Hi guys,
I am trying to compute the dot product of matrices A & B but getting different results compared to when I use the numpy.dot(A, B).
A*B and A.*B also give different results.
According to numpy docs (numpy.dot — NumPy v1.23 Manual), this is what’s happening in my case: dot(a, b)[i,j,k,m] = sum(a[i,j,:] * b[k,:,m]). How do I do this in Julia? Thanks!
Those are totally different operations in Julia. A * B is general matrix multiplication (or matrix-vector multiplication as appropriate). A .* B is the elementwise product of A and B (equivalent to what * does for numpy.ndarray).
Can you give a minimal example of the data you have, the result you want, and what you get instead?
Welcome. I assume that you are new to Julia and I can’t easily check the numpy.dot function for some example result.
But perhaps you are just looking for:
julia> using LinearAlgebra
julia> a=rand(1:10,(3,3,3));
julia> b=rand(1:10,(3,3,3));
julia> dot(a,b)
973
Here’s an example
Julia
using LinearAlgebra
A = [1. 0. 0. 0. 2.; 0. 0. 3. 0. 0.; 0. 0. 0. 0. 0.; 0. 2. 0. 0. 0.; 0. 2. 0. 0. 0.]
F = svd(A)
u, s, v = F;
h = v'.*(diagm(s)*v)
Python
from scipy import linalg
import scipy as sp
A = sp.array(sp.mat('1. 0. 0. 0. 2.; 0. 0. 3. 0. 0.; 0. 0. 0. 0. 0.; 0. 2. 0. 0. 0.; 0. 2. 0. 0. 0.'))
u, s, v = linalg.svd(A)
h = sp.dot(v.T, sp.dot(sp.diag(s), v))
Ok, thanks. You don’t want .* here at all–it’s not a “dot product” in the linear algebra sense, it’s an elementwise product.
Note also that the v you get from Julia is the adjoint of what you got in Python, but that’s easy to deal with. We can also use Diagonal to create a very cheap reference to s that behaves like a diagonal matrix, without actually filling in all the pointless zeros:
julia> h = v * Diagonal(s) * v'
5×5 Matrix{Float64}:
0.447214 0.0 0.0 0.0 0.894427
0.0 2.82843 0.0 0.0 0.0
0.0 0.0 3.0 0.0 0.0
0.0 0.0 0.0 0.0 0.0
0.894427 0.0 0.0 0.0 1.78885
Ah, thanks!