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:
Parameters: | a : int array
choices : sequence of arrays
out : array, optional
mode : {‘raise’ (default), ‘wrap’, ‘clip’}, optional
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Returns: | merged_array : array
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Raises: | ValueError: shape mismatch :
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See also
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]]])