我认为如果两个列表的大小不同或包含不同的项目，上面的大多数解决方案都是无效的。下面的解决方案很简单，应该可以解决这些问题:

import pandas as pd

list1 = ['B', 'A', 'C']  # Required sort order
list2 = ['C', 'A']       # Items to be sorted according to list1

result = pd.merge(pd.DataFrame(list1), pd.DataFrame(list2))
print(list(result[0]))

输出:

['A', 'C']

注意:任何不在list1中的项都将被忽略，因为算法不知道要使用的排序顺序。

More_itertools有一个并行排序可迭代对象的工具:

鉴于

from more_itertools import sort_together


X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0,   1,   1,    0,   1,   2,   2,   0,   1]

Demo

sort_together([Y, X])[1]
# ('a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g')

Zip，按第二列排序，返回第一列。

zip(*sorted(zip(X,Y), key=operator.itemgetter(1)))[0]

我创建了一个更通用的函数，它根据另一个列表对两个以上的列表进行排序，灵感来自@Whatang的答案。

def parallel_sort(*lists):
    """
    Sorts the given lists, based on the first one.
    :param lists: lists to be sorted

    :return: a tuple containing the sorted lists
    """

    # Create the initially empty lists to later store the sorted items
    sorted_lists = tuple([] for _ in range(len(lists)))

    # Unpack the lists, sort them, zip them and iterate over them
    for t in sorted(zip(*lists)):
        # list items are now sorted based on the first list
        for i, item in enumerate(t):    # for each item...
            sorted_lists[i].append(item)  # ...store it in the appropriate list

    return sorted_lists

下面是Whatangs的答案，如果你想获得两个排序的列表(python3)。

X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0,   1,   1,    0,   1,   2,   2,   0,   1]

Zx, Zy = zip(*[(x, y) for x, y in sorted(zip(Y, X))])

print(list(Zx))  # [0, 0, 0, 1, 1, 1, 1, 2, 2]
print(list(Zy))  # ['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']

记住Zx和Zy是元组。 我也在想是否有更好的方法来做到这一点。

警告:如果你用空列表运行它，它会崩溃。

X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0,   1,   1,   0,   1,   2,   2,   0,   1 ]

你可以用一行写出来:

X, Y = zip(*sorted(zip(Y, X)))