numpy.choose()

Construct an array from an index array and a set of arrays to choose from.

First of all, if confused or uncertain, definitely look at the Examples - in its full generality, this function is less simple than it might seem from the following code description (below ndi = numpy.lib.index_tricks):

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

But this omits some subtleties. Here is a fully general summary:

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. Then, a new array with shape Ba.shape is created as follows:

• if mode=raise (the default), then, first of all, each element of a (and thus Ba) must be in the range [0, n-1]; now, 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;
• if mode=wrap, values in a (and thus Ba) may be any (signed) integer; modular arithmetic is used to map integers outside the range [0, n-1] back into that range; and then the new array is constructed as above;
• if mode=clip, values in a (and thus Ba) may be any (signed) integer; negative integers are mapped to 0; values greater than n-1 are mapped to n-1; and then the new array is constructed as above.

Notes

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.

Examples

>>> choices = [[0, 1, 2, 3], [10, 11, 12, 13],
...   [20, 21, 22, 23], [30, 31, 32, 33]]
>>> np.choose([2, 3, 1, 0], choices
... # the first element of the result will be the first element of the
... # third (2+1) "array" in choices, namely, 20; the second element
... # will be the second element of the fourth (3+1) choice array, i.e.,
... # 31, etc.
... )
array([20, 31, 12,  3])
>>> np.choose([2, 4, 1, 0], choices, mode='clip') # 4 goes to 3 (4-1)
array([20, 31, 12,  3])
>>> # because there are 4 choice arrays
>>> np.choose([2, 4, 1, 0], choices, mode='wrap') # 4 goes to (4 mod 4)
array([20,  1, 12,  3])
>>> # i.e., 0

A couple examples illustrating how choose broadcasts:

>>> a = [[1, 0, 1], [0, 1, 0], [1, 0, 1]]
>>> choices = [-10, 10]
>>> np.choose(a, choices)
array([[ 10, -10,  10],
       [-10,  10, -10],
       [ 10, -10,  10]])

>>> # With thanks to Anne Archibald
>>> a = np.array([0, 1]).reshape((2,1,1))
>>> c1 = np.array([1, 2, 3]).reshape((1,3,1))
>>> c2 = np.array([-1, -2, -3, -4, -5]).reshape((1,1,5))
>>> np.choose(a, (c1, c2)) # result is 2x3x5, res[0,:,:]=c1, res[1,:,:]=c2
array([[[ 1,  1,  1,  1,  1],
        [ 2,  2,  2,  2,  2],
        [ 3,  3,  3,  3,  3]],
       [[-1, -2, -3, -4, -5],
        [-1, -2, -3, -4, -5],
        [-1, -2, -3, -4, -5]]])