比较两个包含数字的 python 字符串

Question

UPDATE: I should have specified this sooner, but not all of the names are simply floats. For example, some of them are "prefixed" with "YT". So for example" YT1.1. so, you have the same problem YT1.9 < YT1.11 should be true. I'm really surprised that the string comparison fails....

hello, this should be a pretty simple question but I can't seem to find the answer. I'd like to sort a bunch of XL worksheets by name. Each of the names are numbers but in the same way that textbook "sections" are numbered, meaning section 4.11 comes after 4.10 which both come after 4.9 and 4.1. I thought simply comparing these numbers as string would do but I get the following:

>>> s1 = '4.11'
>>> s2 = '4.2'
>>> s1> s2
False
>>> n1 = 4.11
>>> n2 = 4.2
>>> n1 > n2
False

how can I compare these two values such that 4.11 is greater than 4.2?

Answer 1

Convert the names to tuples of integers and compare the tuples:

def splittedname(s):
    return tuple(int(x) for x in s.split('.'))

splittedname(s1) > splittedname(s2)

Update : Since your names apparently can contain other characters than digits, you'll need to check for ValueError and leave any values that can't be converted to ints unchanged:

import re

def tryint(x):
    try:
        return int(x)
    except ValueError:
        return x

def splittedname(s):
    return tuple(tryint(x) for x in re.split('([0-9]+)', s))

To sort a list of names, use splittedname as a key function to sorted :

>>> names = ['YT4.11', '4.3', 'YT4.2', '4.10', 'PT2.19', 'PT2.9']
>>> sorted(names, key=splittedname)
['4.3', '4.10', 'PT2.9', 'PT2.19', 'YT4.2', 'YT4.11']

Answer 2

This is not a built-in method, but it ought to work:

>>> def lt(num1, num2):
...     for a, b in zip(num1.split('.'), num2.split('.')):
...         if int(a) < int(b):
...             return True
...         if int(a) > int(b):
...             return False
...     return False
... 
... lt('4.2', '4.11')
0: True

That can be cleaned up, but it gives you the gist.

Answer 3

What you're looking for is called "natural sorting". That is opposed to "lexicographical sorting". There are several recipes out there that do this, since the exact output of what you want is implementation specific. A quick google search yields this (note* this is not my code, nor have I tested it):

import re

def tryint(s):
    try:
        return int(s)
    except:
        return s

def alphanum_key(s):
    """ Turn a string into a list of string and number chunks.
        "z23a" -> ["z", 23, "a"]
    """
    return [ tryint(c) for c in re.split('([0-9]+)', s) ]

def sort_nicely(l):
    """ Sort the given list in the way that humans expect.
    """
    l.sort(key=alphanum_key)

http://nedbatchelder.com/blog/200712.html#e20071211T054956

Answer 4

use s1.split(".") to create a list of the items before and after the decimal then sort the list of lists, example:

import random
sheets = list([str(x), str(y)] for x in xrange(1, 5) for y in xrange(0,99))
print sheets
#sheets in order
random.shuffle(sheets)
print sheets
#sheets out of order
sheets.sort()
print sheets
#sheets back in order

So, you implementation might be:

#assume input sheets is a list of the worksheet names
sheets = list(x.split(".") for x in input_sheets)
sheets.sort()

Answer 5

If you know they are real numbers [*] , simply:

>>> float(s1) > float(s2)
True

[*] Otherwise, be ready to handle a raised ValueError .