在python中首先可能进行最高长度的字典排序吗？

Question

Let's say I have a list

list = ['aa', 'bb', 'aa', 'aaa', 'bbb', 'bbbb', 'cc']

if you do list.sort() you get back ['aa', 'aa', 'aaa', 'bb', 'bbb', 'bbbb', 'cc']

Is there a way in python 3 we can get

['aaa', 'aa', 'aa', 'bbbb', 'bbb', 'bb', 'cc']

So within the same lexicographical group order, pick the one with the bigger length first.

Thanks a ton for your help!

Answer 1

You can redefine what less-than means for the strings with a custom class. Use that class as the key for list.sort or sorted .

class C:

    def __init__(self, val):
        self.val = val

    def __lt__(self, other):
        min_len = min((len(self.val), len(other.val)))
        if self.val[:min_len] == other.val[:min_len]:
            return len(self.val) > len(other.val)
        else:
            return self.val < other.val

lst = ['aa', 'bb', 'aa', 'aaa', 'bbb', 'bbbb', 'cc']
slist = sorted(lst, key=C)
print(slist)

Answer 2

元组按字典顺序排列，因此您可以使用（字符串的第一个字符，负长度）的元组作为排序键：

list.sort(key=lambda s: (s[0], -len(s)))

Answer 3

You were able to explain in words how to compare two strings, so you can write this comparison function in python. However, since python 3, .sort() and sorted() both expect a key , rather than a comparison function.

• You can turn the comparison function into a key by using a class and defining its method .__lt__ , as explained in tdelaney's answer ;

• Or you can use functools.cmp_to_key , which was designed specifically for this purpose.

cmp_to_key expects a comparison function which returns -1, 0 or +1 respectively for less than, equal or greater than.

def custom_compare_strings(s1, s2):
  length = min(len(s1), len(s2))
  if s1[:length] == s2[:length]:
    return len(s2) - len(s1)
  else:
    return -1 if s1 < s2 else +1

lst = ['aa', 'bb', 'aa', 'aaa', 'bbb', 'bbbb', 'cc']

import functools
slist = sorted(lst, key=functools.cmp_to_key(custom_compare_strings))
print(slist)
# ['aaa', 'aa', 'aa', 'bbbb', 'bbb', 'bb', 'cc']