查找一个列表中的任何元素出现在另一个列表中的索引 [英] Find the indices at which any element of one list occurs in another

问题描述

获取列表haystack和needles

haystack = ['a', 'b', 'c', 'V', 'd', 'e', 'X', 'f', 'V', 'g', 'h']
needles = ['V', 'W', 'X', 'Y', 'Z']

我需要生成在haystack中出现needles任何元素的索引列表.在这种情况下，这些索引分别是3、6和8

I need to generate a list of the indices at which any element of needles occurs in haystack. In this case those indices are 3, 6, and 8 thus

result = [3, 6, 8]

我发现的这个问题非常相似，并且可以通过

This question I found is very similar and was rather elegantly solved with

result = [haystack.index(i) for i in needles]

不幸的是，在我的情况下，此解决方案给出了ValueError: 'W' is not in list.这是因为此处的区别在于needles的元素可能在haystack中多次出现或根本不出现.

Unfortunately, this solution gives ValueError: 'W' is not in list in my case. This is because the difference here is that an element of needles may occur in haystack a number of times or not at all.

换句话说，haystack可能不包含针或可能包含许多针.

In other words, haystack may contain no needles or it may contain many.

推荐答案

haystack = ['a', 'b', 'c', 'V', 'd', 'e', 'X', 'f', 'V', 'g', 'h']
needles = ['V', 'W', 'X', 'Y', 'Z']
st = set(needles)
print([i for i, e in enumerate(haystack) if e in st])
[3, 6, 8]

即使您使用[haystack.index(i) for i in needles if i in haystack]，它也会不工作，因为您重复了元素.

Even if you used [haystack.index(i) for i in needles if i in haystack] it would not work as you have repeated elements.

进行st = set(needles)意味着我们有一个线性解决方案，因为集合查找为0(1)，对于大输入而言，这将显着提高效率.

Making st = set(needles) means we have a linear solution as set lookups are 0(1) which for large input would be significantly more efficient.

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