[SciPy-User] fast small matrix multiplication with cython?
Skipper Seabold
jsseabold@gmail....
Mon Dec 6 16:34:19 CST 2010
I'm wondering if anyone might have a look at my cython code that does
matrix multiplication and see where I can speed it up or offer some
pointers/reading. I'm new to Cython and my knowledge of C is pretty
basic based on trial and (mostly) error, so I am sure the code is
still very naive.
import numpy as np
from matmult import dotAB, multAB
A = np.array([[ 1., 3., 4.],
[ 5., 6., 3.]])
B = A.T.copy()
timeit dotAB(A,B)
# 1 loops, best of 3: 826 ms per loop
timeit multAB(A,B)
# 1 loops, best of 3: 1.16 s per loop
As you can see my multAB results in a negative speedup of about .75.
I compile the cython code with
cython -a matmult.pyx
gcc -shared -pthread -fPIC -fwrapv -O2 -Wall -fno-strict-aliasing
-I/usr/include/python2.6
-I/usr/local/lib/python2.6/dist-packages/numpy/core/include/ -o
matmult.so matmult.c
Cython code is attached and inlined below.
Profile is here (some of which I don't understand why there are
bottlenecks) http://eagle1.american.edu/~js2796a/matmult/matmult.html
-----------------------------------------------------------
from numpy cimport float64_t, ndarray, NPY_DOUBLE, npy_intp
cimport cython
from numpy import dot
ctypedef float64_t DOUBLE
cdef extern from "numpy/arrayobject.h":
cdef void import_array()
cdef object PyArray_SimpleNew(int nd, npy_intp *dims, int typenum)
import_array()
@cython.boundscheck(False)
@cython.wraparound(False)
cdef inline object matmult(ndarray[DOUBLE, ndim=2, mode='c'] A,
ndarray[DOUBLE, ndim=2, mode='c'] B):
cdef int lda = A.shape[0]
cdef int n = B.shape[1]
cdef npy_intp *dims = [lda, n]
cdef ndarray[DOUBLE, ndim=2] out = PyArray_SimpleNew(2, dims, NPY_DOUBLE)
cdef int i,j,k
cdef double s
for i in xrange(lda):
for j in xrange(n):
s = 0
for k in xrange(A.shape[1]):
s += A[i,k] * B[k,j]
out[i,j] = s
return out
def multAB(ndarray[DOUBLE, ndim=2] A, ndarray[DOUBLE, ndim=2] B):
for i in xrange(1000000):
C = matmult(A,B)
return C
def dotAB(ndarray[DOUBLE, ndim=2] A, ndarray[DOUBLE, ndim=2] B):
for i in xrange(1000000):
C = dot(A,B)
return C
Skipper
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