numpy.choose(a, choices, out=None, mode='raise')
[source]
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:
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;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;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.Parameters: |
|
---|---|
Returns: |
|
Raises: |
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See also
ndarray.choose
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.
>>> 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]]])
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Licensed under the 3-clause BSD License.
https://docs.scipy.org/doc/numpy-1.17.0/reference/generated/numpy.choose.html