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.

Padraic Cunningham · Answer 1 · 2015-04-05T00:25:19.523

23

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.

edited Apr 05 '15 at 00:25

answered Apr 05 '15 at 00:14

Padraic Cunningham

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definitely a better solution to put `needles` in a set. The hash table lookup definitely improves performance. – Anzel Apr 05 '15 at 00:17
@Anzel, defo, obviously won't make much difference here but on a large dataset it would be significant. – Padraic Cunningham Apr 05 '15 at 00:22

Saksham Varma · Answer 2 · 2015-04-05T01:04:05.223

4

needles_set = set(needles)
print [i for i, val in enumerate(haystack) if val in needles_set]

edited Apr 05 '15 at 01:04

answered Apr 05 '15 at 00:58

Saksham Varma

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How to find the indices where the needles_set is not found in haystack? – thepunitsingh Jan 11 '19 at 07:58

score 1 · Answer 3 · answered Apr 05 '15 at 00:19

1

You could try something like the following.

[Haystack.index(x) for x in needles if x in Haystack]

If x is not in haystack then haystack.index(x) will not be called and no error should be thrown.

answered Apr 05 '15 at 00:19

Daniel Robertson

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won't actually work because you would always get the first index of any repeated elements – Padraic Cunningham Apr 05 '15 at 00:26

score 0 · Answer 4 · answered Apr 05 '15 at 00:12

In addition to failing if your needle is not in the haystack, the index method will return only the first position of the element you're looking for, even if that element appears more than once (as in the 'V' in your example). You could just do this:

result = [idx for idx, val in enumerate(haystack) if val in needles]

The enumerate function produces a generator that yields tuples of values - the first being the index and the second being a value:

>>> print(list(enumerate(['a', 'b', 'c'])))

Just check if each value is in your needles list and add the index if it is.

score 0 · Answer 5 · answered Apr 05 '15 at 00:15

0

Definitely not the most efficient way but you could do something like this:

result = []
i=0
while (i < len(haystack)):
    if (needles.count(haystack[i]) > 0):
        result.append(i)
    i+=1

Which will make result = [3, 6, 8]

answered Apr 05 '15 at 00:15

Jon Warren

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

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