Select One Element in Each Row of a Numpy Array by Column Indices [duplicate]

Is there a better way to get the "output_array" from the "input_array" and "select_id" ?

Can we get rid of range( input_array.shape[0] ) ?

>>> input_array = numpy.array( [ [3,14], [12, 5], [75, 50] ] )
>>> select_id = [0, 1, 1]
>>> print input_array
[[ 3 14]
 [12  5]
 [75 50]]

>>> output_array = input_array[  range( input_array.shape[0] ), select_id ]
>>> print output_array
[ 3  5 50]

Solution 1:

You can choose from given array using numpy.choose which constructs an array from an index array (in your case select_id) and a set of arrays (in your case input_array) to choose from. However you may first need to transpose input_array to match dimensions. The following shows a small example:

In [101]: input_array
Out[101]: 
array([[ 3, 14],
       [12,  5],
       [75, 50]])

In [102]: input_array.shape
Out[102]: (3, 2)

In [103]: select_id
Out[103]: [0, 1, 1]

In [104]: output_array = np.choose(select_id, input_array.T)

In [105]: output_array
Out[105]: array([ 3,  5, 50])

Solution 2:

(because I can't post this as a comment on the accepted answer)

Note that numpy.choose only works if you have 32 or fewer choices (in this case, the dimension of your array along which you're indexing must be of size 32 or smaller). Additionally, the documentation for numpy.choose says

To reduce the chance of misinterpretation, even though the following "abuse" is nominally supported, choices should neither be, nor be thought of as, a single array, i.e., the outermost sequence-like container should be either a list or a tuple.

The OP asks:

1. Is there a better way to get the output_array from the input_array and select_id?
   • I would say, the way you originally suggested seems the best out of those presented here. It is easy to understand, scales to large arrays, and is efficient.
2. Can we get rid of range(input_array.shape[0])?
   • Yes, as shown by other answers, but the accepted one doesn't work in general so well as what the OP already suggests doing.