CODEWITHSUNDEEP
pythonlist
Sanchit
Sanchit

Reputation: 3289

Find indices of elements in a list in Python

I have a list with elements, while some of these elements can be repeated. For example, a = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]. I want to find indices of all of these elements. Output should be like: For element 1, indices are [1, 5, 9]. For element 2, indices are [2, 6, 10] etc...

Can somebody please tell me how to do that? Note, code should be general as much as possible.

Upvotes: 1

Views: 157

Answers (5)

user2555451
user2555451

Reputation:

Here is a pretty general means:

>>> lst = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]
>>> dct = {x:[] for x in lst}
>>> for x,y in enumerate(lst, 1):
...     dct[y].append(x)
...
>>> dct
{1: [1, 5, 9], 2: [2, 6, 10], 3: [3, 7, 11], 4: [4, 8, 12]}
>>>

Note however that Python indexes start at 0, so the list for 1 should be [0, 4, 8], the list for 2 [1, 5, 9], etc. However, since you want the indexes to be +1, I set enumerate to start at 1.

The above solution uses pure Python without any imports. However, if you import collections.defaultdict, you can increase the performance:

>>> from collections import defaultdict
>>> dct = defaultdict(list)
>>> for x,y in enumerate(lst, 1):
...     dct[y].append(x)
...
>>> dct
{1: [1, 5, 9], 2: [2, 6, 10], 3: [3, 7, 11], 4: [4, 8, 12]}
>>>

Upvotes: 3

Jon Clements
Jon Clements

Reputation: 142106

As long as the item is hashable, then:

from collections import defaultdict

data = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]
dd = defaultdict(list)
for idx, item in enumerate(data):
    dd[item].append(idx)

# defaultdict(<type 'list'>, {1: [0, 4, 8], 2: [1, 5, 9], 3: [2, 6, 10], 4: [3, 7, 11]})

Upvotes: 2

DanielB
DanielB

Reputation: 2958

You can try using something like this:

def get_indexes(my_array, item):
    return [i for i, e in enumerate(my_array) if e == item]

Using one of your examples:

>>> print get_indexes([1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4], 2)
[1, 5, 9]

Upvotes: 1

Akavall
Akavall

Reputation: 86128

numpy can be useful for something like this:

>>>> import numpy as np
>>> a
[1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]
>>> np.where(np.array(a) == 1)[0]
array([0, 4, 8])

Upvotes: 0

Kobi K
Kobi K

Reputation: 7931

Simple Example Using enumerate

list = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]    

myIndexes = [i for i,value in enumerate(list) if value == 1]

print myIndexs

[0, 4, 8]

At your example you said:

For element 1, indices are [1, 5, 9]

You actually wanted index + 1! pay attention! Lists start from 0.

So to get index + 1 you can do:

myIndexes = [i+1 for i,value in enumerate(list) if value == 1]

print myIndexs

[1, 5, 9]

Upvotes: 1

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