How to return all the minimum indices in numpy

Question

I am a little bit confused reading the documentation of argmin function in numpy. It looks like it should do the job:

Reading this

Return the

Answer 1

Assuming that you want the indices of a list, not a numpy array, try

my_list = [5, 3, 2, 1, 1, 1, 6, 1]
np.where(my_list == min(my_list))[0]

The index [0] is because numpy returns a tuple of your answer and nothing (answer as a numpy array). Don't ask me why.

Answer 2

I would like to quickly add that as user grofte mentioned, np.where returns a tuple and it states that it is a shorthand for nonzero which has a corresponding method flatnonzero which returns an array directly.

So, the cleanest version seems to be

my_list = np.array([5, 3, 2, 1, 1, 1, 6, 1])
np.flatnonzero(my_list == my_list.min())
=> array([3, 4, 5, 7])

Answer 3

That documentation makes more sense when you think about multidimensional arrays.

>>> x = numpy.array([[0, 1],
...                  [3, 2]])
>>> x.argmin(axis=0)
array([0, 0])
>>> x.argmin(axis=1)
array([0, 1])

With an axis specified, argmin takes one-dimensional subarrays along the given axis and returns the first index of each subarray's minimum value. It doesn't return all indices of a single minimum value.

To get all indices of the minimum value, you could do

numpy.where(x == x.min())

Answer 4

See the documentation for numpy.argmax (which is referred to by the docs for numpy.argmin):

In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned.

The phrasing of the documentation ("indices" instead of "index") refers to the multidimensional case when axis is provided.

So, you can't do it with np.argmin. Instead, this will work:

np.where(arr == arr.min())