如何对字符串中的混淆数字列表进行排序？

Question

 list=[a1b,a100b,a2b,a99b]

I would like to transform by comparing the digit sandwitched the letter a and b like below.

 [a1b,a2b,a99b,a100b]

I tried

 list.sort()

But it didnt work well.

How can I sort?

Answer 1

Option 1
natsort.natsorted
The natsort module works nicely here -

>>> from natsort import natsorted
>>> natsorted(['a1b','a100b','a2b','a99b'])
['a1b', 'a2b', 'a99b', 'a100b']

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Option 2
sorted + re.search
With regex , I'd recommend defining a function that calls re.search to find and extract numbers, with a little checking to ensure that no exceptions are thrown when the pattern is not found in the string.

import re
def f(x):
     m = re.search('\d+', x)
     return int(m.group()) if m else x

>>> sorted(['a1b','a100b','a2b','a99b'], key=f)
['a1b', 'a2b', 'a99b', 'a100b']

You can achieve some speed gain if you have a preexisting list on which you call list.sort . list.sort performs an in-place sort and is going to be a bit faster than sorted because it operates in place and does not generate a copy of the data.

Another thing to note is that this version of a regex based sort is more robust than a lambda . It becomes possible to catch and handle exceptions, and you aren't constrained by the single line requirement of a lambda .

---

Performance

l = ['a1b','a100b','a2b','a99b'] * 10000

%timeit natsorted(l)
1 loop, best of 3: 437 ms per loop

%timeit sorted(l, key=f)
10 loops, best of 3: 92.4 ms per loop

Note that actual timings differ by versions, environment, and data. I have not benchmarked the other answers as they do not generalise well to arbitrarily structured input.

Answer 2

You can use regular expressions to isolate the digits for the key function that you pass to list.sort or sorted :

import re

pat = re.compile(r'a(\d+)b')  # capture group of digits between a and b
lst = ['a1b', 'a100b', 'a2b', 'a99b']
sorted(lst, key=lambda s: int(pat.search(s).group(1)))
# ['a1b', 'a2b', 'a99b', 'a100b']

Answer 3

You can simply extract middle value by int(s[1:-1]) as a key to compare:

>>> L = ['a1b','a100b','a2b','a99b']
>>> L.sort(key=lambda s: int(s[1:-1]))
>>> L
['a1b', 'a2b', 'a99b', 'a100b']