Python - Array of Indexes of all appearances of a certain element in an array

Question

This question is very similar to this question, Question, except for the fact that I need an answer for Python.

I'll explain the question anyway. If I have an array of characters, for example ["a", "b", "a", "c"], then it needs to return an array that includes the indexes of all the appearances of a certain character.

This means that if I input ['a', 'b', 'a', 'c', 'c', 'a'], and I need to find all appearances of 'a', then it would output [0, 2, 5].

I also need the program to be pretty fast, as I need to mass generate my final product, so the whole program needs to be quick. I will probably be running this piece of code millions of times in my code so the quicker it is the better, but I am accepting anything right now.

Right now my code does not work since it simply output [0, 0, 0] (I am running through a for loop so it isn't fast either)

Could someone please help? Thanks!

Use Numpy if you need speed fir these functions. They are built in. — Ethan
– Ethan, Commented Feb 16, 2020 at 22:04
Why is it a list of characters instead of a string? String might allow faster solutions. — Kelly Bundy
– Kelly Bundy, Commented Feb 16, 2020 at 22:19
@ethan: would would this be? I am already using a lot of numpy in my code, so what would this build-in function be? — Mike Smith
– Mike Smith, Commented Feb 16, 2020 at 22:33
@Bob Robert. If you create a masked array you can use np.where(myMaskedArray==‘a’) and it will return an array of indexes. — Ethan
– Ethan, Commented Feb 16, 2020 at 23:05

Dominic D · Accepted Answer · 2020-02-16 22:27:30Z

4

A simple option, not optimised for speed:

list_of_characters = ['a', 'b', 'a', 'c', 'c', 'a']
character = 'a'
index_list = [i for i,v in enumerate(list_of_characters) if v == character]

Edit: If you're doing this calculation a lot and the list doesn't change between each run, you should pre-compute the solutions and store them in an easily accessible format.

edited Feb 16, 2020 at 22:27

answered Feb 16, 2020 at 22:06

Dominic D

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1 Comment

Alexander Over a year ago

Never use list as a variable name, as it clobbers the builtin list command, otherwise good solution to find the index locations of a single character.

Alexander · Accepted Answer · 2020-02-16 22:22:25Z

2

You can use a defaultdict to default the index locations of each character to a list, then just enumerate over your list and append the index locations. This allows you to later index the location of any character in the list.

from collections import defaultdict

my_list = ['a', 'b', 'a', 'c', 'c', 'a']

dd = defaultdict(list)
for n, c in enumerate(my_list):
    dd[c].append(n)

>>> dd['a']
[0, 2, 5]

>>> dict(dd)
{'a': [0, 2, 5], 'b': [1], 'c': [3, 4]}

edited Feb 16, 2020 at 22:22

answered Feb 16, 2020 at 22:17

Alexander

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1 Comment

Mike Smith Over a year ago

This would be nice, except for the fact that it would use unnecessary memory and time (I know, very slight, but every millisecond here makes a difference).

Ethan · Accepted Answer · 2020-02-16 23:23:21Z

1

Just use the np.where function.

>>> l=np.array(['a','b','a','c','c','a'])
>>> np.where(l=='a')
(array([0, 2, 5]),)

answered Feb 16, 2020 at 23:23

Ethan

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