Is there a Numpy function to return the first index of something in an array?

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]),)

you can also convert a Numpy array to list in the air and get its index . for example

l = [1,2,3,4,5] #python list
a = numpy.array(l) #numpy array
i = a.tolist().index(2) # i will return index of 2
print i

Will print 1.

If you're going to use this as an index into something else, you can use boolean indices if the arrays are broadcastable; you don't need explicit indices. The absolute simplest way to do this is to simply index based on a truth value.

other_array[first_array == item]

Any boolean operation works:

a = numpy.arange(100)
other_array[first_array > 50]

The nonzero method takes booleans, too:

index = numpy.nonzero(first_array == item)[0][0]

The two zeros are for the tuple of indices (assuming first_array is 1D) and then the first item in the array of indices.

to index on any criteria, you can so something like the following:

In [1]: from numpy import *
In [2]: x = arange(125).reshape((5,5,5))
In [3]: y = indices(x.shape)
In [4]: locs = y[:,x >= 120] # put whatever you want in place of x >= 120
In [5]: pts = hsplit(locs, len(locs[0]))
In [6]: for pt in pts:
   .....:         print(', '.join(str(p[0]) for p in pt))
4, 4, 0
4, 4, 1
4, 4, 2
4, 4, 3
4, 4, 4

[edit] and here's a quick function to do what list.index() does, except doesn't raise an exception if it's not found. beware -- this is probably very slow on large arrays. you can probably monkeypatch this on to arrays if you'd rather use it as a method.

def ndindex(ndarray, item):
    if len(ndarray.shape) == 1:
        try:
            return [ndarray.tolist().index(item)]
        except:
            pass
    else:
        for i, subarray in enumerate(ndarray):
            try:
                return [i] + ndindex(subarray, item)
            except:
                pass
In [1]: ndindex(x, 103)
Out[1]: [4, 0, 3]

There are lots of operations in numpy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:

numpy.nonzero(array - item)

You could then take the first elements of the lists to get a single element.

An alternative to selecting the first element from np.where() is to use a generator expression together with enumerate, such as:

>>> import numpy as np
>>> x = np.arange(100)   # x = array([0, 1, 2, 3, ... 99])
>>> next(i for i, x_i in enumerate(x) if x_i == 2)
2

For a two dimensional array one would do:

>>> x = np.arange(100).reshape(10,10)   # x = array([[0, 1, 2,... 9], [10,..19],])
>>> next((i,j) for i, x_i in enumerate(x) 
...            for j, x_ij in enumerate(x_i) if x_ij == 2)
(0, 2)

The advantage of this approach is that it stops checking the elements of the array after the first match is found, whereas np.where checks all elements for a match. A generator expression would be faster if there's match early in the array.