if I have a list of strings e.g. ["a143.txt", "a9.txt", ] how can I sort it in ascending order by the numbers in the list, rather than by the string. I.e. I want "a9.txt" to appear before "a143.txt" since 9 < 143.
thanks.
5 Answers
It's called "natural sort order", From http://www.codinghorror.com/blog/2007/12/sorting-for-humans-natural-sort-order.html
Try this:
import re
def sort_nicely( l ):
""" Sort the given list in the way that humans expect.
"""
convert = lambda text: int(text) if text.isdigit() else text
alphanum_key = lambda key: [ convert(c) for c in re.split('([0-9]+)', key) ]
l.sort( key=alphanum_key )
4 Comments
kindall
I'd use
text.lower() at the end of the convert = line to make it case-insensitive.
ʇsәɹoɈ
+1. You might want to replace the lambda with a proper function definition, for readability. Incidentally, Debian package version numbers are compared more or less like this. debian.org/doc/debian-policy/ch-controlfields.html#s-f-Version
Maciej Ziarko
+1 Nice answer. The only thing I didin't like are extra white-spaces. I mean here:
[ convert(c) for c in re.split('([0-9]+)', key) ] and l.sort( key=alphanum_key ) and sort_nicely( l )
PaulMcG
+1, nicely done! I redid alphanum_key as
alphanum_key = lambda key: map(convert, re.split('([0-9]+)', key)).Use list.sort() and provide your own function for the key argument. Your function will be called for each item in the list (and passed the item), and is expected to return a version of that item that will be sorted.
See http://wiki.python.org/moin/HowTo/Sorting/#Key_Functions for more information.
Comments
If you want to completely disregard the strings, then you should do
import re
numre = re.compile('[0-9]+')
def extractNum(s):
return int(numre.search(s).group())
myList = ["a143.txt", "a9.txt", ]
myList.sort(key=extractNum)
Comments
>>> paths = ["a143.txt", "a9.txt"]
>>> sorted(paths, key=lambda s: int(re.search("\d+", s).group()))
['a9.txt', 'a143.txt']
More generic, if you want it to work also for files like: a100_32_12 (and sorting by numeric groups):
>>> paths = ["a143_2.txt", "a143_1.txt"]
>>> sorted(paths, key=lambda s: map(int, re.findall("\d+", s)))
['a143_1.txt', 'a143_1.txt']
Comments
list.sort() is deprecated (see Python.org How-To) . sorted(list, key=keyfunc) is better.
import re
def sortFunc(item):
return int(re.search(r'[a-zA-Z](\d+)', item).group(1))
myList = ["a143.txt", "a9.txt"]
print sorted(myList, key=sortFunc)
5 Comments
tokland
list.sort() is deprecated? "Usually it's less convenient than sorted()" is the only thing in this direction I found. I have to say, though, that I'd be more than happy to see the in-place sorting go away, but it seems unlikely.
Maciej Ziarko
It is not deprecated. docs.python.org/library/stdtypes.html#mutable-sequence-types
Prydie
It may not be technically depreciated but it is considered the "old" method and is labeled as such on Python.org.
Prydie
True
list.sort() is certainly slightly more memory efficient but the difference is negligible as far sensible sized lists are concerned. I can't find a good explanation for why but as far as I am aware the preferred and more 'pythonic' way of doing sorting is using sorted().Maciej Ziarko
It says
sorted() returns a NEW list, not that's it's NEW. It's definitely less efficient if you really interested in modifying your list. sorted() is good for tuples.
scipyornumpy. If this is the case, please remove those tags.