Python根据另一个列表对一个列表进行排序 [英] Python Sort One List According to Another List

问题描述

我有两个列表，第一个列表是键顺序，第二个列表是元组列表.

I have two lists, the first list is the key order, the second list is a tuple list.

colorOrder = ['red', 'blue', 'yellow', 'green']
tupleList = [(111,'red'),(222,'pink'),(333,'green')]

请注意，这两个列表不是一对一的关系. colorOrder中没有某些颜色，colorOrder中的某些颜色从未出现在tupleList中.因此，它不同于其他类似的重复问题.

Please notice the two lists are not one-to-one relationship. Some colors are not in colorOrder, and some colors in colorOrder never appear in tupleList. So It is different from other similiar duplicate problems.

我需要根据colorOrder对tupleList进行排序.

I need to Sort the tupleList according to the colorOrder.

我可以使用两个嵌套的for循环来解决此问题，但需要更有效的解决方案.

I can solve this problem using two nested for loops, but need a more efficient solution.

#First sort according to the color order
    for aColor in colorOrder:
        for aTuple in tupleList:
            if aTuple[1] == aColor:
                ResultList.append(aTuple)
#Second add the tuples to the ResultList, whose color is not in the colorOrder
    for aTuple in tupleList:
        if aTuple[1] not in colorOrder:
            ResultList.append(aTuple)

推荐答案

首先，我将colorOrder映射为:

colorMap = {c: i for i, c in enumerate(colorOrder)}

现在，使用colorMap.get

sorted(tupleList, key=lambda tup: colorMap.get(tup[1], -1))

这会将事物不在地图中首先.如果您希望将它们添加为 last ，只需使用一个非常大的数字即可:

This puts things not in the map first. If you'd rather add them last, just use a really big number:

sorted(tupleList, key=lambda tup: colorMap.get(tup[1], float('inf')))

这篇关于Python根据另一个列表对一个列表进行排序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！