python - Sorting a list and dict keys -


i'm using python 2.7 , have list:

photos = ['arthur20.jpg', 'arthur7.jpg', 'arthur11.jpg', 'arthur3.jpg', 'arthur5.jpg', 'arthur17.jpg', 'arthur15.jpg', 'arthur2.jpg', 'arthur13.jpg', 'arthur8.jpg', 'arthur9.jpg', 'arthur18.jpg', 'arthur4.jpg', 'arthur6.jpg', 'arthur10.jpg', 'arthur12.jpg', 'arthur14.jpg', 'arthur19.jpg', 'arthur16.jpg', 'arthur1.jpg']

how can organize list? tried use sort() didn't work, returned me this:

['arthur1.jpg', 'arthur10.jpg', 'arthur11.jpg', 'arthur12.jpg', 'arthur13.jpg', 'arthur14.jpg', 'arthur15.jpg', 'arthur16.jpg', 'arthur17.jpg', 'arthur18.jpg', 'arthur19.jpg', 'arthur2.jpg', 'arthur20.jpg', 'arthur3.jpg', 'arthur4.jpg', 'arthur5.jpg', 'arthur6.jpg', 'arthur7.jpg', 'arthur8.jpg', 'arthur9.jpg']

and strings of list keys dict , have put keys in order too, this:

dict = {"arthur1.jpg":1, "arthur2.jpg":2 ...}

sort() sort list in natural ordering, in case - lexicographical ordering. can sort number following "arthur" custom sort function. easiest way this, imho, extract number using regular expression:

import re photos.sort(key = lambda x : int(re.findall('\d+', x)[0]))

once have list sorted, creating dict want matter of ziping:

from collections import ordereddict dict = ordereddict(zip(photos, range(1, len(photos) + 1)))

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