Reference list item as key in sorted for loop

Question

Sorry for boring question but I can't figure this out:

for f in sorted(os.listdir('.')): print f

Output:

p1.html
p10.html
p11.html
p12.html
p13.html
p14.html
p15.html
p16.html
p17.html
p18.html
p19.html
p2.html
p20.html
p21.html
p22.html
p3.html
p4.html
...

Obviously I want to sort by number and I can do it with this key: f.split('.')[0][1:] but how to reference that key in this for loop?

I tried for f in sorted(os.listdir('.'), key=f.split('.')[0][1:]) but it does not work of course

TIA

Related: http://stackoverflow.com/questions/4836710/does-python-have-a-built-in-function-for-string-natural-sort — Sven Marnach, Apr 02 '11 at 21:42
@otrov, its there to make your life easy. Not wanting to load it is silly. — Winston Ewert, Apr 02 '11 at 22:30
@Winston Ewert: Well, as you can see from the provided answer it would be overkill to do so. Yeah, I could have solved it with couple of extra lines myself, but then I wouldn't go here and ask for help. BTW re module is one of my favorites regularly, but not here — otrov, Apr 03 '11 at 00:54
@otrov, ok. Just as long as you don't have a silly superstition about loading the module. — Winston Ewert, Apr 03 '11 at 01:18

score 1 · Accepted Answer · answered Apr 02 '11 at 21:42

1

You'd need a lambda expression:

sorted(os.listdir('.'), key=lambda f: int(f.split('.')[0][1:]))

answered Apr 02 '11 at 21:42

Sven Marnach

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1

He's probably looking for a numerical sort... so IMO it should be `lambda f:int(f.split('.')[0][1:])` – 6502 Apr 02 '11 at 21:45

Reference list item as key in sorted for loop

1 Answers1