如何在列表中找到首先出现在另一个给定列表中的元素的索引？ [英] How to find the index of the element in a list that first appears in another given list?

问题描述

a = [3, 4, 2, 1, 7, 6, 5]
b = [4, 6]

答案应该是1.因为在a中，4在列表b中首先出现，并且它的索引是1。

The answer should be 1. Because in a, 4 appears first in list b, and it's index is 1.

问题是python中有没有快速代码来实现这个目标？

The question is that is there any fast code in python to achieve this?

PS：实际上a是随机排列而b是子集a，但它表示为一个列表。

PS: Actually a is a random permutation and b is a subset of a, but it's represented as a list.

推荐答案

如果 b 是被视为子集（顺序无关紧要，所有值都出现在 a 中），然后使用 min（）带 map（）：

If b is to be seen as a subset (order doesn't matter, all values are present in a), then use min() with a map():

min(map(a.index, b))

这将返回最低的索引。这是一个O（NK）解决方案（其中N是 a 的长度，K是 b 的K），但是所有循环都在C代码中执行。

This returns the lowest index. This is a O(NK) solution (where N is the length of a, K that of b), but all looping is executed in C code.

另一种选择是将 a 转换为集合并使用 next（）循环超过 enumerate（）：

Another option is to convert a to a set and use next() on a loop over enumerate():

bset = set(b)
next(i for i, v in enumerate(a) if v in bset)

这是一个O（N）解决方案，但具有更高的常量成本（要执行的Python字节码）。它在很大程度上取决于 a 和 b 的大小，哪一个会更快。

This is a O(N) solution, but has higher constant cost (Python bytecode to execute). It heavily depends on the sizes of a and b which one is going to be faster.

对于问题中的小输入示例， min（map（...）） wins：

For the small input example in the question, min(map(...)) wins:

In [86]: a = [3, 4, 2, 1, 7, 6, 5]
    ...: b = [4, 6]
    ...:

In [87]: %timeit min(map(a.index, b))
    ...:
608 ns ± 64.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [88]: bset = set(b)
    ...:

In [89]: %timeit next(i for i, v in enumerate(a) if v in bset)
    ...:
717 ns ± 30.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

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