Python排序字​​符串以数字开头

Question

I have the next list:

a = ['1th Word', 'Another Word', '10th Word']
print a.sort()
>>> ['10th Word', '1th Word', 'Another Word']

But I need:

['1th Word', '10th Word','Another Word']

Is there an easy way to do this?

I tried:

r = re.compile(r'(\d+)')
def sort_by_number(s):
    m = r.match(s)
    return m.group(0)

x.sort(key=sort_by_number)

But some strings do not have numbers and this leads to an errors. Thanks.

Answer 1

This is commonly called "natural sort".

See Does Python have a built in function for string natural sort? and Python analog of natsort function (sort a list using a "natural order" algorithm) ; also http://code.activestate.com/recipes/285264-natural-string-sorting/ and http://www.codinghorror.com/blog/2007/12/sorting-for-humans-natural-sort-order.html .

Answer 2

Here's a function that works for the general case

import re
def natkey(s):
    return [int(p) if p else q for p, q in re.findall(r'(\d+)|(\D+)', s)]

x = ['1th Word', 'Another Word 2x', 'Another Word 20x', '10th Word 10', '2nd Word']

print sorted(x)
print sorted(x, key=natkey)

Result:

['10th Word 10', '1th Word', '2nd Word', 'Another Word 20x', 'Another Word 2x']
['1th Word', '2nd Word', '10th Word 10', 'Another Word 2x', 'Another Word 20x']

Answer 3

r = re.compile(r'(\d+)')
def sort_by_number(s):
    m = r.match(s)
    return m and m.group(0) or s

x.sort(key=sort_by_number)

Key is that if there was not match, return the string as is