CODEWITHSUNDEEP
pythonintersection
MarkokraM
MarkokraM

Reputation: 990

Returning indexes of intersected lists with Python

I have two lists:

1. ['a', 'b', 'c', 'd', 'e', 'c', 'd', 'f']
2. ['c', 'd']

and I'd like to get indexes of the intersection a, b:

3. [[2, 3], [5, 6]]

How would you do that with Python?

Also these inputs:

1. ['263', '9', '470', '370', '576', '770', '800', '203', '62', '370', '576', '370', '25', '770', '484', '61', '914', '301', '550', '770', '484', '1276', '108']
2. ['62', '370', '576']

should give:

3. [[8, 9, 10]]

Upvotes: 1

Views: 108

Answers (3)

Daniel Holden
Daniel Holden

Reputation: 88

For your given example this will work

>>> x = ['a', 'b', 'c', 'd', 'e', 'c', 'd', 'f']
>>> y = ['c', 'd']
>>> z = [[i for i, xi in enumerate(x) if xi == yi] for yi in y]
>>> z
[[2, 5], [3, 6]]
>>> zip(*z)
[(2, 3), (5, 6)]

It makes uses of the enumerate function to get the indices of x along with the values and then transposes the result using zip(*z). You can convert from tuples to lists afterward.

Edit: transposed result.

Upvotes: 1

devnull
devnull

Reputation: 123708

One way would be:

>>> l1 = ['a', 'b', 'c', 'd', 'e', 'c', 'd', 'f']
>>> l2 = ['c', 'd']
>>> [range(i,i+len(l2)) for i in xrange(len(l1)-len(l2)+1) if l2 == l1[i:i+len(l2)]]
[[2, 3], [5, 6]]
>>>

Upvotes: 2

martijnn2008
martijnn2008

Reputation: 3640

Maybe a little too much code, but it works. 

def indexes(list, element):
    c = 0
    output = []
    for e in list:
        if e == element:
            output.append(c)
        c += 1
    return output

a = ['a', 'b', 'c', 'd', 'a']
b = ['a', 'd']
output = []
for el in b:
    output.append(indexes(a, el))

print(output)

Upvotes: 0

Related Questions