Find indices of elements equal to zero in a NumPy array

numpy.where() is my favorite.

>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> numpy.where(x == 0)[0]
array([1, 3, 5])

The method where returns a tuple of ndarrays, each corresponding to a different dimension of the input. Since the input is one-dimensional, the [0] unboxes the tuple's only element.

You can search for any scalar condition with:

>>> a = np.asarray([0,1,2,3,4])
>>> a == 0 # or whatver
array([ True, False, False, False, False], dtype=bool)

Which will give back the array as an boolean mask of the condition.

You can also use nonzero() by using it on a boolean mask of the condition, because False is also a kind of zero.

>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])

>>> x==0
array([False, True, False, True, False, True, False, False, False, False, False], dtype=bool)

>>> numpy.nonzero(x==0)[0]
array([1, 3, 5])

It's doing exactly the same as mtrw's way, but it is more related to the question ;)

If you are working with a one-dimensional array there is a syntactic sugar:

>>> x = numpy.array([1,0,2,0,3,0,4,5,6,7,8])
>>> numpy.flatnonzero(x == 0)
array([1, 3, 5])