Remove duplicates from list and order it by count? (python 3)

Question

How do I remove all duplicates from list and order it based on count?

s = ["foo", "this", "lem", "foo", "lem"]
s = ["foo", "ebh", "wosv", "wosv", "ebh", "4whcu"]
#should return
>>> ["foo", "lem", "this"]
>>> ["ebh", "wosv", "foo", "4whcu"]

Rules:

• Must be shorter than the below code in terms of byte size.
• Order must be based on the count of in descending order.
• If there are two strings with the same count, the first appearance comes first
• All duplicates must be removed, only one of each string should be in the final list

I currently have

list({}.fromkeys(sorted(s,key=lambda y:-s.count(y))))

I've noticed answers like set but that arbitrarily changes the order, which I do not want.

Edit: Sorry for the poor quality before. I had whipped this right before I slept and I was quite tired. This is not for StackOverflow because I'm trying to golf/shorten the size of this code as much as possible. I've tried looking for answers but I haven't been able to find anything.

Answer 1

44 bytes

sorted({}.fromkeys(s),key=s.count,reverse=1)

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An improvement to the below, replacing the key with a built-in rather than a lambda, and using reverse to swap the comparisons. We'd like to do [::-1] on the final result instead, but that doesn't do the right stable tiebreaks.

We could also use dict(zip(s,s)) in place of {}.fromkeys(s) for the same length.

---

47 bytes

sorted({}.fromkeys(s),key=lambda y:-s.count(y))

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A simple transposition of your code, moving your {}.fromkeys de-duplication trick before the sorting. Since sorting converts to a list, extracting the keys from a dictionary, this saves the list() call of the original.

For ease of reference, the original code is:

53 bytes

list({}.fromkeys(sorted(s,key=lambda y:-s.count(y))))

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Answer 2

APL, 11 chars

∪⌷⍨∘⊂∘⍒⊢∘≢⌸

⊢∘≢⌸ right arg: number of apperance of each element, in order of first appearance

∪ left arg: unique elements, in order of first appearance`

⌷⍨∘⊂∘⍒ sorts elements of the left arg based on the grade-down of the right arg`

Try it online: https://tio.run/##SyzI0U2pTMzJT///v9jwUdsE9bT8fHUF9ZKMzGIglZOaCyQhQiA2V7ERQk1qUgaQLM8vLkNQEDGT8ozkUnUurjSg4kcdqx71bH/Uu@JRx4xHXU0gsnfSo65FIEbnokc9O7i4HvVNBSpMUyg2RDCN/v8HAA