🔒 Make the code run faster

[elfin breach]

I have this code who work well but is there a way to make my code go faster ? ```py
def count_number_of_letter_in_a_word(word):
return sum(letter.isalpha() for letter in word.lower())

def count_occurrences_in_text(word, text):
SplitString = text.lower().replace("."," ").split()
WordOccurences = 0
if " " in word and len(word) > 1 :
return(text.lower().count(word.lower()))

for words in SplitString:
    if " " not in word:
        if word.lower() in words and len(word) > 1 and count_number_of_letter_in_a_word(word) == count_number_of_letter_in_a_word(words) and "'" not in words:
            WordOccurences+=1
        elif word.lower() in words and len(word) > 1 and count_number_of_letter_in_a_word(word) == count_number_of_letter_in_a_word(words) and text[0] == "'" and text[-1] == "'":
            WordOccurences+=1
        elif word.lower() == words:
            WordOccurences+=1

return WordOccurences

[wary hearth]

@elfin breach

Python help channel opened

Remember to:

• Ask your Python question, not if you can ask or if there's an expert who can help.
• Show a code sample as text (rather than a screenshot) and the error message, if you've got one.
• Explain what you expect to happen and what actually happens.

:warning: Do not pip install anything that isn't related to your question, especially if asked to over DMs.

Closes after a period of inactivity, or when you send !close.

[quasi skiff]

Your variable names leave a lot to be desired

But I highly suspect this is suboptimal

Also, it's not clear what the intent is

[quick sequoia]

count_occurrences_in_text could just be text.count(word)

[quasi skiff]

The confusing part is they're using the .count method... But only if a space is in word?

And then later they're comparing the length of the alpha chars

[elfin breach]

Let me send yout the whole code

https://paste.pythondiscord.com/4VUQ

here it is

[quasi skiff]

That... Doesn't help at all.

Can you explain the goal of the program?

The assignment perhaps?

[elfin breach]

Ok I need to pass all the unit test

the goal is just to count the number of time a word is on a string

but since there is different case I can't just do a .count()

it won't work

and I need to solve all the test as fast as possible

[dusty wolf]

why not
what goes wrong if you use .count()?

[quasi skiff]

I don't think your code currently accomplishes that

[elfin breach]

it should be even faster

count function won't work well for some cases

[quasi skiff]

OP is correct about this

[elfin breach]

Did like this as you said

[elfin breach]

Could you tell me what is wrong in my code and how should I fix it

[dusty wolf]

is the red (2 when it should be 10) one of the cases that .count() doesn't work?

[elfin breach]

yeah

and there is several of them where it doesn't work with just a "count"

I know that the part that goes slow is this py for words in SplitString: if " " not in word: if word.lower() in words and len(word) > 1 and count_number_of_letter_in_a_word(word) == count_number_of_letter_in_a_word(words) and "'" not in words: WordOccurences+=1 elif word.lower() in words and len(word) > 1 and count_number_of_letter_in_a_word(word) == count_number_of_letter_in_a_word(words) and text[0] == "'" and text[-1] == "'": WordOccurences+=1 elif word.lower() == words: WordOccurences+=1 but I don't know how to do the same withouth using list looping

[dusty wolf]

idk what you're on because it works for me

>>> def cnt(word, text):
...     return text.lower().count(word.lower())
...
>>> cnt('I', 'Georges is my name and I like python. Oh ! your name is georges? And you like Python!\nYes is is true, I like PYTHON\nand my name is GEORGES')
10
>>>

or do you mean it should be 2

[elfin breach]

it should be 2

[dusty wolf]

ok, reason about why it should be 2

[elfin breach]

I guess that we should only look for the word "I" and not word that have I like "is or like" etc

And I think the count function search for the letter I in any word that got it

[dusty wolf]

o I c

what about, you split the sentence into words, then just check each word to see if it's an exact match

[elfin breach]

Like there there is only 1 I

but for python there is 3

[dusty wolf]

or if a word is separated by spaces, instead of matching "I", why not match " I "

[elfin breach]

Was going to say that

let me try it

[dusty wolf]

and cover some edge cases like "I like apples." by adding a ' ' to the beginning & end of the string

[elfin breach]

I need now to replace every special character with " "

[dusty wolf]

hopefully the task has given you what should & shouldn't be considered a special character

if so, you can use a loop to change the original string until all special characters have become ' '

[elfin breach]

Only said that ...

Like there you see that maley is not inside the string

[dusty wolf]

well just don't replace "'" with ' '

[elfin breach]

because it's in a word

Well the count technique won't work

because I'm "hardcoding" result for each case that doesn't work

it's even worse than the one I did at first

[dusty wolf]

which part are you hardcoding?

[elfin breach]

first result

then know I will have other problem with the rest

[dusty wolf]

what's your code right now

[elfin breach]

it doesn't even work lol

[dusty wolf]

that's not hardcoding it, that's just removing punctuations

[dusty wolf]

doesn't work because edgecase, see this

[elfin breach]

yeah missword but like removing part for just the first test isn't great

What do you mean by edgecase?

[dusty wolf]

you're counting ' I ', so something like 'I like apples.' will fail because I is the first word in the sentence

[elfin breach]

yeah

[dusty wolf]

you can use a loop to do it if it looks ugly

punctuation = ... # string containing all special characters
for char in punctuations:
    text = text.replace(char, '')

[dusty wolf]

and I proposed a fix already, see above

by adding a ' ' to the beginning & end of the string

and another edge case might be something like

>>> ' I I I '.count(' I ')
2
# .count() doesn't count overlaps, so
# ' I I I '
#  ^^^
#      ^^^
# those are the 2 it counts
```to fix that you can replace every `' '` with `'  '`

[elfin breach]

code py def count_occurrences_in_text(word, text): text = text.lower() punctuations = [".", ",", ";", ":", "!", "?", "...", "«", "»", "“", "”", "(", ")", "[", "]", "{", "}", "'", "-", "-", "*", "#", "/", "\\", "&", "@"] for char in punctuations: text = text.replace(char, '') if "george" in text: print("New string after removing punctiation " , text) return (text.lower().count(word.lower() + " "))

[dusty wolf]

.

[elfin breach]

oh

that ? text = " " + text + " "

Fixed

like this

def count_occurrences_in_text(word, text):
    text = text.lower()
    text = " " + text + " "
    punctuations = [".", ",", ";", ":", "!", "?", "...", "«", "»", "“", "”", "(", ")", "[", "]", "{", "}", "'", "-", "-", "*", "#", "/", "\\", "&", "@", "\n"]
    for char in punctuations:
        text = text.replace(char, ' ')
    return (text.lower().count(word.lower() + " "))
    ```

but new problem

[dusty wolf]

word.lower() + " "

[dusty wolf]

.

[elfin breach]

ok ok

Did it

now fails here:

[dusty wolf]

cause you counted ' as a special character and not what the test told you to do?

[elfin breach]

yeah fixed it

I know the problem

it's because it's the first word

[dusty wolf]

you didn't include " in your special characters

[elfin breach]

and we are doing " " + text + " "

[dusty wolf]

you included these 2 “ ” which idk what those are tbh

[elfin breach]

We should do like if it's first word then just text + " " else " " + text + " "

[elfin breach]

asked chatgpt to make the list 😂

[dusty wolf]

chatgpt sucks

[elfin breach]

passed way more test

with your technique

[dusty wolf]

and gpt also missed this one

[dusty wolf]

ah, it's getting into the annoying af territory I see
why do I feel like I've seen this exact task before actually (I specifically rmb seeing "Reflexion Mirror" somewhere)

[elfin breach]

yep

it's a "." problem

but if I remove "." from the list of char the first test failed ...

[dusty wolf]

I'm thinking if there's a way so you don't have to rewrite again due to stupid test cases

[elfin breach]

I may have found something

maybe we should also remove special char from the word too

let me try

[dusty wolf]

no
cnt('c.v', 'cv') should be 0, your solution will make it 1

[elfin breach]

well passed 1 test

[dusty wolf]

" # regard ' as text:"
nice one, that was a fucking lie

[elfin breach]

😂 😂 😂

[dusty wolf]

and also the parameter word is also a lie at this point because you're given entire sentences now

[elfin breach]

yep

looking into a solution for linguist

[dusty wolf]

I'm assuming you can't import anything?

if you can, this task's a job for re

[dusty wolf]

I guess you can strip '

[elfin breach]

that's all they said

[dusty wolf]

if I were you, I'd ignore that and just get something working first

[elfin breach]

Find a solution

[dusty wolf]

and also this statement might be completely false if you only need a few regexes to solve it, but a lot of native str operations if you didn't re

[elfin breach]

punctuations = [".", ",", ";", ":", "!", "?", "...", "«", "»", "“", "”", "(", ")", "[", "]", "{", "}", "*", "#", "/", "\", "&", "@", "\n" ,'"',"'''"]
just added ''' to the list

and it worked

New problem

[elfin breach]

yes

[dusty wolf]

e.g.

for char in punctuations:
    text = text.replace(char, ' ')
```most likely slower than
```py
re.sub(f'[{punctuations}]', '', text)
```because you were looping a bunch of times and making a lot of strings

[elfin breach]

I will first finish with the way we are doing now

[dusty wolf]

so without re?
I asked because at this point I think re will be way simpler

[elfin breach]

I you think I wil do it

anything faster is better

[dusty wolf]

like the entire problem is

• for text like apple. you should ignore .
• for text like bapple, that's not an apple
• for text like my c.v. to, you should not ignore . (because it's part of the word)

so my idea is if you just cooked up a regex that's
<detect start of word><the word you're supposed to match><any remaining special characters>then everything's solved

[elfin breach]

yes

Like this idea

[dusty wolf]

well let's get regex-ing then

• start of word is probably just ' ' (again you can add a ' ' to the text for edge cases like I like apples.)
• any remaining special characters that can be done easily with character sets [] and *
  e.g. [abc] is a regex that matches either a b or c; a* is a regex that matches a 0 or 1 time

[elfin breach]

Wait

All the test work with this ```py

def count_occurrences_in_text(word, text):
text = text.lower()
text = " " + text + " "
punctuations = [".", ",", ";", ":", "!", "?", "...", "«", "»", "“", "”", "(", ")", "[", "]", "{", "}", "*", "#", "/", "\", "&", "@", "\n" ,'"',"'''","''","__"]
for char in punctuations:
text = text.replace(char, ' ')
word = word.replace(char, ' ')
if word.lower() == "linguist":
print("sentences: " , text)
return (text.lower().count(" " + word.lower() + " "))
```

Now will just transform all of this into a regex

[dusty wolf]

ehh I feel like you passed because the test case didn't cover it

[elfin breach]

But isn't what we have done to pass the code similar to "hardcoding"?

[elfin breach]

Well 😂

[dusty wolf]

yeah checking for linguist is 100% hardcoding

well well but the regex will actually work w/o hardcoding

hopefully

[elfin breach]

I hope

[elfin breach]

But the test aren't well made too

[dusty wolf]

like what if you just counted the regex fr' {word}[{punctuations}]*'

[elfin breach]

you can't have a generic function

but I hope the regex will do the work

[elfin breach]

How would I put it in my code?

forgot how to regex don't do it that often

[dusty wolf]

like literally what if you just did

def cnt(word, text):
  punctuations = ...
  matches = re.findall(fr' {word}[{punctuations}]* ', ' ' + text.replace(' ', '  ') + ' ')
  return len(matches)

well that wouldn't pass for '''linguist ig

[elfin breach]

thanks

[dusty wolf]

right, ignorecase

re.findall(..., flags=re.IGNORECASE)

[elfin breach]

Dumb

forgot one thing

didn't redeclare the ponctuation list

[dusty wolf]

hm

[elfin breach]

I think my new code is good

it go 10 times faster than the previous one

Thanks for you help

[dusty wolf]

maybe this regex works?

import re, string
from string import punctuation
def cnt(word, text):
    matches = re.findall(fr"[{punctuation} ]{word}[{punctuation} ]", ' ' + text.replace(' ', '  ') + ' ', re.IGNORECASE)
    return len(matches)

[elfin breach]

well

yeah

you texted me 2 minutes after I sended the code

but I will send it

[dusty wolf]

oh

[elfin breach]

there is no limitation

[dusty wolf]

well no, test that it actually works

[elfin breach]

it does even faster than my code:

[dusty wolf]

"regex is slower"
though tbf if you replaced the for char in punctuations: text = text.replace(char, '') with a regex I wouldn't expect too much a speed difference then

[elfin breach]

so it's the same speed ?

[dusty wolf]

maybe, you can try it

[elfin breach]

let me try a larger groupe of test

[dusty wolf]

re.sub(f'[{punctuations}]', '', text) removes all punc characters at once, probably faster than the loop

[elfin breach]

it's the same speed

because It loop trough a list of 9 elem

so it's not that big

[dusty wolf]

eh ok then
anyways, you got your answers now so problem solved?

[elfin breach]

yes

thanks

let's hope they like this code better

[dusty wolf]

I mean you can just tell them that you tried both (re and no re) and the former was faster, despite the task saying otherwise

[elfin breach]

yeah

Will say find another way to make the code

and they will have 2 code to see

maybe they will be like he tried even after something faster

[dusty wolf]

gl to you

[elfin breach]

thanks

[elfin breach]

Worked wellll

[dusty wolf]

ey yo nice
good luck to you

[elfin breach]

thanks

[wary hearth]

Python help channel closed

This help channel has been closed and it's no longer possible to send messages here. If your question wasn't answered, feel free to create a new post in #1035199133436354600. To maximize your chances of getting a response, check out this guide on asking good questions.