如何根据另一个列表对列表进行排序? [英] How to sort a list according to another list?

问题描述

有一个列表:

a = [("ax", 1), ("ec", 3), ("bk", 5)]

另一个列表:

b = ["ec", "ax", "bk"]

我想根据b对a进行排序:

sort_it(a, b)

a = [("ec", 3), ("ax", 1), ("bk", 5)]

该怎么做?

推荐答案

a.sort(key=lambda x: b.index(x[0]))

此操作使用a中每个元组的第一个元素的b中的索引作为其排序依据的值对a进行就地排序.

This sorts a in-place using the the index in b of the first element of each tuple from a as the values it sorts on.

另一种可能更简洁的书写方式是:

Another, possibly cleaner, way of writing it would be:

a.sort(key=lambda (x,y): b.index(x))

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如果您有大量项目，则做一些不同的事情可能会更有效率，因为.index()在很长的列表上可能是一项昂贵的操作，并且您实际上不需要进行完整的排序，因为您已经知道订单了:

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If you had large numbers of items, it might be more efficient to do things a bit differently, because .index() can be an expensive operation on a long list, and you don't actually need to do a full sorting since you already know the order:

mapping = dict(a)
a[:] = [(x,mapping[x]) for x in b]

请注意，这仅适用于2元组的列表.如果您希望它适用于任意长度的元组，则需要对其进行一些修改:

Note that this will only work for a list of 2-tuples. If you want it to work for arbitrary-length tuples, you'd need to modify it slightly:

mapping = dict((x[0], x[1:]) for x in a)
a[:] = [(x,) + mapping[x] for x in b]

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