Descending order unique squares

[OmegaRed94]

Good day, Dear Pythonistas

I have the following problem: Given a list, we should write a function that returns sorted in descending order unique squares of numbers contained in this list

I have got the solution in the following way, it works correctly:

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def squares_list(l):     for elem in l:         empty_list.extend(elem)     new_list=list(set(empty_list))     new_list.sort(reverse=True)     squares=[i**2 for i in new_list]     return squares print(squares_list([[1, 2], [3], [4], [3, 1]]))

Could you, please, suggest, how I could write this code or the solution in the shortest possible optimal way (a smaller number of characters)?

[deanhystad]

Search for a short way to flatten the list. Maybe itertools.chain?

Do not use variables to save intermediate results. Chain operations together using return values

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return [i**2 for i in new_list] # instead of squares=[i**2 for i in new_list] return squares

Use sorted instead of sort.

1	sorted_unique_list=sorted(set(unsorted_list))

I can write your function as a 1 liner, and I didn't have to be clever. For a one liner it doesn't look too bad

[OmegaRed94]

Thank you very much) By now it looks like this:

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import itertools def process(l):     return [i**2 for i in sorted(set(list(itertools.chain(*l))))[::-1]]

[ndc85430]

Why do you need the call to list there?

[deanhystad]

Leave out the list() call. Not needed. Have sorted reverse the order

[OmegaRed94]

Leave out the list() call. Not needed. Have sorted reverse the order

Thanks again

[ndc85430]

If the nesting of all the function calls bothers you, you can probably flatten it out with thread_first or thread_last from the third-party Toolz library.