🔒 Help with sorted()

[clever wolf]

I'm having trouble figuring out why this isn't functioning as expected.

customer_data = [["Ainsley", "Small", True], ["Ben", "Large", False], ["Chani", "Medium", True], ["Depak", "Medium", False]]
customer_data_final = customer_data + [
["Amit", "Large", True],
["Karim", "X-Large", False]]

Provides list of customers

print ("Here is a list of all of your customers:" + "\n" )
for customer in customer_data_final:

print(sorted(customer[0]))

[hearty raft]

@clever wolf

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[clever wolf]

Particularly the last three lines. this is the output. I expected each name to be sorted, not the letters of each name. what can i do?

[split tulip]

print(sorted (customer)[0])

[torn meteor]

hm. you can't compare str with bool.

[split tulip]

I was a bit brain dead

Wait

customer_list.sort()

That should do my the trick

Noo

I am stupid what am I doing, sorry

[torn meteor]

no why? that sounds good. but it's in-place.

[split tulip]

Nope the name is in the first item of list

I am really tired, sorry

[rotund hull]

Then you need to pass a key to your sorting function that fetches the first item in your lists.

[torn meteor]

it's not necessary when all names are different.

[rotund hull]

ah yeah, right

[clever wolf]

in simple terms please 🙏

[split tulip]

You should use a dictionary, instead

Really

[mellow raven]

u need sorted(customer) probably

[torn meteor]

sorted(customer_data_final)

[mellow raven]

and what u where probably trying is expressed as
sorted(customer, key= lambda x: x[0])

[rotund hull]

And if you only want the names then retrieve them afterwards by indexing your sublists.

[torn meteor]

for customer in sorted(customer_data_final):
    print(customer)```

[clever wolf]

yeah i saw that earlier but i didn't get what that meant so i decided to not use it

[split tulip]

it just means it will sort by the first item in each list

[mellow raven]

it defines a key
what it does is instead of comparing the elements it compares key(element)

[clever wolf]

and a key meaining the person's name in this case?

[mellow raven]

lambda x: x[0] is just a short way to describe a function returning the first element of an iterable

[split tulip]

!eval```py
test=[["a", "z"],
["b", "y"]
]
print(sorted(test, key=lambda x: x[0]))
print(sorted(test, key=lambda x: x[1]))

[hearty raft]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | [['a', 'z'], ['b', 'y']]
002 | [['b', 'y'], ['a', 'z']]

[mellow raven]

key is a function that is aplied before sorting

basically u sort
[key(i) for i in list]

[clever wolf]

that worked 👍

[clever wolf]

and if you wanted it otherwise you'd just change the index?

[mellow raven]

just get the corresponding list values not the key(element) values

[mellow raven]

if u change the index to example 1 it vill compare the second letter of the names

[clever wolf]

ohhh

ok

[split tulip]

Only if it's x[0][1]?

[mellow raven]

if u want to sort based on length can u tell me what key u need? (just to check if u got these keys)

[split tulip]

If the key is x[1] it will be sorted by Small, Medium, Large ect. Alphabetically

[torn meteor]

If no key is specified, it will sort by name, then by size (when name is equal), then by expedited (when size is equal)

[mellow raven]

no it will be sorted
Large Medium Small
based on the [1]th char

[split tulip]

Yes I said alphabetically

Not literally

It will not sort by literally meaning but the alphabetical order

Apologies for the ambiguity

[mellow raven]

alphabetically but based on the second letter

oh wait i got it we arnt talking about single lists but nested ones right?

[split tulip]

If it's x[1] it will be sorted by the second item, not the first item's seconds letter

[mellow raven]

yes just i didnt get that its list[list[str]] not list[str]

[split tulip]

I see, don't worry

[torn meteor]

[["Ainsley", "Small",   True ],
 ["Ben",     "Large",   False],
 ["Chani",   "Medium",  True ],
 ["Depak",   "Medium",  False],
 ["Amit",    "Large",   True ],
 ["Karim",   "X-Large", False]]```

[split tulip]

I said this assuming it is list[list[str]], if you want the first item's second letter that is

[split tulip]

Yup, that's what I was referring to

[mellow raven]

u could sort by size with the key
lambda x: ["Small","Medium","Large"].index(x[1])

[split tulip]

I don't see why op isn't using dictionary, as using dictionary allow indexing with name

You can then skip looping through the whole list to find a person's data

[split tulip]

e.g.

{"Ainsley": {"size": "Small", "thing": True},
"Ben": {"size": "Large", "thing": False}
}

[clever wolf]

i'm not familiar with dictionaries yet

[split tulip]

I see, that's ok, you can keep using list for the time being, but try out dictionary when you can

[clever wolf]

i will thanks :)

[torn meteor]

the thing is expedited shipping.

[split tulip]

I see, I didn't know what the code is for, so I threw in random key name I can think up with

[hearty raft]

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