查找一个列表的任何元素出现在另一个列表中的索引

问题描述

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Take 列表 haystack 和 needles

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

我需要生成一个索引列表，其中 needles 的任何元素出现在 haystack 中.在这种情况下，这些索引是 3、6 和 8，因此

result = [3, 6, 8]

我发现的这个问题 非常相似，用

解决得相当优雅

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

不幸的是，在我的例子中，这个解决方案给出了 ValueError: 'W' is not in list.这是因为这里的区别在于 needles 的元素可能会在 haystack 中出现多次或根本不出现.

换句话说，haystack 可能没有针，也可能有很多.

解决方案

haystack = ['a', 'b', 'c', 'V', 'd', 'e', 'X', 'f', 'V', 'g', 'h']针 = ['V', 'W', 'X', 'Y', 'Z']st = 设置(针)打印([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 Needs if i in haystack] 它也不起作用，因为您有重复的元素.

使 st = set(needles) 意味着我们有一个线性解决方案，因为集合查找是 0(1)，这对于大输入会显着提高效率.>

Take lists haystack and needles

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

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]

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

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

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.

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

解决方案

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]

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

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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