如何根据另一个列表对元组列表进行排序 [英] How to sort a list of tuples 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)]

如何做到这一点?

推荐答案

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

这使用 a 中每个元组的第一个元素的 b 中的索引作为它排序的值就地对 a 进行排序

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写法是:

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

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

<小时>

如果你有大量的项目，做一些不同的事情可能会更有效，因为 .index() 可能是一个长列表上的昂贵操作，而你没有由于您已经知道顺序，因此实际上需要进行完整排序:

---

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 元组列表.如果您希望它适用于任意长度的元组，则需要稍微修改它:

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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