How to custom sort an alphanumeric list?

Question

I have the following list

l = ['SRATT', 'SRATW', 'CRAT', 'CRA0', 'SRBTT', 'SRBTW', 'SRAT0', 'SRBT0']

which I would like to sort alphabetically, with the added rule that a string containing a number at the end (actually always 0) must come after the last fully alphabetical string (last letter is at most W).

How can I do that? (using, if possible, a simple method like sorted)

For this example list, the desired result would be

['CRAT', 'CRA0', 'SRATT', 'SRATW' , 'SRAT0', 'SRBTT', 'SRBTW', 'SRBT0']

e.g. the following doesn't work

sorted(l, key=lambda x: x[-1].isdigit())

since it puts the strings containing a final number at the end, like so

['SRATT', 'SRATW', 'CRAT', 'SRBTT', 'SRBTW', 'CRA0', 'SRAT0', 'SRBT0']

Do the strings contain other characters than alphanumerical ones? — Willem Van Onsem, Jan 27 '17 at 12:59

vishes_shell · Accepted Answer · 2017-01-27T15:23:33.427

6

Working solution at the bottom!

First attempt:

>>> l = ['SRATT', 'SRATW', 'CRAT', 'CRA0', 'SRBTT', 'SRBTW', 'SRAT0', 'SRBT0']
>>> sorted(l, key=lambda x: (x[:-1], x[-1].isdigit()))
['CRAT', 'CRA0', 'SRATT', 'SRATW', 'SRAT0', 'SRBTT', 'SRBTW', 'SRBT0']

UPDATE

@StefanPochmann said that this will fail with same beginning and different last non-digit character.

We can add additional element in the end of key, which would be element itself

>>> l = ['SRATT', 'SRATW', 'CRAT', 'CRA0', 'SRBTT', 'SRBTW', 'SRAT0', 'SRBT0', 'B', 'A']
>>> sorted(l, key=lambda x: (x[:-1], x[-1].isdigit(), x))
                                                      ^
                                             additional element
['A', 'B', 'CRAT', 'CRA0', 'SRATT', 'SRATW', 'SRAT0', 'SRBTT', 'SRBTW', 'SRBT0']

UPDATE(final, i hope so)

@Demosthene noted that the second attempt is not working, and that's true

So working solution would be to pick any digit at the end of element(if it exist) and change to symbol that is out of letters and digits scope, e.g. '{':

sorted(l, key=lambda x: ''.join((x[:-1], '{')) if x[-1].isdigit() else x)

or

sorted(l, key=lambda x: x[:-1] + '{' if x[-1].isdigit() else x)

as @StefanPochmann noted. Which is probably faster.

edited Jan 27 '17 at 15:23

answered Jan 27 '17 at 12:59

vishes_shell

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Thanks! will accept answer in a few minutes (SO won't let me do it right now...) – Demosthene Jan 27 '17 at 13:01
@StefanPochmann i've updated answer, it's now working with single characters. Thanks. – vishes_shell Jan 27 '17 at 13:20
@vishes_shell It didn't just fail one character elements, that was just a minimal example. It also failed for example `['ABCDEG', 'ABCDEF']`. But your fix fixed that as well. – Stefan Pochmann Jan 27 '17 at 13:23
@StefanPochmann you're right, sorry, for putting your example in wrong problem explanation. – vishes_shell Jan 27 '17 at 13:26
1

@vishes_shell I've actually spotted another issue; adding an element `'SRE'` to `l` and using `sorted(l, key=lambda x: (x[:-1], x[-1].isdigit(), x))` actually puts `'SRE'` before `'SRATT'`. How would you do "absolute" alphabetical sorting? – Demosthene Jan 27 '17 at 13:30
@Demosthene thanks, you're right, i've updated answer.:) – vishes_shell Jan 27 '17 at 14:39
Why `''.join((x[:-1], '{'))` instead of just `x[:-1] + '{'`? And just to clarify: `['SRE0', 'SREZ']` can't happen (because of *"last letter is at most W"*). – Stefan Pochmann Jan 27 '17 at 15:19
@StefanPochmann my apologies, haven't noted that at most 'W' letter. – vishes_shell Jan 27 '17 at 15:24
One more thing: If the list contains the empty string, then you crash. Though maybe that can't happen, either. – Stefan Pochmann Jan 27 '17 at 20:03

Jean-François Fabre · Answer 2 · 2017-01-27T13:21:48.443

4

You have to retain the alphabetic criterion on the string (minus the last element), and introduce the other criterion: ends by a digit.

sorted(l, key=lambda x: (x[:-1] ,x[-1].isdigit()))

a more complex but robust way:

sorted(l, key=lambda x: (x[:-1] if len(x)>1 and not x[-1].isdigit() else x,x[-1].isdigit() if x else False))

(fixed the corner case noted by Stefan where the list is made of 1 or 0 sized elements or the ['AB', 'AA'] case)

edited Jan 27 '17 at 13:21

answered Jan 27 '17 at 12:59

Jean-François Fabre

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Your fix didn't really fix it, now it still fails for example `['AB', 'AA']`. – Stefan Pochmann Jan 27 '17 at 13:19
you are right. The "ends by 0 only" makes this answer overkill. Would be good if any number of digits. – Jean-François Fabre Jan 27 '17 at 13:22

Stefan Pochmann · Answer 3 · 2017-01-27T13:33:21.750

2

Here's another simple way, just treat 0 as Z:

>>> sorted(l, key=lambda x: x.replace('0', 'Z'))
['CRAT', 'CRA0', 'SRATT', 'SRATW', 'SRAT0', 'SRBTT', 'SRBTW', 'SRBT0']

(I'm assuming there are no zeros earlier in the string, let me know if that's wrong.)

edited Jan 27 '17 at 13:33

answered Jan 27 '17 at 13:15

Stefan Pochmann

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it's true that OP said "number in the end" and then said "it's always 0". – Jean-François Fabre Jan 27 '17 at 13:19
@Jean-FrançoisFabre Yeah but that doesn't rule out zeros elsewhere. – Stefan Pochmann Jan 27 '17 at 13:21
in that case do: `re.sub(r"\d$","Z",x)` – Jean-François Fabre Jan 27 '17 at 13:23
@Jean-FrançoisFabre Yes, or `re.sub(r"0$","Z",x)`. But if my assumption is correct, I prefer the non-regex solution. – Stefan Pochmann Jan 27 '17 at 13:27

How to custom sort an alphanumeric list?

3 Answers3

UPDATE

UPDATE(final, i hope so)