在矩阵分解中。 有类问题比較常见，即矩阵的元素仅仅有0和1。 相应实际应用中的场景是：用户对新闻的点击情况，对某些物品的购买情况等。
基于graphchi里面的矩阵分解结果不太理想。调研了下相关的文献，代码主要实现了基于PLSA的分解方法，具体请參考后面的參考文献
#!/usr/local/bin/python
#-*-coding:utf-8-*-
import sys
import math
import numpy as np
import string
import random
"""
reimplement algorithm of aspect_bernoulli.m
which may be usefull in one class matrix factorization
"""
"""
get matrix that all elements are 1
"""
def get_unit_matrix(row, col):
return np.ones((row, col), dtype=int)
def norm_zero(array):
col, row = array.shape
for c in xrange(col):
for r in xrange(row):
if array[c][r] < 0.000001:
array[c][r] = 0.000001
return array
"""
X: binary data matrix, each column is an observation
K: number of aspects(components) to estimate
iter: max number of iterations to do
ufile: save matrix result u
vfile: save martix result v
"""
def run_bernoulli(X, K, iter, ufile, vfile):
X = np.array(X)
T,N = X.shape
S = np.random.rand(K, N)
S_SUM = np.add.reduce(S)
S_SUM_EXT = np.tile(S_SUM, (K, 1))
S = S/S_SUM_EXT
A = np.random.rand(T, K)
UM_A = get_unit_matrix(T, K)
A_TEMP = UM_A - A
UM_X = get_unit_matrix(T, N)
X_TEMP = UM_X - X
l = []
for i in xrange(iter):
#STEP1
AS = np.dot(A, S) #A*S
AS_NORM = norm_zero(AS) #max(eps, A*S)
X_AS_NORM = X/AS_NORM #(X./max(eps, A*S))
FIRST = np.dot(A.T, X_AS_NORM) #A‘*(X./max(eps, A*S))
S_1 = S*FIRST #S.*(A‘*(X./max(eps, A*S)))
AS_TEMP = UM_X - AS #1-A*S
AS_TEMP_NORM = norm_zero(AS_TEMP) #max(eps, 1-A*S)
X_AS_TEMP_NORM = X_TEMP/AS_TEMP_NORM #(1-x)./max(eps, 1-A*S)
SECOND = np.dot(A_TEMP.T, X_AS_TEMP_NORM) #(1-A)‘*((1-x)./max(eps, 1-A*S))
S_2 = S*SECOND #S.*((1-A)‘*((1-x)./max(eps, 1-A*S)))
S = S_1 + S_2 #S.*(A‘*(X./max(eps, A*S))) + S.*((1-A)‘*((1-x)./max(eps, 1-A*S)))
S_SUM = np.add.reduce(S)
S_SUM_EXT = np.tile(S_SUM, (K, 1))
S = S/S_SUM_EXT
#STEP2
AS = np.dot(A, S) #A*S
AS_NORM = norm_zero(AS) #max(eps, A*S)
X_AS_NORM = X/AS_NORM #(X./max(eps, A*S))
A = A* np.dot(X_AS_NORM, S.T)
#STEP3
A_TEMP_S = np.dot(A_TEMP, S) #A1*S
A_TEMP_S_NORM = norm_zero(A_TEMP_S) #max(eps, A1*S)
A_TEMP_S_NORM_X = X_TEMP/A_TEMP_S_NORM #(1-X)./max(eps, A1*S)
A_TEMP = A_TEMP * (np.dot(A_TEMP_S_NORM_X, S.T)) #A1.*(((1-x)./max(eps, A1*S))*S‘)
A = A/(A + A_TEMP) #A./(A+A1)
A_TEMP = UM_A - A #1-A
#STEP4 compute loss function
AS = np.dot(A, S) #A*S
AS_NORM = norm_zero(AS) #max(eps, A*S)
AS_NORM = np.log(AS_NORM) #log(max(eps, A*S))
LOSS_FIRST = X * AS_NORM # X.*log(max(eps, A*S))
AS_TEMP = UM_X - AS #1-A*S
AS_TEMP_NORM = norm_zero(AS_TEMP) #max(eps, 1-A*S)
AS_TEMP_NORM = np.log(AS_TEMP_NORM) #log(max(eps, 1-A*S))
LOSS_SECOND = X_TEMP * AS_TEMP_NORM #(1-X).*log(max(eps, 1-A*S))
LOSS = LOSS_FIRST + LOSS_SECOND #X.*log(max(eps, A*S)) + (1-X).*log(max(eps, 1-A*S))
LOSS_SUM = np.add.reduce(LOSS) #sum(X.*log(max(eps, A*S)) + (1-X).*log(max(eps, 1-A*S)))
loss = np.sum(LOSS_SUM)/(N*T) #sum(sum(X.*log(max(eps, A*S)) + (1-X).*log(max(eps, 1-A*S))))/(N*T)
l.append(loss)
rmse = estimate_rmse(X, AS)
if i > 1:
if math.fabs(l[i] - l[i-1]) < 0.00001:break
print "iter = %d, loss = %f, rmse = %f"%(i, loss, rmse)
np.savetxt(ufile, A)
np.savetxt(vfile, S)
參考文献：
1. The aspect Bernoulli model: multiple causes of presences and absences (文章算法相应文章)
2. One-Class Matrix Completion with Low-Density Factorizations