How to get a list of indexes selected by a specific value efficiently with numpy arrays?

Question

I have a numpy array like this:

import numpy as np
arr = np.array([9, 6, 3, 8, 2, 3, 3, 4, 4, 9, 5, 6, 6, 6, 6, 7, 8, 9])

And I want to get

Answer 1

This should not be too slow. The array is iterated only once. The result (ind) is a dictionary value -> list of indexes.

import numpy as np
arr = np.array([2, 3, 3, 4, 4, 9, 5, 6, 6, 6, 6, 7, 8, 9])

ind = dict()
for i, val in enumerate(arr):
  ind.setdefault(val, []).append(i)

Answer 2

You can use numpy.unique to find all the unique values and numpy.where to find their indexes:

import numpy as np
arr = np.array([2, 3, 3, 4, 4, 9, 5, 6, 6, 6, 6, 7, 8, 9])

# get the unique values
unique_arr = np.unique(arr)

# loop through the unique numbers and find the indeces
indexes_value = {}
for num in unique_arr:
    indexes = np.where(arr == num)[0]
    indexes_value[num] = indexes  # or list(indexes) if you prefer

Now you have a dictionary of indexes of each value and you can assign what you want to your index_list_* lists.

Answer 3

Might not be the fastes but a oneliner with numpy would be:

index_dict = {v: np.flatnonzero(arr == v) for v in np.unique(arr)}