# Matrix Multiplication with Python 3.5

Only recently I have started to use Python 3. It's been out for good 8+ years and all these excuses about *incompatibility *with some packages were just lazy. Most packages I use are already ported and if I ever find that something is incompatible... well, I'll think then. But for now let me pat myself on the back for this great leap, because:

In Python 3.5.3 (released today) there is an operator for matrix multiplication! Check out: PEP 465 -- A dedicated infix operator for matrix multiplication. The choice of operator, **@**, is a bit unfortunate, because of the decorators and general association with reference/internet, but seeing how few possibilities are left it's probably the best choice.

Yes, this is big news for me. The number of times I confused myself with my own matrix operations is just too damn high! I cannot agree more with the author of the PEP 465, so let my shamelessly copy&paste (paraphrased) his reasoning. Behold!

(...) encounter many mathematical formulas that look like:Again: yes!Here the various variables are all vectors or matrices (details for the curious: [5] ).S= (Hβ−r)^{T }(HVH^{T })^{− 1 }(Hβ−r)Now we need to write code to perform this calculation. In current numpy, matrix multiplication can be performed using either the function or method call syntax. Neither provides a particularly readable translation of the formula:

import numpy as np from numpy.linalg import inv, solve # Using dot function: S = np.dot((np.dot(H, beta) - r).T, np.dot(inv(np.dot(np.dot(H, V), H.T)), np.dot(H, beta) - r)) # Using dot method: S = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r)With the

@operator, the direct translation of the above formula becomes:S = (H @ beta - r).T @ inv(H @ V @ H.T) @ (H @ beta - r)Notice that there is now a transparent, 1-to-1 mapping between the symbols in the original formula and the code that implements it.

Of course, an experienced programmer will probably notice that this is not the best way to compute this expression. The repeated computation of

Hβ−rshould perhaps be factored out; and, expressions of the formdot(inv(A), B)should almost always be replaced by the more numerically stablesolve(A, B). When using@, performing these two refactorings gives us:# Version 1 (as above) S = (H @ beta - r).T @ inv(H @ V @ H.T) @ (H @ beta - r) # Version 2 trans_coef = H @ beta - r S = trans_coef.T @ inv(H @ V @ H.T) @ trans_coef # Version 3 S = trans_coef.T @ solve(H @ V @ H.T, trans_coef)Notice that when comparing between each pair of steps, it's very easy to see exactly what was changed. If we apply the equivalent transformations to the code using the .dot method, then the changes are much harder to read out or verify for correctness:

# Version 1 (as above) S = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r) # Version 2 trans_coef = H.dot(beta) - r S = trans_coef.T.dot(inv(H.dot(V).dot(H.T))).dot(trans_coef) # Version 3 S = trans_coef.T.dot(solve(H.dot(V).dot(H.T)), trans_coef)Readability counts! The statements using

@are shorter, contain more whitespace, can be directly and easily compared both to each other and to the textbook formula, and contain only meaningful parentheses. This last point is particularly important for readability: when using function-call syntax, the required parentheses on every operation create visual clutter that makes it very difficult to parse out the overall structure of the formula by eye, even for a relatively simple formula like this one. Eyes are terrible at parsing non-regular languages. I made and caught many errors while trying to write out the 'dot' formulas above. I know they still contain at least one error, maybe more. (Exercise: find it. Or them.) The@examples, by contrast, are not only correct, they're obviously correct at a glance.