I was trying to figure out the fastest way to do matrix multiplication and tried 3 different ways:
- Pure python implementation: no surprises here.
- Numpy implementation using numpy.dot(a, b)
- Interfacing with C using ctypesmodule in Python.
This is the C code that is transformed into a shared library:
#include <stdio.h>
#include <stdlib.h>
void matmult(float* a, float* b, float* c, int n) {
    int i = 0;
    int j = 0;
    int k = 0;
    /*float* c = malloc(nay * sizeof(float));*/
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            int sub = 0;
            for (k = 0; k < n; k++) {
                sub = sub + a[i * n + k] * b[k * n + j];
            }
            c[i * n + j] = sub;
        }
    }
    return ;
}
And the Python code that calls it:
def C_mat_mult(a, b):
    libmatmult = ctypes.CDLL("./matmult.so")
    dima = len(a) * len(a)
    dimb = len(b) * len(b)
    array_a = ctypes.c_float * dima
    array_b = ctypes.c_float * dimb
    array_c = ctypes.c_float * dima
    suma = array_a()
    sumb = array_b()
    sumc = array_c()
    inda = 0
    for i in range(0, len(a)):
        for j in range(0, len(a[i])):
            suma[inda] = a[i][j]
            inda = inda + 1
        indb = 0
    for i in range(0, len(b)):
        for j in range(0, len(b[i])):
            sumb[indb] = b[i][j]
            indb = indb + 1
    libmatmult.matmult(ctypes.byref(suma), ctypes.byref(sumb), ctypes.byref(sumc), 2);
    res = numpy.zeros([len(a), len(a)])
    indc = 0
    for i in range(0, len(sumc)):
        res[indc][i % len(a)] = sumc[i]
        if i % len(a) == len(a) - 1:
            indc = indc + 1
    return res
I would have bet that the version using C would have been faster ... and I'd have lost ! Below is my benchmark which seems to show that I either did it incorrectly, or that numpy is stupidly fast:

I'd like to understand why the numpy version is faster than the ctypes version, I'm not even talking about the pure Python implementation since it is kind of obvious.
 
     
     
     
     
     
     
     
     
    