Posted By: Anonymous
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, a (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]
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. Instead, you could try using
* will be treated like matrix multiplication.
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.dotwith two major exceptions: no scalar multiplication but it works with stacks of matrices.
>>> np.matmul(a, b) array([16, 6, 8])
numpy.innerfunctions the same way as
numpy.dotfor 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.tensordotwith the optional argument
>>> np.tensordot(a, b, axes=1) array([16, 6, 8])
numpy.vdotif 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