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I want to take the indices of each element in the list a=[2,4,5,2] when I enter a.index(2) I get 0 as output. How can I take the index of the fourth element like that way in Python?

Taku
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Sarith Wasitha
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    `a.index(2)` returns the index of the first `2` in `a`. Did you mean you want to access the element `4` (different from the fourth element)? That would simply be `a[1]` (python indexing starts at 0). Also try `[(i, x) in enumerate(a)]`. – pault Jan 27 '18 at 06:11
  • Well, there are 2 twos, so it gets the first one – whackamadoodle3000 Jan 27 '18 at 06:11
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    `index()` only gets the index of the first occurrence. Do you want all the indexes for each item? – RoadRunner Jan 27 '18 at 06:11

3 Answers3

2

You could recover all the indices like so

indices = [i for i in xrange(len(a)) if a[i] == 2]
Olivier Melançon
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1

You could try this to get them all:

[print i for i,v in enumerate(a) if v==2] # or any number
whackamadoodle3000
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1

You could also store the indexes for each item in a collections.defaultdict():

from collections import defaultdict

a=[2,4,5,2]

indexes = defaultdict(list)
for i, e in enumerate(a):
    indexes[e].append(i)

print(indexes)

Which gives:

defaultdict(<class 'list'>, {2: [0, 3], 4: [1], 5: [2]})

Then you could access the indexes for each item like this:

>>> indexes[2]
[0, 3]
>>> for i in indexes[2]:
...     print(i, '-->', a[i])
... 
0 --> 2
3 --> 2

Or as @Idlehands pointed out in the comments, you can use dict.setdefault() here:

indexes = {}
for i, e in enumerate(a):
    indexes.setdefault(e, []).append(i)

print(indexes)

Which also gives:

{2: [0, 3], 4: [1], 5: [2]}
RoadRunner
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  • You can also do this with the `built-in dict`. `indexes.setdefault(e, list()).append(i)` would work just the same. – r.ook Jan 27 '18 at 06:39
  • @Idlehands Yep that would work just as great. Even more efficient: `indexes.setdefault(e, []).append(i)`. I believe using `[]` is less costly than `list()`, I could be mistaken though. – RoadRunner Jan 27 '18 at 06:53