cp分解实现_matlab实现CP分解 代码
使用tensor-box，在matlab中实现cp分解
1、X = sptenrand([5 4 3], 10)
//随机生成一个稀疏三维张量5*4* 3，其中有10条数据
X is a sparse tensor of size 5 x 4 x 3 with 10 nonzeros
(1,4,1) 0.4966
(2,2,3) 0.8998
(3,2,3) 0.8216
(3,3,1) 0.6449
(3,3,3) 0.8180
(3,4,1) 0.6602
(4,1,2) 0.3420
(4,1,3) 0.2897
(5,2,2) 0.3412
(5,3,2) 0.5341
2、P = parafac_als(X,2)
//用ALS交替最小二乘来分解张量，分解参数R=2，即分解为两个秩一张量；其中分解参数最多可以设置为min(5*4；5* 3；4*3)
CP_ALS:
Iter 1: fit = 3.219563e-001 fitdelta = 3.2e-001
Iter 2: fit = 3.645517e-001 fitdelta = 4.3e-002
Iter 3: fit = 3.732887e-001 fitdelta = 8.7e-003
Iter 4: fit = 3.809608e-001 fitdelta = 7.7e-003
Iter 5: fit = 4.021826e-001 fitdelta = 2.1e-002
Iter 6: fit = 4.427524e-001 fitdelta = 4.1e-002
Iter 7: fit = 4.734919e-001 fitdelta = 3.1e-002
Iter 8: fit = 4.848760e-001 fitdelta = 1.1e-002
Iter 9: fit = 4.890031e-001 fitdelta = 4.1e-003
Iter 10: fit = 4.907726e-001 fitdelta = 1.8e-003
Iter 11: fit = 4.916244e-001 fitdelta = 8.5e-004
Iter 12: fit = 4.920996e-001 fitdelta = 4.8e-004
Iter 13: fit = 4.924246e-001 fitdelta = 3.2e-004
Iter 14: fit = 4.926962e-001 fitdelta = 2.7e-004
Iter 15: fit = 4.929575e-001 fitdelta = 2.6e-004
Iter 16: fit = 4.932285e-001 fitdelta = 2.7e-004
Iter 17: fit = 4.935198e-001 fitdelta = 2.9e-004
Iter 18: fit = 4.938385e-001 fitdelta = 3.2e-004
Iter 19: fit = 4.941904e-001 fitdelta = 3.5e-004
Iter 20: fit = 4.945813e-001 fitdelta = 3.9e-004
Iter 21: fit = 4.950178e-001 fitdelta = 4.4e-004
Iter 22: fit = 4.955072e-001 fitdelta = 4.9e-004
Iter 23: fit = 4.960583e-001 fitdelta = 5.5e-004
Iter 24: fit = 4.966814e-001 fitdelta = 6.2e-004
Iter 25: fit = 4.973882e-001 fitdelta = 7.1e-004
Iter 26: fit = 4.981921e-001 fitdelta = 8.0e-004
Iter 27: fit = 4.991075e-001 fitdelta = 9.2e-004
Iter 28: fit = 5.001490e-001 fitdelta = 1.0e-003
Iter 29: fit = 5.013282e-001 fitdelta = 1.2e-003
Iter 30: fit = 5.026502e-001 fitdelta = 1.3e-003
Iter 31: fit = 5.041052e-001 fitdelta = 1.5e-003
Iter 32: fit = 5.056587e-001 fitdelta = 1.6e-003
Iter 33: fit = 5.072418e-001 fitdelta = 1.6e-003
Iter 34: fit = 5.087490e-001 fitdelta = 1.5e-003
Iter 35: fit = 5.100586e-001 fitdelta = 1.3e-003
Iter 36: fit = 5.110745e-001 fitdelta = 1.0e-003
Iter 37: fit = 5.117692e-001 fitdelta = 6.9e-004
Iter 38: fit = 5.121888e-001 fitdelta = 4.2e-004
Iter 39: fit = 5.124165e-001 fitdelta = 2.3e-004
Iter 40: fit = 5.125308e-001 fitdelta = 1.1e-004
Iter 41: fit = 5.125856e-001 fitdelta = 5.5e-005
Final fit = 5.125856e-001
P is a ktensor of size 5 x 4 x 3
P.lambda = [ 1.3189 1.1109 ]
P.U{1} =
0.0019 0.2743
0.6406 -0.0177
0.7679 0.9615
-0.0000 0.0000
-0.0000 -0.0000
P.U{2} =
-0.0000 0.0000
0.9413 -0.0855
0.2693 0.7083
-0.2036 0.7007
P.U{3} =
0.0402 0.8828
-0.0000 -0.0000
0.9992 0.4698