What's the difference between numpy.take and numpy.choose?

Question

It seems that numpy.take(array, indices) and numpy.choose(indices, array) return the same thing: a subset of array indexed by indice

Answer 1

They are certainly not equivalent, as you can see by giving the same arguments (switched) to both methods:

>>> a = np.array([[1, 2, 3, 4], 
                  [5, 6, 7, 8], 
                  [9, 10, 11, 12], 
                  [13, 14, 15, 16]])
>>> np.choose([0, 2, 1, 3], a)
array([ 1, 10,  7, 16]) # one from each row
>>> np.take(a, [0, 2, 1, 3])
array([1, 3, 2, 4]) # all from same row

I suggest you read the documentation on take and choose.

Answer 2

Explanation of np.choose() behavior in the documentation.

• numpy.choose(a, choices, out=None, mode='raise')[source]

np.choose(a,c) == np.array([c[a[I]][I] for I in ndi.ndindex(a.shape)])

index array and choices

import numpy as np

print("--------------------------------------------------------------------------------")
print("Index array 'a' and array of choice(s).")
print("--------------------------------------------------------------------------------")
a = np.array([4, 3, 2, 1])
print("'a' is {} shape is {}\n".format(
    a, a.shape
))
    
c = np.arange(60).reshape((5, 3, 4))
_choice = c[0]
print("An example choice is \n{} \nshape is {}\n".format(
    _choice,
    _choice.shape
))

--------------------------------------------------------------------------------
Index array 'a' and array of choice(s).
--------------------------------------------------------------------------------
'a' is [4 3 2 1] shape is (4,)

An example choice is 
[[ 0  1  2  3]
 [ 4  5  6  7]
 [ 8  9 10 11]] 
shape is (3, 4)

Broadcast

Given an “index” array (a) of integers and a sequence of n arrays (choices), a and each choice array are first broadcast, as necessary, to arrays of a common shape; calling these Ba and Bchoices[i], i = 0,…,n-1 we have that, necessarily, Ba.shape == Bchoices[i].shape for each i.

Ba.shape == Bchoices[i].shape.

print("--------------------------------------------------------------------------------")
print("np.choose() broadcasts 'a' and 'choice' to match (here only on a)")
print("--------------------------------------------------------------------------------")
a = np.vstack([
    np.array(a) for i in range(_choice.shape[0])
])
print("Broadcast shape of 'a' is {}\n 'a' is \n{}\n.".format(a.shape, a))

--------------------------------------------------------------------------------
np.choose() broadcasts 'a' and 'choice' to match (here only on a)
--------------------------------------------------------------------------------
Broadcast shape of 'a' is (3, 4)
 'a' is 
[[4 3 2 1]
 [4 3 2 1]
 [4 3 2 1]]

Choose values from choices

c[a[I]][I] for I in ndi.ndindex(a.shape)

suppose that i (in that range) is the value at the (j0, j1, …, jm) position in Ba - then the value at the same position in the new array is the value in Bchoices[i] at that same position;

print("--------------------------------------------------------------------------------")
print("Simulate np.choose() behavior of choosing elements c[a[I]][I]]]:")
print("np.array([ c[a[I]][I] for I in np.lib.index_tricks.ndindex(a.shape) ])")
print("--------------------------------------------------------------------------------")
result = np.array([]).astype(int)
for I in np.lib.index_tricks.ndindex(a.shape):
    print("Selecting the element {} at c[{},{}]".format(
        c[a[I]][I].astype(int), a[I], I
    ))
    result = np.concatenate([result, [c[a[I]][I].astype(int)]])
    print("chosen items: {}".format(result))
    
print("\nResult is \n{}\n".format(
    result.reshape(a.shape)
))

--------------------------------------------------------------------------------
Simulate np.choose() behavior of choosing elements c[a[I]][I]]]:
np.array([ c[a[I]][I] for I in np.lib.index_tricks.ndindex(a.shape) ])
--------------------------------------------------------------------------------
Selecting the element 48 at c[4,(0, 0)]
chosen items: [48]
Selecting the element 37 at c[3,(0, 1)]
chosen items: [48 37]
Selecting the element 26 at c[2,(0, 2)]
chosen items: [48 37 26]
Selecting the element 15 at c[1,(0, 3)]
chosen items: [48 37 26 15]
Selecting the element 52 at c[4,(1, 0)]
chosen items: [48 37 26 15 52]
Selecting the element 41 at c[3,(1, 1)]
chosen items: [48 37 26 15 52 41]
Selecting the element 30 at c[2,(1, 2)]
chosen items: [48 37 26 15 52 41 30]
Selecting the element 19 at c[1,(1, 3)]
chosen items: [48 37 26 15 52 41 30 19]
Selecting the element 56 at c[4,(2, 0)]
chosen items: [48 37 26 15 52 41 30 19 56]
Selecting the element 45 at c[3,(2, 1)]
chosen items: [48 37 26 15 52 41 30 19 56 45]
Selecting the element 34 at c[2,(2, 2)]
chosen items: [48 37 26 15 52 41 30 19 56 45 34]
Selecting the element 23 at c[1,(2, 3)]
chosen items: [48 37 26 15 52 41 30 19 56 45 34 23]

Result is 
[[48 37 26 15]
 [52 41 30 19]
 [56 45 34 23]]

np.choose(a,c)

print("--------------------------------------------------------------------------------")
print("Run np.choose(a, c):")
print("--------------------------------------------------------------------------------")
print(np.choose(a, c))

--------------------------------------------------------------------------------
Run np.choose(a, c):
--------------------------------------------------------------------------------
[[48 37 26 15]
 [52 41 30 19]
 [56 45 34 23]]

Answer 3

numpy.take(array, indices) and numpy.choose(indices, array) behave similarly on 1-D arrays, but this is just coincidence. As pointed out by jonrsharpe, they behave differently on higher-dimensional arrays.

numpy.take

numpy.take(array, indices) picks out elements from a flattened version of array. (The resulting elements are of course not necessarily from the same row.)

For example,

numpy.take([[1, 2], [3, 4]], [0, 3])

returns

array([1, 4])

numpy.choose

numpy.choose(indices, set_of_arrays) plucks out element 0 from array indices[0], element 1 from array indices[1], element 2 from array indices[2], and so on. (Here, array is actually a set of arrays.)

For example

numpy.choose([0, 1, 0, 0], [[1, 2, 3, 4], [4, 5, 6, 7]])

returns

array([1, 5, 3, 4])

because element 0 comes from array 0, element 1 comes from array 1, element 2 comes from array 0, and element 3 comes from array 0.

More Information

These descriptions are simplified – full descriptions can be found here: numpy.take, numpy.choose. For example, numpy.take and numpy.choose behave similarly when indices and array are 1-D because numpy.choose first broadcasts array.