numpy matrix vector multiplication
Question:
When I multiply two numpy
arrays of sizes (n x n)*(n x 1), I get a matrix of size (n x n). Following normal matrix multiplication rules, an (n x 1) vector is expected, but I simply cannot find any information about how this is done in Python’s Numpy module.
The thing is that I don’t want to implement it manually to preserve the speed of the program.
Example code is shown below:
a = np.array([[5, 1, 3], [1, 1, 1], [1, 2, 1]])
b = np.array([1, 2, 3])
print a*b
>>
[[5 2 9]
[1 2 3]
[1 4 3]]
What I want is:
print a*b
>>
[16 6 8]
Answers:
Simplest solution
Use numpy.dot
or a.dot(b)
. See the documentation here.
>>> a = np.array([[ 5, 1 ,3],
[ 1, 1 ,1],
[ 1, 2 ,1]])
>>> b = np.array([1, 2, 3])
>>> print a.dot(b)
array([16, 6, 8])
This occurs because numpy arrays are not matrices, and the standard operations *, +, -, /
work element-wise on arrays.
Note that while you can use numpy.matrix
(as of early 2021) where *
will be treated like standard matrix multiplication, numpy.matrix
is deprecated and may be removed in future releases.. See the note in its documentation (reproduced below):
It is no longer recommended to use this class, even for linear algebra. Instead use regular arrays. The class may be removed in the future.
Thanks @HopeKing.
Other Solutions
Also know there are other options:
-
As noted below, if using python3.5+ the @
operator works as you’d expect:
>>> print(a @ b)
array([16, 6, 8])
-
If you want overkill, you can use numpy.einsum
. The documentation will give you a flavor for how it works, but honestly, I didn’t fully understand how to use it until reading this answer and just playing around with it on my own.
>>> np.einsum('ji,i->j', a, b)
array([16, 6, 8])
-
As of mid 2016 (numpy 1.10.1), you can try the experimental numpy.matmul
, which works like numpy.dot
with two major exceptions: no scalar multiplication but it works with stacks of matrices.
>>> np.matmul(a, b)
array([16, 6, 8])
-
numpy.inner
functions the same way as numpy.dot
for matrix-vector multiplication but behaves differently for matrix-matrix and tensor multiplication (see Wikipedia regarding the differences between the inner product and dot product in general or see this SO answer regarding numpy’s implementations).
>>> np.inner(a, b)
array([16, 6, 8])
# Beware using for matrix-matrix multiplication though!
>>> b = a.T
>>> np.dot(a, b)
array([[35, 9, 10],
[ 9, 3, 4],
[10, 4, 6]])
>>> np.inner(a, b)
array([[29, 12, 19],
[ 7, 4, 5],
[ 8, 5, 6]])
Rarer options for edge cases
-
If you have tensors (arrays of dimension greater than or equal to one), you can use numpy.tensordot
with the optional argument axes=1
:
>>> np.tensordot(a, b, axes=1)
array([16, 6, 8])
-
Don’t use numpy.vdot
if you have a matrix of complex numbers, as the matrix will be flattened to a 1D array, then it will try to find the complex conjugate dot product between your flattened matrix and vector (which will fail due to a size mismatch n*m
vs n
).
When I multiply two numpy
arrays of sizes (n x n)*(n x 1), I get a matrix of size (n x n). Following normal matrix multiplication rules, an (n x 1) vector is expected, but I simply cannot find any information about how this is done in Python’s Numpy module.
The thing is that I don’t want to implement it manually to preserve the speed of the program.
Example code is shown below:
a = np.array([[5, 1, 3], [1, 1, 1], [1, 2, 1]])
b = np.array([1, 2, 3])
print a*b
>>
[[5 2 9]
[1 2 3]
[1 4 3]]
What I want is:
print a*b
>>
[16 6 8]
Simplest solution
Use numpy.dot
or a.dot(b)
. See the documentation here.
>>> a = np.array([[ 5, 1 ,3],
[ 1, 1 ,1],
[ 1, 2 ,1]])
>>> b = np.array([1, 2, 3])
>>> print a.dot(b)
array([16, 6, 8])
This occurs because numpy arrays are not matrices, and the standard operations *, +, -, /
work element-wise on arrays.
Note that while you can use numpy.matrix
(as of early 2021) where *
will be treated like standard matrix multiplication, numpy.matrix
is deprecated and may be removed in future releases.. See the note in its documentation (reproduced below):
It is no longer recommended to use this class, even for linear algebra. Instead use regular arrays. The class may be removed in the future.
Thanks @HopeKing.
Other Solutions
Also know there are other options:
-
As noted below, if using python3.5+ the
@
operator works as you’d expect:>>> print(a @ b) array([16, 6, 8])
-
If you want overkill, you can use
numpy.einsum
. The documentation will give you a flavor for how it works, but honestly, I didn’t fully understand how to use it until reading this answer and just playing around with it on my own.>>> np.einsum('ji,i->j', a, b) array([16, 6, 8])
-
As of mid 2016 (numpy 1.10.1), you can try the experimental
numpy.matmul
, which works likenumpy.dot
with two major exceptions: no scalar multiplication but it works with stacks of matrices.>>> np.matmul(a, b) array([16, 6, 8])
-
numpy.inner
functions the same way asnumpy.dot
for matrix-vector multiplication but behaves differently for matrix-matrix and tensor multiplication (see Wikipedia regarding the differences between the inner product and dot product in general or see this SO answer regarding numpy’s implementations).>>> np.inner(a, b) array([16, 6, 8]) # Beware using for matrix-matrix multiplication though! >>> b = a.T >>> np.dot(a, b) array([[35, 9, 10], [ 9, 3, 4], [10, 4, 6]]) >>> np.inner(a, b) array([[29, 12, 19], [ 7, 4, 5], [ 8, 5, 6]])
Rarer options for edge cases
-
If you have tensors (arrays of dimension greater than or equal to one), you can use
numpy.tensordot
with the optional argumentaxes=1
:>>> np.tensordot(a, b, axes=1) array([16, 6, 8])
-
Don’t use
numpy.vdot
if you have a matrix of complex numbers, as the matrix will be flattened to a 1D array, then it will try to find the complex conjugate dot product between your flattened matrix and vector (which will fail due to a size mismatchn*m
vsn
).