使用特定方法对字符串列表进行排序 [英] Sorting a list of strings with a specific method

问题描述

假设我有一个这样的字符串列表

let's say i have a list of strings like this

L = ['5', '3', '4', '1', '2', '2 3 5', '2 4 8', '5 22 1 37', '5 22 1 22', '5 22 1 23', ....]

我该如何对列表进行排序，这样我就可以得到这样的东西:

How can i sort this list so that i would have something like this:

L = ['1', '2', '3','4', '5', '2 3 5', '2 4 8', '5 22 1 22', ' 5 22 1 23', '5 22 1 37', ...]

基本上，我需要根据2个字符串之间的第一个不同数字对列表进行排序

basically i need to order the list based on the first different number between 2 strings

推荐答案

您可以使用元组进行排序:

You could sort using a tuple:

L = ['5', '3', '4', '1', '2', '2 3 5', '2 4 8', '5 22 1 37', '5 22 1 22', '5 22 1 23']

result = sorted(L, key=lambda x: (len(x.split()),) + tuple(map(int, x.split())))

print(result)

输出

['1', '2', '3', '4', '5', '2 3 5', '2 4 8', '5 22 1 22', '5 22 1 23', '5 22 1 37']

这个想法是使用一个元组作为键，其中第一个元素是字符串中数字的数量，其余元素是数字的元组.例如对于'2 3 5'，键是(3, 2, 3, 5)

The idea is to use as key a tuple where the first element is the amount of numbers in the string and the rest is the tuple of numbers. For example for '2 3 5' the key is (3, 2, 3, 5)

如@ PM2Ring所建议，您可以使用def函数代替lambda:

As suggested by @PM2Ring you could use a def function instead of a lambda:

def key(x):
    numbers = tuple(map(int, x.split()))
    return (len(numbers),) + numbers

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