python - sort strings in list by skipping first 4 lines

Question

My list looks this:

l = ['abcdbcex', 'abcdbc1fz', 'abcdcd11', 'abcd11a5']

I want to sort it in a way that it will become:

l = ['abcd11a5', 'abczbc1fz', 'abcfbcex', 'abcecd11']

So sort in ascending order, but skipping first 4 letters, also skipping everything after the 7th letter.

I have tried doing: sorted(list, key=lambda x:x[:4])

but this does not skip the first 4 letters, because I don't know how to do it.

You confused the slice, it's `x[4:]` – Tomerikoo Dec 16 '20 at 13:25 — Tomerikoo, Dec 16 '20 at 13:25
@Tomerikoo typo in the question – Margarita Weiß Dec 16 '20 at 13:26 — Margarita Weiß, Dec 16 '20 at 13:26
Still, notice the location of the `:`... – Tomerikoo Dec 16 '20 at 13:27 — Tomerikoo, Dec 16 '20 at 13:27
@Tomerikoo ah yes thank you. – Margarita Weiß Dec 16 '20 at 13:28 — Margarita Weiß, Dec 16 '20 at 13:28

score 2 · Answer 1 · answered Dec 16 '20 at 13:26

2

You can't call it list, it's a reserved word. THis does work though:

sorted(xlist,key=lambda s: s[4:])

['abcd11a5', 'abcdbc1fz', 'abcdbcex', 'abcdcd11']

answered Dec 16 '20 at 13:26

Sergio Lucero

1 Answers1