data_left(X, Y, feature_index, split_value):
    """Split the data `X` and `Y`, at the feature indexed by `feature_index`.
    If the value is less than `split_value` then return it as part of the left group.
    
    # Arguments
        X: np.array of size `(n_objects, n_in)`
        Y: np.array of size `(n_objects, 1)`
        feature_index: index of the feature to split at 
        split_value: value to split between
    # Output
        (XY_left): np.array of size `(n_objects_left, n_in + 1)`
    """
    X_left, Y_left = None, None
    XY_left = []
    XY = np.append(X, Y)
    n = len(XY) 
    for row in range(n):
        if row[feature_index] < split_value:
            XY_left.append(row[feature_index])
            
    return XY_left
#

Hi, can anyone help me with this problem? I don't understand what I do wrong.

arctic wedgeBOT
#

Hey @bold timber!

It looks like you tried to attach file type(s) that we do not allow (.ipynb). We currently allow the following file types: .gif, .jpg, .jpeg, .mov, .mp4, .mpg, .png, .mp3, .wav, .ogg, .webm, .webp, .flac, .m4a, .csv, .json.

Feel free to ask in #community-meta if you think this is a mistake.

mint palm
#

UserWarning: User provided device_type of 'cuda', but CUDA is not available. Disabling
warnings.warn('User provided device_type of 'cuda', but CUDA is not available. Disabling')

#

I am training YOLOv7 but it is not using GPU

hoary wigeon
#

Anyone here have used KModes? I want to automate the procedure of finding the value of n_cluster in KModes.

lapis sequoia
#

Does anyone have a working example of using Grad Cam with PyTorch or tensorflow?

#

Activation maps*

silent stump
#

Hi guys does anyone know, is the coefficients essentially the slope in linear regression? and intercept the location? just trying to understand this scikit module. thx

dusty valve
#

I'm just wondering if my regressor is predicting wrong, I got 40 years of Walmart stock price data, split it into 1 week and the day after the week is the label. So the shape of train data is (2080, 7, 1)). 2080 weeks of prices. I take the last year worth of data for testing and verifying, so it's (52, 7, 1) as a shape. And I just do model.predict(x_test). But it seems too accurate. Did I mess up somewhere? ( I can paste code if required )

WMT_7.png
#

I trained for 10 epochs only

serene scaffold
dusty valve
#

It gets similar results with other stocks ( predicting them not training )

#

So is the prediction data wrong?

serene scaffold
vestal siren
serene scaffold
mint palm
serene scaffold
mint palm
#

torch

serene scaffold
vestal siren
serene scaffold
mint palm
serene scaffold
# mint palm

great, what is your python version and your os? please do not show any more screenshots--just text.

mint palm
# 
import torch
print(torch.__version__)
print(torch.cuda.is_available())
#

3.10

#

is it a cuda mismatch?

serene scaffold
#

please say your OS

vestal siren
lapis sequoia
mint palm
#

window 11 insider 22623.891

serene scaffold
#

one moment.

#

@mint palm please run pip install https://download.pytorch.org/whl/cu117/torch-1.13.0%2Bcu117-cp310-cp310-win_amd64.whl

mint palm
#

do i need to remove previous cuda

serene scaffold
#

no

mint palm
#

ok

serene scaffold
serene scaffold
mint palm
# 
PS C:\Users\rahul> pip install https://download.pytorch.org/whl/cu117/torch-1.13.0%2Bcu117-cp310-cp310-win_amd64.whl
WARNING: pip is being invoked by an old script wrapper. This will fail in a future version of pip.
Please see https://github.com/pypa/pip/issues/5599 for advice on fixing the underlying issue.
To avoid this problem you can invoke Python with '-m pip' instead of running pip directly.
Defaulting to user installation because normal site-packages is not writeable
ERROR: torch-1.13.0+cu117-cp310-cp310-win_amd64.whl is not a supported wheel on this platform.```
serene scaffold
#

if you have the C++ build tools, you might be able to build it from source

vestal siren
#

I now have this:

# 
#At each node in the tree, the data is split according to a split criterion and each split is passed
#onto the left/right child respectively. Implement the following function to return all rows in X and Y 
#such that the left child gets all examples that are less than the split value and vice versa.
def w1_tree_split_data_left(X, Y, feature_index, split_value):
    """Split the data `X` and `Y`, at the feature indexed by `feature_index`.
    If the value is less than `split_value` then return it as part of the left group.
    
    # Arguments
        X: np.array of size `(n_objects, n_in)`
        Y: np.array of size `(n_objects, 1)`
        feature_index: index of the feature to split at 
        split_value: value to split between
    # Output
        (XY_left): np.array of size `(n_objects_left, n_in + 1)`
    """
    X_left, Y_left = None, None
    XY_left = np.empty(0)
    XY = np.append(X, Y)
    n = len(XY) 
    for row in range(n):
        if row[feature_index] < split_value:
            XY_left.append(row[feature_index])
            
    return XY_left
serene scaffold
#

you will not be able to append to XY_left if it is an array, since the shape of arrays cannot be changed.

vestal siren
#

Ooooo

serene scaffold
#

if you keep it as a list, you might be able to convert it to an array at the very end, with return np.array(XY_left)

#

though that might cause an error depending on what is in XY_left

#

you can also do

return np.array([
    row[feature_index] for row in range(len(XY))
    if row[feature_index] < split_value)
])

and delete the for loop. But that assumes that your current for loop does what you need it to do. It might not.

vestal siren
serene scaffold
vestal siren
#

O true sorry, TypeError: 'int' object is not subscriptable

#

This is about the if statement line

serene scaffold
#

that probably means that row is an int. what type do you expect row to be?

vestal siren
#

I though int?

serene scaffold
#

what would 5[3] mean?

vestal siren
#

Not sure exactly? Is that possible?

serene scaffold
#

it's not.

#

the [3] part is a subscript, and the error message told you that ints aren't subscriptable.

#

remember that [ ] is a list when it's by itself, and a subscript when it's on the end of something.

#

anyway, you have for row in range(n). range(n) is an int iterator. but you named it row, so you probably want a sequence of some kind.

vestal siren
#

Yes, I wanted to return all rows in X and Y

vestal siren
serene scaffold
steady basalt
#

u referring to numpy array vs python list?

serene scaffold
#

you create the XY variable and never use it.

serene scaffold
vestal siren
#

I used that to get the size, so I could iterate over it

steady basalt
#

what are the main differences?

serene scaffold
vestal siren
serene scaffold
# steady basalt what are the main differences?

they are both sequences where you can overwrite elements, and they are both implemented as contiguous blocks of memory on the hardware side. but those are the only similarities.

arrays have to be homogenous, they can have more than one dimension, and the shape cannot be changed (though elements can be).

lists can be heterogenous, and the length is dynamic. but there is no such thing as a "2d list", because nested lists are entirely separate objects from their container.

serene scaffold
vestal siren
#

Yes that is what I thought

serene scaffold
# vestal siren Yes that is what I thought 
    for row in range(n):
        if row[feature_index] < split_value:
            XY_left.append(row[feature_index])

at what point in this loop are you getting values from X or Y to go in XY_left?

vestal siren
#

Oooo now I see

#

Nowhere

#

Should it be ```py
for row in X, Y

#

That's what I had first, but that didn't work either

serene scaffold
#

not quite. you would need to use zip for that

#

!zip

arctic wedgeBOT
#

The zip function allows you to iterate through multiple iterables simultaneously. It joins the iterables together, almost like a zipper, so that each new element is a tuple with one element from each iterable.

letters = 'abc'
numbers = [1, 2, 3]
# list(zip(letters, numbers)) --> [('a', 1), ('b', 2), ('c', 3)]
for letter, number in zip(letters, numbers):
    print(letter, number)

The zip() iterator is exhausted after the length of the shortest iterable is exceeded. If you would like to retain the other values, consider using itertools.zip_longest.

For more information on zip, please refer to the official documentation.

steady basalt
#

arrays more efficient than nested lists then

vestal siren
#

Oooo okay

steady basalt
#

i wonder how they get the computer to treat arrays not like nested lists

serene scaffold
# steady basalt ah ok

novices often use "list" and "array" interchangeably, but this is actually just as coherent as using "list" and "tuple" interchangeably.

vestal siren
#

But what I also didn't understand was that there is hint that says: you can append the output (Y) to the X vector of features so XY will have n_in+1 columns

#

Could you maybe clarify that that?

serene scaffold
#

and the computer can get every 5th element efficiently with pointers and some basic math

steady basalt
#

why does python not want to do that

serene scaffold
serene scaffold
vestal siren
serene scaffold
vestal siren
#

feature_index = 1
and split_value = 1.9751321507065613

#

I don't know what the expected output is, because my code doesn't work

#

(XY_left): np.array of size (n_objects_left, n_in + 1)

serene scaffold
vestal siren
#

Unfortunately, he won't provide them😓

#

I can only run my code code and see if it works

dusk tide
#

Is balancing the dataset Is necessary to implement transfer learning??

vernal anchor
#

hey can someone help me?

#

I'm trying to create something like this

#

I have some values in a list and I want to see how many clusters are created and what are the boundary values of each other

#

I could really use some help cause I have a deadline coming up
thanks in advance!

#

oh btw I already asked several times at the help channels but didn't receive much help

steady basalt
#

pure python?

#

no package?

silver widget
#

Hi all . I need advice on code performance

for col in self.df.select_dtypes(exclude='object').columns:
     self.table.loc[self.table.variable_name == col, f"{self.category}: {cat} \n mean ± sd "] = \
                    f"{self.df[self.df[self.category] == cat][col].mean():.2f} ± " \
                    f"{self.df[self.df[self.category] == cat][col].std():.2f}"

This is my code to create a table(pd.dataframe object) which includes n counts and % of each category of each column
but streamlit raises performance error to change the code to pd.concat(axis=1)
can anyone suggest how to improve the performance here?

zenith shard
#

You're doing self.df[self.df[self.category] == cat] multiple times

#

you can do it once and cache the result

silver widget
#

Can you show me an example how to do it in one iteration?

#

or the idea

desert oar
arctic wedgeBOT
#

Here's how to format Python code on Discord:

```py
print('Hello world!')
```

These are backticks, not quotes. Check this out if you can't find the backtick key.

silver widget
desert oar
desert oar
#

also just for your own sanity, i'd avoid doing computations inside f-strings. makes the code hard to follow

# 
for col in self.df.select_dtypes(exclude='object').columns:
     colname = f"{self.category}: {cat} \n mean ± sd "
     mask = self.df[self.category] == cat
     m = self.df[mask, col].mean()
     s = self.df[mask, col].std()
     self.table.loc[self.table.variable_name == col, colname] = f"{m:.2f} ± {s:.2f}"

here's how i'd rewrite your code @silver widget . but again i think we can make this even more efficient

#

also, this looks questionable self.table.variable_name == col

fallen crown
#

Hi, i don't understand something with my model, I used the same dataset for training and validation so my score should be equal to 1 ? But it is actually equal to 0.83, does somebody have an idea of what happens ?

desert oar
fallen crown
#

okkey thanks !

desert oar
#

@fallen crown as always, i suggest working through some actual equations and convincing yourself of this, instead of taking my word for it

#

consider the model y = ax + b with parameters a,b, but try to fit that to data generated from y = x^2 + 3

#

@silver widget self.table probably should be using variable_name as its index, unless you already have another index that you are using

fallen crown
desert oar
#

also @silver widget what are self.category and cat? i assume the former is supposed to be a column name and the latter is supposed to be a specific value thereof?

river sapphire
#

how does backpropagation work with 2 or more hidden layers?

desert oar
river sapphire
silver widget
river sapphire
#

and then sum it up?

#

for multiple outputs ofc

desert oar
#

@river sapphire the chain rule uses multiplication, not addition. but yes, you end up with a separate partial derivative for every single weight in the model

river sapphire
#

the article i'm reading says to add the orange and blue essentially

desert oar
#

can you post the article?

river sapphire
#

this is what they said

unknown.png
#

but from what i'm understanding if I have say a weight^1 subscript 11 with 2 hidden layers and 2 hidden nodes per layer with 2 outputs

#

that weight would affect the 1st node in the 1st hidden layer but also all the nodes in the second hidden layer

#

which would affect all the outputs so i'm wondering how you would calculate the chain derivative path

#

or I mean better stated how would you calculate the partial derivative of the total error with respect to that weight

desert oar
#

@river sapphire i see, is this a network with two outputs?

river sapphire
desert oar
river sapphire
#

so do you calculate all the chain derivative paths then add them?

desert oar
#

remember that we are interested in the vector of partial derivatives with respect to every individual weight

river sapphire
#

i'm kinda confused

desert oar
#

don't guess at the math here

#

go back to one output

#

rather, start there

#

and look at how the article builds this up from the loss function. remember that we are computing the gradient of the loss function, so we must start there

#

the sum arises because, in the words of the article:

Because our cost function is a summation of individual costs for each output, we can calculate the derivative chain for each path and simply add them together.

#

you also need to sum up over observations in the dataset/batch

#

but that's because the derivative of a sum is just a sum of the derivatives. the important part is understanding each of those summed-up derivatives, and that's the chain rule.

wooden sail
#

hmm

#

ah for a moment i thought linearity was being conflated with chain rule, but no. all good 😛

desert oar
#

phew, i get nervous when i'm talking about neural network math and you or emyrs or squiggle show up to correct me 😆

river sapphire
#

wait ok so

image.png
serene scaffold
#

I feel excluded. but you're right 😢

desert oar
serene scaffold
#

that's my point pepe_laugh

river sapphire
#

shoot i'm still confused

desert oar
#

it seems like you're focused on the latter too much

#

i also am going to suggest not reading past the "Generalizing a Method" headline until you get the first part comfortably. because understanding the first part is essential to making sense of the "generalizing" part

#

it sounds incredibly tedious, but sit down with a notepad and try to re-derive the equations that the author derived

river sapphire
#

alright

desert oar
#

maybe not all of them because that's a lot of writing, but derive one of the "layer 1" partial derivatives and one of the "layer 2" partial derivatives

#

remember: you're looking for the partial derivative of the loss function with respect to one of those parameters. stay focused on that and your calculus fundamentals, and you should be OK.

wooden sail
#

in the same link you shared, you should really focus on the case with a single "hidden layer"

#

the rest kinda telescopes from there quite naturally

river sapphire
#

ok I'll try that

wooden sail
#

this is basically all you'll need

image.png
river sapphire
#

for the notation of the biases do you say b^1 or b^2

wooden sail
#

you can if you want

#

probably with parentheses to not confuse it with exponentiation

desert oar
#

notation whenever you have multiple "indexes" always gets messy

river sapphire
desert oar
#

when i write these things by hand, i do use superscript indexes, but i try to write them a bit lower than i would write an exponent

wooden sail
#

the best kept secret in math is that the notation means nothing alone 😛 you can define whatever notation you want, and that is also what everyone else does. so the first thing to do when reading something is figure out what the specific notation in what you're reading means

river sapphire
#

the problem is when the notation everyone else uses in their articles is one that you don't understand

desert oar
#

you can also put a , in the subscript to separate different indexes, or another option is to write each item as if it were a function, using () and no subscripts. or you can even use numpy-style array indexing

desert oar
wooden sail
#

it helps if you take a moment with pen and paper to translate the expressions into the notation you prefer

#

(and it's essentially necessary as you move into more complex reads)

fallen crown
#

@desert oar I use a linear regression model and fit it with data generated from y = x**2+3 like that ```python
x = np.arange(1,50)
y = np.array(list(map(lambda x : x**2+3, dataset)))

model = LinearRegression()
X = x
Y = y

X = X.reshape(49,1)
Y = Y.reshape(49,1)

# 
model.fit(X,Y)
model.score(X,Y)```
#

score is 0.93

desert oar
fallen crown
#

But i still don't get it.... What is the meaning of 0.93

#

okkey

desert oar
#

as for what each score function means, that depends on the score function

fallen crown
#

Ok i understand

#

i will read the docs

wooden sail
#

in general though, the value of the score alone means nothing 😛

#

you don't actually care about the minimum, you care about the minimizer (the parameters)

desert oar
#

the objective function is treated like an implementation detail in sklearn

wooden sail
#

even so, it's useful to keep in mind that no single metric or score can tell you the whole story

desert oar
#

very true

#

r-squared is also called the "coefficient of determination" which other people have explained better than i can https://stats.stackexchange.com/questions/tagged/r-squared https://en.wikipedia.org/wiki/Coefficient_of_determination

Cross Validated
Newest 'r-squared' Questions

Q&A for people interested in statistics, machine learning, data analysis, data mining, and data visualization

Coefficient of determination

In statistics, the coefficient of determination, denoted R2 or r2 and pronounced "R squared", is the proportion of the variation in the dependent variable that is predictable from the independent variable(s).
It is a statistic used in the context of statistical models whose main purpose is either the prediction of future outcomes or the testing ...

wooden sail
#

r-squared is also an affine transformation of the usual 2-norm error/euclidean distance/mean squared error

#

just scaled and shifted

desert oar
#

hm, that's one way to think about it. but the scale and shift is dependent on the data

wooden sail
#

yes, but those things don't affect the gradient. the constant vanishes and the scale is multiplied to all the gradients

#

so in that sense it's equivalent

#

you could choose r-squared as your cost func and you'd get exactly the same parameters as output

desert oar
#

true, so it's an affine transformation of the loss function when fitting with least squares

#

the statistical interpretation though is that it's the % of variance in y explained by variance in x

wooden sail
#

indeed

vestal siren
# 
def w2_flatten_forward(x_input):
    """Perform the reshaping of the tensor of size `(K, L, M, N)` 
        to the tensor of size `(K, L*M*N)`
    # Arguments
        x_input: np.array of size `(K, L, M, N)`
    # Output
        output: np.array of size `(K, L*M*N)`
    """
    vol_shape = x_input.shape[:-1]
    n_voxels = np.prod(vol_shape)
    output = x_input.reshape(n_voxels, x_input.shape[-1])
    return output
#

Question: The Flatten layer receives a 4-dimensional tensor of size (n_obj, n_channels, h, w) as its input and reshapes it into a 2-dimensional tensor (matrix) of size (n_obj, n_channels * h * w).

#

Can anyone help my why I get a wrong output?

wooden sail
#

you're doing something weird

#

also it seems they want you to unfold in fortran order

#

my suggestion would be return x_input.reshape(x_input.shape[0], -1, order='F')

#

what's your thinking when doing vol_shape = x_input.shape[:-1] ?

#

lookie here

#

!e

import numpy as np
x = np.zeros((2,3,4,5))
print(x.shape[:-1])
arctic wedgeBOT
#

@wooden sail :white_check_mark: Your 3.11 eval job has completed with return code 0.

(2, 3, 4)
wooden sail
#

what that does is ignore the last index, when what you want is to only keep the first

#

if anything you could've used 1:

vestal siren
wooden sail
#

that's the same as i did, but you counted from the wrong direction :p

vestal siren
#

Oh wait, if I replaced it by 1, I get this error: ValueError: cannot reshape array of size 76800 into shape (100,16)

vestal siren
#

are there maybe some edge cases that are missing?

wooden sail
#

no

#

it does exactly what you asked for for any 4D array

#

keep the 1st dimension untouched and unfold the other 3

#

if you want a special unfolding order different from fortran order/matlab unfolding order, some changes are needed. but given that you said it worked on the test case, it should be fine

#

what are you testing it on

vestal siren
#

Automark in notebook

wooden sail
#

i mean on what array lol

#

like what exactly are you doing? show some code

vestal siren
#

test_input = np.zeros((100, 3, 16, 16))

print(w2_flatten_forward(test_input).shape)

#

This was in the description: You can use test data and compare the final shape. It should be (100, 768) for the following example.
Please ignore the use of np.zeros in this case. We are just interested in transforming shapes.
Be aware: This test will fail if you do not return an array like object!

wooden sail
#

sure, so

#

!e

import numpy as np
x = np.zeros((100,3,16,16))
x = x.reshape(100, -1, order='F')
print(x.shape)
vestal siren
arctic wedgeBOT
#

@wooden sail :white_check_mark: Your 3.11 eval job has completed with return code 0.

(100, 768)
wooden sail
#

yes

vestal siren
#

Hmm, then I'm not sure why it would fail

wooden sail
#

it could be that the test cases ignore the ordering. you can remove the order='F' part and try it out

bitter wasp
#

@wooden sail

wooden sail
#

though that's technically wrong

bitter wasp
#

Hey Edd

vestal siren
#

Weird

wooden sail
#

yeah well, that's also transposing

bitter wasp
#

Is there anyone have Discord bot py system code?

wooden sail
#

i would argue it's incorrect because the size of the first dimension SEEMS the same as in the original array, but the samples are now different!

#

look at this example

#

!e

import numpy as np
x = np.random.rand(2,3)
print(x)
print(x.reshape(6, order='F'))
print(x.reshape(6))
arctic wedgeBOT
#

@wooden sail :white_check_mark: Your 3.11 eval job has completed with return code 0.

001 | [[0.59967224 0.86778639 0.86594742]
002 |  [0.50774334 0.52727131 0.88028635]]
003 | [0.59967224 0.50774334 0.86778639 0.52727131 0.86594742 0.88028635]
004 | [0.59967224 0.86778639 0.86594742 0.50774334 0.52727131 0.88028635]
wooden sail
#

there we go

vestal siren
#

Aha

wooden sail
#

look at the first column in the matrix. it has 0.599 etc, then 0.5077 etc

#

so if i do something that preserves the size of this first dimension (the columns), i would also expect the samples themselves to remain the same

vestal siren
#

Ooo I see

wooden sail
#

so with order='F' we stack the columns on top of each other. but if you remove that, numpy uses C ordering by default, which stacks ROWS, not columns

#

that preserves the entries of the rows, and in doing so shuffles the columns

#

i would claim reshaping this way makes no sense 😛 but anyway. ask your lecturer to be clear with the ordering because it matters

vestal siren
#

Yes I will for sure ask about this

#

Thanks for your help!

#

I also had another question about decision trees

wooden sail
#

i have to go sleep :x someone else will help you out

vestal siren
#

Okay no worries

#

Good night

soft badge
#

Guys what do you recomend for learn machine learning

serene scaffold
arctic wedgeBOT
#
Resources

The Resources page on our website contains a list of hand-selected learning resources that we regularly recommend to both beginners and experts.

soft badge
#

!resources data-science

arctic wedgeBOT
#
Resources

The Resources page on our website contains a list of hand-selected learning resources that we regularly recommend to both beginners and experts.

serene scaffold
soft badge
#

ok, i clicked

#

do you guess dificult learn this?

serene scaffold
soft badge
#

i talk about make models i domine this

#

dont be searched right?

serene scaffold
#

I'm sorry, but I do not understand what you are saying.

soft badge
#

I say to develop applications with machine learning, do you need to have a high level of knowledge?

serene scaffold
soft badge
#

professional

serene scaffold
river sapphire
#

according to wolfram it says:

image.png
#

and this is the cost function they used in the article i'm reading

image.png
#

but the article i'm reading says this:

unknown.png
#

wouldn't it be a - y?

desert oar
river sapphire
#

they're saying you take the derivative of the cost function with respect to the output

desert oar
#

a being the model output?

river sapphire
#

yes

desert oar
#

what's f'?

river sapphire
#

that one's just the partial derivative of the output with respect to the activation function

desert oar
#

is f the activation function?

river sapphire
#

wait

desert oar
#

can you share the actual article?

river sapphire
#

now i'm confused lol

#

it's the same article

desert oar
#

oh

#

good on you for digging into the equations btw

river sapphire
#

a would be the activated version of z

desert oar
#

i don't particularly like their notation here

river sapphire
#

yeah it's kinda confusing

desert oar
#

the (3) and a are both somewhat nonstandard and therefore mysterious

#

i suppose they mean a to be mnemonic for "activation", which makes sense

#

i see, a(3) is the activation after layer 3, treating the input as layer 1

river sapphire
#

i have no clue what the f'(a(3)) means

desert oar
#

i don't know why they didn't call the input layer 0

#

but okay, fine. a(3) is the output of the 3rd layer including activation function, which in this case is the output layer, and therefore the output of the entire model

river sapphire
#

a(3) would be the final output wouldn't it

desert oar
#

yep

river sapphire
#

so why do they do f'(a(3))?

desert oar
#

i'm getting there 😛

river sapphire
#

wouldn't it be like a'(3)

desert oar
#

hang on, still getting there 😛

#

ah, you're getting into the "Generalizing a method" part now

river sapphire
#

yeah the part where the combine first and second column

desert oar
#

i think this is meant to sketch out the general form of these δ things

#

i think it's meant to be a placeholder for "the derivative of some other function, evaluated at a(3)"

#

meaning that δ(3) has this very general format, no matter what else is in layers 1 and 2

river sapphire
#

no it says they're combining the first and second column so

#

that means combining these

image.png
desert oar
#

yeah. let me read this more carefully

#

no, still think that's valid

river sapphire
#

my first question was what mysterious function is f being performed on a

#

or is that just to indicate the partial derivative of a with respect to z?

desert oar
#

yes, i think that's what f'(a(3)) is supposed to be

#

and i think the author is trying to de-emphasize what exactly f' looks like, in order to make the rest of the "shape" of the expression clearer

river sapphire
#

maybe

#

it was confusing to me

desert oar
#

it was confusing to me too, they didn't explain it

river sapphire
#

also second question say m=2 therefore the cost would be 1/2 * (y - a)

desert oar
#

i only understood what it meant because i know that the first factor must be (1/m) (y - a(3)), hence the other factor must be the f'(a(3))

river sapphire
#

the partial derivative of the cost with respect to a should be a - y according to wolfram

desert oar
river sapphire
#

oh so it would be 1/2 * ((y[i] - a[i])^2 + (y[i2] - a[i2])^2)?

desert oar
#

oh sorry. the (1/2m) is in front

river sapphire
#

but then they say 1/2m sigma (y[i] - a[i])^2

desert oar
#

so the loss for 2 data points is (1 / 2) * (1 / 2) * ((y[1] - a[1])^2 + (y[2] - a[2])^2)

river sapphire
#

yes then the partial derivative would be what then

#

of the cost function with respect to the neural network outputs

desert oar
#

with respect to a[i]? that's a good homework assignment 😛

#

don't use wolfram alpha! this is power rule & chain rule, you should be able to eyeball this with some practice

river sapphire
#

I know it's just i'm confused because other websites said it's a - y but when I did it it was different

desert oar
#

well you can write a - y or y - a and it's the same, because you're squaring it

#

stick to whatever the author does, to keep your life simple

river sapphire
#

no the derivative

desert oar
#

just take the derivative of the loss function as written, don't worry about what other websites say

river sapphire
#

ok one last question so in your calculation of the loss for 2 data points

#

you said 1/2 * 1/2 right

#

oh wait ok so it's 1/4 * the entire squared error loss?

desert oar
#

then you results will match the author's

river sapphire
#

why do some people use sigma 1/2(target-output)^2

desert oar
#

that's what the author here uses too

#

oh, why do people not divide by m?

river sapphire
#

yeah

#

no like there's a difference because if you put 1/2 outside the sigma

#

it's saying 1/2 times the squared error loss

#

but when you put it inside it's saying 1/2 * each loss

#

this is confusing to explain i'll just show you an image

desert oar
#

i see your confusion

river sapphire
desert oar
#

you know that the big sigma means "summation" right?

river sapphire
#

yes

desert oar
#

do you agree that these two are equivalent @river sapphire ?

c*a[1] + c*a[2] + c*a[3]

c * (a[1] + a[2] + a[3])
river sapphire
#

yeah

#

why does this person drop the m though

desert oar
#

well there's your answer 🙂

desert oar
river sapphire
#

oh

desert oar
#

the 1/2 just makes the output nicer when you apply the power rule

#

(which you will hopefully see when you work through it)

#

it's a general fact about convex optimization that a monotonic transformation of the function being maximized does not change the argmax

#

draw some pictures and convince yourself of that

river sapphire
#

ok

rugged comet
soft badge
#

guys about study of ia, is most theoric or practice?

iron basalt
#

*And any other bugs.

rugged comet
iron basalt
#

I would not have written it this way though.

#

But this kind of loose notation with pattern matching required by the reader is common in ML.

iron basalt
rugged comet
#

Thank you.

soft badge
#

squiggle, for learn machine learning what do you recommend?

serene scaffold
#

@soft badge I don't think you answered my earlier question. or you didn't ping me when you did.

serene scaffold
soft badge
soft badge
serene scaffold
# soft badge IA is math pure right?

it's "AI" in English--some people might not infer that you mean "inteligencia artifical", or what have you.

AI is closer to pure math than it is to software development. and it can be described using only mathematical constructs. but I don't know where something stops being "pure math".

desert oar
#

@rugged comet that chart is such a bad example of overfitting that i suspect there's a bug in how you're using the validation data

#

like you're using the wrong dataframe, or it contains all the same value, or something else weird

soft badge
#

for example i an studying on course of IBM

#

but have concepts that i search other content for explain

iron basalt
# soft badge squiggle, for learn machine learning what do you recommend?

Reading books, papers, and implementing things from scratch, even computing by hand on paper. Make sure your mathematical fundamentals are solid and try applying that math to problems other than ML for practice (random fun example: https://massaioli.wordpress.com/2013/01/12/solving-minesweeper-with-matricies/ (I spend my free time on random problems like this / I do math for fun)). It's hard to really recommend something since a lot of what I know just came from random problem solving over a long period of time. Books are great, but they are a starting point or for when i'm in the mood for the "pure" math / want to do some formal proofs (which I also enjoy).

Programming by Robert Massaioli
Robert Massaioli
Solving Minesweeper with Matrices

What is your motivation for writing this? Note: skip to the next section if you don’t care about the back-story and want to get straight to the actual algorithm. Back in 2008 I was starting C…

#

As for AI, that is even more difficult, as it's even more scattered knowledge, much of which came from finding the right people online and in person / connecting with the lineage of ideas (some dating back to the late 1800s).

soft badge
#

ok, cool, I start to study but I seem to not learn or forget, do you have any tips?

iron basalt
serene scaffold
strange elbowBOT
soft badge
#

for example is it better to focus on one topic eg decorators know everything or give an introduction at the beginning and focus on the most important things at the beginning?

iron basalt
#

The problems found in many text books are for one specific aspect (e.g. "now prove that this is continuous at x0"), but the more interesting problems are ones that require multiple ideas at the same time. Such problems are more difficult, but because they kind of have a "journey" to them, they are memorable, like a story.

#

It's hard to remember specific little things without a bigger picture / story.

#

It needs to fit in with something else, for your associative memory (ML concept too) to be fully utilized.

#

(*It's also why traditionally things such as a tribe's morals and which animals (such as snakes) to stay away from, etc, come in the form of stories)

soft badge
#

understood, make sense

#

i was studying programming, hacking and IA in same day

#

do you guess interisting focus in one?

river sapphire
#

confused on what this equation means

image.png
#

so it mentioned proportional error meaning if we multiplied columns 1 and 2 (δ) by column 3 it would give us a measure of proportional error

unknown.png
#

then it says

We'll also redefine δ for all layers excluding the output layer to include this combination of weighted errors.

#

i'm not sure what to say i'm confused on what's going on in that equation

river sapphire
#

it said excluding the output so this equation wouldn't work on the output layer correct?

#

wait what's n again

#

n is the number of nodes in the layer I think

#

not sure they didn't define what n is

rugged comet
river sapphire
iron basalt
wooden sail
#

though to derive matrix calculus identities in the first place one usually has to go indexwise anyway :p it just using them tho, yea

rugged comet
fervent hatch
#

is it possible to predict the result of 3 dice based from how high I drop those dice from and the side of the dice that is on top?

wooden sail
#

statistically the chance is the same regardless of that if it's a fair die. if not, you could possibly do it with several thousands of examples. but what you're asking is really more in the direction of a physical simulation, where you'd solve differential equations to model the die

fervent hatch
wooden sail
#

2k is probably too few

#

and the amount of things you would be able to model would be very limited

#

i wouldn't trust such a model to work in general unless you can exactly reproduce the throwing conditions

fervent hatch
wooden sail
#

6, you mean

#

what you'll get is effectively the same as a histogram

#

there are already optimal strategies to pick the outcome without using machine learning

fervent hatch
#

we just used the 4 sides as our features for the dice initial state and we are like required to use machine learning to predict the outcome

fallen crown
#

Hi i am trying to add to add a legend to my iris dataset but i don't know how to do it....

finite knoll
#

Hi everyone. I seem to be having a very basic problem with manipulating data in pandas.

I wrote this line of code which, in theory, should act as a simple if statement when creating new columns. Create columns A2, B2, C2. If column D has value "HRK" then take values from A, B, C and divide them by 7.

I get a successfull load of data, but the new columns are not dividede by 7.

When it's out of the where syntax, then it works normally. I know that I am missing a step here as probably Python is not sure how to divide the columns. I just don't know how to tell him to divide each column by 7.

df[['A2','B2','C2']] = df[['A','B','C']].where(df['D'] == 'HRK', df[['A','B','C']] / 7 )

mighty patio
# fallen crown Hi i am trying to add to add a legend to my iris dataset but i don't know how to...

Easiest way is to slice your dataset before you plot it, and then plot each category with its corresponding label
I assume you have some category variable in addition to the color

import numpy as np
import matplotlib.pyplot as plt
x = np.random.random(25)
y = np.random.random(25)
categories = np.random.random_integers(0,2, 25)
def graphique():
    fig, ax = plt.subplots()
    for cat in range(3):
        ax.scatter(x[categories == cat], y[categories == cat], label = f"category: {cat}")
    ax.legend()
graphique()
image.png
river sapphire
#

but when I multiplied the matrices I got

#

do I add the 1st row and 1st column and the 2nd row and 1st column?

#

and do the same for the 2nd column?

#

how do they get a matrix of 1 * 2?

lapis sequoia
arctic wedgeBOT
#
pandas.DataFrame.divide 

DataFrame.divide(other, axis='columns', level=None, fill_value=None)```
Get Floating division of dataframe and other, element-wise (binary operator truediv).

Equivalent to `dataframe / other`, but with support to substitute a fill\_value for missing data in one of the inputs. With reverse version, rtruediv.

Among flexible wrappers (add, sub, mul, div, mod, pow) to arithmetic operators: +, -, *, /, //, %, **.
lapis sequoia
#

This would prob work, tho I guess you may have some different issue.

#

Why don't you share the exact error?

finite knoll
lapis sequoia
#

why do we need df.where here tho? np.where should be sufficient.

finite knoll
#

I will give it a go. I might have skipped some fundamentals concerning code structure.

lapis sequoia
#

see this part is good.

image.png
#

I'll tell you how I usually deal this.

lapis sequoia
finite knoll
lapis sequoia
#

Oh right. hold on.

lapis sequoia
#

you are making new cols, lemme see if that works.

#

That is working fine as well.

torn hull
#

hey guys so

#

I was checking out this code can someone help me to understand it?

finite knoll
finite knoll
lapis sequoia
spice edge
#

Hello, I have a question, how can i convert nested list into numpy to prevent the Value error : failed to convert a Numlpy array to a Tensor

lapis sequoia
spice edge
#

but then i still have this error :/

lapis sequoia
#

Because I think your error is saying you have a numpy array and you cant convert it to tensor.

spice edge
#

do you know what can be the possible reason for this problem of conversion to a tensor ?

lapis sequoia
#

The data, gotta see what the data is.

spice edge
#

hm can i share the data here or should i open a help session ?

lapis sequoia
#

You can share here.

finite knoll
spice edge
# lapis sequoia You can share here.

The data are generated by different function so it not like a panda dataframe or csv. I posted the code on stackoverflow so it's easier to copy paste it. Can I share the link here ?

lapis sequoia
spice edge
#

Thanks :)

#

Tell me if there is any errors.

lapis sequoia
spice edge
#

it's says (200,) for both of them

lapis sequoia
#

Given your model shouldn't it be like (200, 64) and (200,) ?

#

And given this code

# 
RUL = []
for i in range(len(finalDays)):
    rulTemp = [np.max(finalDays[i]) - x for x in finalDays[i]]
    RUL.append(rulTemp)

I expect RUL to be a 2D array

spice edge
#

Okay i see

#

So then the error come from the shape of my X and the RUL 1D?

lapis sequoia
#

yep.

spice edge
#

Thanks

robust cliff
#

it would be dope if someone figured out what the hell is happening

#

it's trying to build binaries for some reason i don't know

#

python: 3.11 (tried with 3.8 and does the same)
os: windows 11

#

wheel is on the latest version

#

so is pip

tidal bough
robust cliff
#

that does the same exact thing on python 3.8 tho

tidal bough
#

Now that's somewhat strange, it does have 3.8 wheels. For python3.8, did you update pip and setuptools? That might help.

robust cliff
#

let me try rq

tidal bough
#

that's not building the same thing, though

#

this one tries building scipy

robust cliff
#

ah

#

just so you know

#

i have no idea what any of this means

tidal bough
#

well, it's a different package from scikit-image, which is the one that's failing to build for you on 3.11

#

why would it build scipy, though, it has wheels for 3.8?..

#

try with --only-binary, perhaps?

robust cliff
#

like this pip install --only-binary scikit-image

tidal bough
#

yeah

robust cliff
#

:((

tidal bough
#

"pip install --only-binary scikit-image scikit-image"? did you put scikit-image twice?

#

though strange either way

#

try maybe pip install --only-binary scipy and see if that works

robust cliff
#

i put it twice because it would not work with just one

#

ERROR: You must give at least one requirement to install (see "pip help install")

tidal bough
#

ooh, I see

#

it's pip install --only-binary :all: scikit-image

robust cliff
#

it's also downloading scipy now

#

let's see

#

WOOOOOOOOOOOOOOOOOOO

#

it fixed

tidal bough
#

great, still don't see why it didn't just work normall though - pip is supposed to prefer wheels by default

#

What's your pip --version?

robust cliff
#

pip 22.3.1 from c:\users\dbuon\appdata\local\programs\python\python38-32\lib\site-packages\pip (python 3.8)

#

did i somehow mess with pip settings without noticing?

tidal bough
#

ooh. That's the latest pip version, but you're apparently using 32-bit python.

robust cliff
#

OHZAM

tidal bough
#

That might be the reason, I guess?

robust cliff
#

ig i downloaded the 32 bit installer without noticing

#

still would you mind explaining the difference of pip being used by python 32 bit vs 64 bit?

tidal bough
#

I'm not sure still why 32-bit python would cause that problem... did pip mention the exact name of the wheel it downloaded for scipy?

robust cliff
#

Collecting scipy>=1.4.1
Downloading scipy-1.9.1-cp38-cp38-win32.whl (34.5 MB)
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 34.5/34.5 MB 6.4 MB/s eta 0:00:00

tidal bough
#

ooooh

#

okay, that explains everything

robust cliff
#

i don't know what's going on

tidal bough
#

so, here's the thing

robust cliff
#

shoot

tidal bough
#

The normal way pip works is as follows:

  • first, it finds the last version of a library matching the constraints. Here, the only constraint scikit-image wants is >=1.4.1, so it just grabs the last version, 1.9.3.
  • then, it checks if there's a compatible wheel for that version. If there is, it uses it. If there's not, it tries to compile from source (and for you, compiling from source doesn't work)
#

And for 1.4.3, there aren't any win32 wheels! Hence, compiling from source it is.

robust cliff
#

i see

tidal bough
#

But when you require --only-binary, it works differently - it finds the last version matching the requirements which also has compatible wheels. And as you can see by the name, here it actually grabbed scipy-1.9.1-cp38-cp38-win32.whl - that's a wheel for 1.9.1, not the latest 1.9.3. So it had to downgrade two versions to be able to use a wheel.

robust cliff
#

ahhhhh

#

alright i get it

tidal bough
#

So pip isn't really at fault here, it was just trying to get you the latest version even if that meant building from source. If anyone's at fault, it's scipy maintainers for, I'm guessing, no longer posting win32 wheels.

robust cliff
#

so all the problems came from me installing the 32 bit version of python

#

ong

tidal bough
#

sure, nowadays many things don't support 32bit. really, it's nice there were any wheels for it at all.

robust cliff
#

i actually ignored the -32 in the path thinking it was something related to the version, not the architecture ;_;

#

btw thanks for all the help, appreciate it

finite knoll
# lapis sequoia My guess would be you may be making some silly mistakes.

So the weirdest thing happened. It ignores Valuta if there are strings, but it's fine if it's an int. Which makes 0 sense since the column is an String not an int. I just don't get it.

df[['Uplata2','Isplata2','Saldo2']] = df[['Uplata','Isplata','Saldo']].where(df['Valuta'] == 1, df[['Uplata','Isplata','Saldo']].div(7))

EDIT:

I did it! Found the solution although am not sure why it works! Does a seperate pd.DataFrame somehow create some new context?

df[['Uplata2','Isplata2','Saldo2']] = df[['Uplata','Isplata','Saldo']].where( pd.DataFrame(df['Valuta']) == 'HRK', df[['Uplata','Isplata','Saldo']].div(7))

random forum
#

When I do model.fit in tensor flow with verbose=0 it still prints epoch information does anyone know how can I fix this

lapis sequoia
finite knoll
iron basalt
bold timber
#

Hello guys, can you give me intuition about why we need to use EarlyStopping and ReduceLROnPlateau at the same time?

Earlier I thought when we've been using EarlyStopping we didn't need ReduceLROnPlateau because EarlyStopping will be stopping the process if the model doesn't improve for x number of epochs.

river sapphire
rancid sorrel
#

anyone know why on earth do you have to install python 3.6/3.7 to get amd tensorflow-directml to install?

serene scaffold
misty flint
#

anyone ever used onnx or torchscript for inference?

desert oar
desert oar
# rugged comet <@389497659087650836> Did you have a chance to look at the notebook? Here is t...

my main complaint is the same as before: no explanation of the data. how was it collected? how were the labels applied? are these tweets a random sample? what date range do they cover? there's also not any exploratory analysis. i don't even know the distribution of classes or the most frequent words. what steps are you taking for string cleaning? how do you know some of the tweets aren't misclassified?

#

also this doesn't look great

results___47_0.png
#

oh i see the counts of both classes. that's barely imbalanced.

#

if anything that's surprisingly well balanced

#

since you're including hashtags in here, you might want to either remove the # (treat the hashtag like a word) or use character n-grams instead of words (more or less entirely avoiding the issue but maybe making the model work harder to find good features)

#

but without any discussion or analysis of the raw data this is all speculation

#

i wouldn't trust these results at all tbh, sorry to say

#

hopefully my reasons make sense

idle urchin
#

name ID
bill.bob 3
albert.lee 3
sam.robert 2
adam.mason 2
kevin.fong 1 if this is my dataframe how can I sort by id and then under each id sort the names albhabetically

serene scaffold
brave sand
#

why does model.save() not work?

arctic wedgeBOT
#

Hey @charred umbra!

It looks like you tried to attach file type(s) that we do not allow (). We currently allow the following file types: .gif, .jpg, .jpeg, .mov, .mp4, .mpg, .png, .mp3, .wav, .ogg, .webm, .webp, .flac, .m4a, .csv, .json.

Feel free to ask in #community-meta if you think this is a mistake.

serene scaffold
rugged comet
# desert oar my main complaint is the same as before: no explanation of the data. how was it ...

The notebook is not complete yet. So I appreciate your detailed feedback.
Unfortunately, we weren't given much of any information about the dataset. This
https://www.kaggle.com/competitions/nlp-getting-started/overview
and this
https://www.kaggle.com/competitions/nlp-getting-started/data
is really all we know going into the problem. Even the link to website which gathered the data doesn't have the dataset anymore.
We don't know how the data was collected.
We know that the tweets were hand labeled.
We don't know if the tweets are a random sample.
We don't know what date range they cover.

On your comment about exploratory data analysis, I tried to do some starting with the Explore and analyze the data section. However, my impression from you is that it's not sufficient.
On your comment about the most frequent words, I didn't know that this could be important. After all, I'm just getting started with NLP.

On your comment about the graphs, I understand that it is significantly overfitting. Did you notice anything in the code that could cause something so intense?

On your comment about the hashtags, I agree with you that the # character could be removed. I will try this out. Can you talk to me a bit about character n-grams? I haven't heard of this before.

brave sand
serene scaffold
brave sand
serene scaffold
brave sand
serene scaffold
brave sand
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loss is decreasing and accuracy increases

serene scaffold
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how? the print statements only happen once, and are not in a loop.

brave sand
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the numbers are changing

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Test loss: 0.027803506702184677 Test accuracy: 0.991599977016449 Keras weights file (<HDF5 file "variables.h5" (mode r+)>) saving: ...layers\conv2d ......vars .........0 .........1 ...layers\conv2d_1 ......vars .........0 .........1 ...layers\dense ......vars .........0 .........1 ...layers\dropout ......vars ...layers\flatten ......vars ...layers\max_pooling2d ......vars ...layers\max_pooling2d_1 ......vars ...metrics\mean ......vars .........0 .........1 ...metrics\mean_metric_wrapper ......vars .........0 .........1 ...optimizer ......vars .........0 .........1 .........10 .........11 .........12 .........2 .........3 .........4 .........5 .........6 .........7 .........8 .........9 ...vars Keras model archive saving: File Name Modified Size config.json 2022-11-19 20:44:11 2753 metadata.json 2022-11-19 20:44:11 64 variables.h5 2022-11-19 20:44:11 447568

serene scaffold
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okay, I guess that information is displayed by model.fit (line 71) instead of the two print calls in 74-5.

brave sand
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prediction is still 1

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not sure why

serene scaffold
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what did you expect it to be?

brave sand
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the number I draw

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in the canvas

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everytime the number is 1

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the predicted one at least

serene scaffold
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@brave sand I'll try running your program and see what happens.

brave sand
serene scaffold
# brave sand updates?

the model seems to converge after three epochs (so you don't need 15), but for some reason I can't use streamlit.

brave sand
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pip install streamlit == 1.13.0 @serene scaffold

serene scaffold
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editing a message to add a ping has no effect, FYI. you have to do the ping when you first say the message.

brave sand
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ah got it

serene scaffold
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and that version of streamlit doesn't appear to exist. I was using

streamlit==1.12.0
streamlit-drawable-canvas==0.9.2
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I can just make an image manually, I guess.

brave sand
serene scaffold
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@brave sand it worked

In [39]: model.predict(_)
1/1 [==============================] - 0s 46ms/step
Out[39]: array([[5.2142264e-03, 1.4473951e-03, 2.0098801e-01, 7.2605598e-01, 2.1587194e-04, 8.9083081e-03, 2.4537125e-04, 3.1921629e-02, 7.0626062e-04, 2.4296993e-02]], dtype=float32)

In [40]: _.argmax()
Out[40]: 3
brave sand
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then how come it doesn’t work for me lol

serene scaffold
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idk. one of my coworkers created MNIST, so maybe he blessed me.

weary crown
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flex

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@serene scaffold emerso and i are in a hackathon for this rn

brave sand
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very strange behavior

weary crown
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so thx for carrying us

serene scaffold
brave sand
weary crown
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@serene scaffold every time i predict it says "1" with 12.549% acc

serene scaffold
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which means that the argmax is the "real answer"

weary crown
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didnt work 😦

brave sand
brave sand
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this is the prediction array

serene scaffold
brave sand
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yes

serene scaffold
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I got 7 for my seven

image.png
brave sand
# 
if st.button("Predict"):
    img = Image.fromarray(canvas_result.image_data.astype('uint8'), 'RGB')
    img = img.convert('L')
    # preprocess image
    img = img.resize((28, 28))
    img = np.array(img)
    img = img.reshape(1, 28, 28, 1)
    img = img.astype('float32')
    img /= 255
    # predict digit
    prediction = predict_img(img)
    print(prediction)
    st.write("The digit is: ", prediction.argmax())```
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this is so weird. is the conversion datatype wrong?

serene scaffold
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my images are black and white, which means that the empty space is 255. so I do some arithmetic to invert it and to squish everything between 0 and 1

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namely 1 - (np.asarray(Image.open("./seven.png").convert("L").resize((28, 28))) / 255)

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I'm also passing it to the model as a (1, 28, 28) array. I'm not sure why you're doing (1, 28, 28, 1)

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(you could do (3, 28, 28) if you wanted to predict 3 images at the same time, or whatever other number.)

brave sand
serene scaffold
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well fuck

brave sand
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this is witchcraft

serene scaffold
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I would print out img and see what it looks like

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if your empty space isn't 0s, for example, that would mess up the model

brave sand
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printing img doesn't work

serene scaffold
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why not

brave sand
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not sure why

serene scaffold
weary crown
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@serene scaffold img rn is only 0s

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and its not 784

serene scaffold
weary crown
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yes but that cant be

brave sand
weary crown
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wait no it cant

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actually it can tho

serene scaffold
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and if you want to prevent the latter, you have to manually safeguard invalid inputs that don't cause an error.

brave sand
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my output isn't all zeros

serene scaffold
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like, if you make a model that classifies if something is a cat or a dog, and you give it a picture of a turtle, it will still say that it's a cat or a dog.

brave sand
weary crown
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yeah wait its not all 0s i just didnt scroll enough

brave sand
weary crown
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😭

serene scaffold
serene scaffold
weary crown
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like arguments

misty flint
serene scaffold
weary crown
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@serene scaffold should we be loading with rgb or greyscale