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Testing numba's autojit decorator on simple code for matrix factorization.
# -*- coding: utf-8 -*-
#!/usr/bin/python
#
# Modified example of Albert Au Yeung's on matrix factorization:
# http://www.quuxlabs.com/blog/2010/09/matrix-factorization-a-simple-tutorial-and-implementation-in-python/#source-code
import numpy as np
from numba.decorators import autojit, jit
def run():
R = [
[5,3,0,1],
[4,0,0,1],
[1,1,0,5],
[1,0,0,4],
[0,1,5,4],
]
R = np.array(R, dtype=np.float)
N, M = R.shape
K = 2
P = np.random.rand(N,K)
Q = np.random.rand(M,K)
nP, nQ, steps, error = matrix_factorization(R, P, Q, K)
print 'Factorization complete in %i steps with a total erorr of %.3f' \
% (steps, error)
nR = np.dot(nP, nQ.T)
return nR
#@autojit
def matrix_factorization(R, P, Q, K):
steps = 5000
alpha = 0.0002
beta = 0.02 / 2.0
Q = Q.T
n, m = R.shape
step = 0.0
for step in xrange(steps):
for i in xrange(n):
for j in xrange(m):
if R[i][j] > 0:
eij = R[i][j] - np.dot(P[i,:], Q[:,j])
for k in xrange(K):
P[i][k] += alpha * (2 * eij * Q[k][j] - beta * P[i][k])
Q[k][j] += alpha * (2 * eij * P[i][k] - beta * Q[k][j])
e = 0.0
for i in xrange(n):
for j in xrange(m):
if R[i][j] > 0:
e += (R[i][j] - np.dot(P[i,:], Q[:,j]))**2
for k in range(K):
e += beta * (P[i][k]**2 + Q[k][j]**2)
if e < 0.001:
break
return P, Q.T, step, e
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