Find the indices at which any element of one list occurs in another

Question

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.

Answer 1

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.