Yes, given an array, array, and a value, item to search for, you can use np.where as:

itemindex = numpy.where(array==item)

The result is a tuple with first all the row indices, then all the column indices.

For example, if an array is two dimensions and it contained your item at two locations then

array[itemindex[0][0]][itemindex[1][0]]

would be equal to your item and so would be:

array[itemindex[0][1]][itemindex[1][1]]

If you need the index of the first occurrence of only one value, you can use nonzero (or where, which amounts to the same thing in this case):

>>> t = array([1, 1, 1, 2, 2, 3, 8, 3, 8, 8])

>>> nonzero(t == 8)

(array([6, 8, 9]),)

>>> nonzero(t == 8)[0][0]

6

If you need the first index of each of many values, you could obviously do the same as above repeatedly, but there is a trick that may be faster. The following finds the indices of the first element of each subsequence:

>>> nonzero(r_[1, diff(t)[:-1]])

(array([0, 3, 5, 6, 7, 8]),)

Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s:

[1, 1, 1, 2, 2, 3, 8, 3, 8, 8]

So it’s slightly different than finding the first occurrence of each value. In your program, you may be able to work with a sorted version of t to get what you want:

>>> st = sorted(t)

>>> nonzero(r_[1, diff(st)[:-1]])

(array([0, 3, 5, 7]),)