-
numpy.tensordot(a, b, axes=2)
[source] -
Compute tensor dot product along specified axes for arrays >= 1-D.
Given two tensors (arrays of dimension greater than or equal to one),
a
andb
, and an array_like object containing two array_like objects,(a_axes, b_axes)
, sum the products ofa
?s andb
?s elements (components) over the axes specified bya_axes
andb_axes
. The third argument can be a single non-negative integer_like scalar,N
; if it is such, then the lastN
dimensions ofa
and the firstN
dimensions ofb
are summed over.Parameters: a, b : array_like, len(shape) >= 1
Tensors to ?dot?.
axes : int or (2,) array_like
- integer_like If an int N, sum over the last N axes of
a
and the first N axes ofb
in order. The sizes of the corresponding axes must match. - (2,) array_like Or, a list of axes to be summed over, first sequence applying to
a
, second tob
. Both elements array_like must be of the same length.
Notes
- Three common use cases are:
-
axes = 0
: tensor product $aotimes b$axes = 1
: tensor dot product $acdot b$axes = 2
: (default) tensor double contraction $a:b$
When
axes
is integer_like, the sequence for evaluation will be: first the -Nth axis ina
and 0th axis inb
, and the -1th axis ina
and Nth axis inb
last.When there is more than one axis to sum over - and they are not the last (first) axes of
a
(b
) - the argumentaxes
should consist of two sequences of the same length, with the first axis to sum over given first in both sequences, the second axis second, and so forth.Examples
A ?traditional? example:
>>> a = np.arange(60.).reshape(3,4,5) >>> b = np.arange(24.).reshape(4,3,2) >>> c = np.tensordot(a,b, axes=([1,0],[0,1])) >>> c.shape (5, 2) >>> c array([[ 4400., 4730.], [ 4532., 4874.], [ 4664., 5018.], [ 4796., 5162.], [ 4928., 5306.]]) >>> # A slower but equivalent way of computing the same... >>> d = np.zeros((5,2)) >>> for i in range(5): ... for j in range(2): ... for k in range(3): ... for n in range(4): ... d[i,j] += a[k,n,i] * b[n,k,j] >>> c == d array([[ True, True], [ True, True], [ True, True], [ True, True], [ True, True]], dtype=bool)
An extended example taking advantage of the overloading of + and *:
>>> a = np.array(range(1, 9)) >>> a.shape = (2, 2, 2) >>> A = np.array(('a', 'b', 'c', 'd'), dtype=object) >>> A.shape = (2, 2) >>> a; A array([[[1, 2], [3, 4]], [[5, 6], [7, 8]]]) array([[a, b], [c, d]], dtype=object)
>>> np.tensordot(a, A) # third argument default is 2 for double-contraction array([abbcccdddd, aaaaabbbbbbcccccccdddddddd], dtype=object)
>>> np.tensordot(a, A, 1) array([[[acc, bdd], [aaacccc, bbbdddd]], [[aaaaacccccc, bbbbbdddddd], [aaaaaaacccccccc, bbbbbbbdddddddd]]], dtype=object)
>>> np.tensordot(a, A, 0) # tensor product (result too long to incl.) array([[[[[a, b], [c, d]], ...
>>> np.tensordot(a, A, (0, 1)) array([[[abbbbb, cddddd], [aabbbbbb, ccdddddd]], [[aaabbbbbbb, cccddddddd], [aaaabbbbbbbb, ccccdddddddd]]], dtype=object)
>>> np.tensordot(a, A, (2, 1)) array([[[abb, cdd], [aaabbbb, cccdddd]], [[aaaaabbbbbb, cccccdddddd], [aaaaaaabbbbbbbb, cccccccdddddddd]]], dtype=object)
>>> np.tensordot(a, A, ((0, 1), (0, 1))) array([abbbcccccddddddd, aabbbbccccccdddddddd], dtype=object)
>>> np.tensordot(a, A, ((2, 1), (1, 0))) array([acccbbdddd, aaaaacccccccbbbbbbdddddddd], dtype=object)
- integer_like If an int N, sum over the last N axes of
numpy.tensordot()
2017-01-10 18:18:51
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