Sorting List In Python
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 appea
Solution 1:
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
defsort_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 )
Solution 2:
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.
Solution 3:
If you want to completely disregard the strings, then you should do
import re
numre = re.compile('[0-9]+')
defextractNum(s):
returnint(numre.search(s).group())
myList = ["a143.txt", "a9.txt", ]
myList.sort(key=extractNum)
Solution 4:
>>> 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']
Solution 5:
list.sort()
is deprecated (see Python.org How-To) . sorted(list, key=keyfunc)
is better.
import re
defsortFunc(item):
returnint(re.search(r'[a-zA-Z](\d+)', item).group(1))
myList = ["a143.txt", "a9.txt"]
printsorted(myList, key=sortFunc)
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