我试图在C
中使用循环来乘以几个矩阵。我在R
中获得了预期的答案,但无法在C
中获得预期的答案。我怀疑这个问题与+=
函数有关,它似乎在循环的第一次迭代后使产品的值加倍。
我对C
不是很熟悉,并且无法将+=
函数替换为将返回预期答案的函数。
感谢您的任何建议。
首先,这是返回预期答案的R
代码:
B0 = -0.40
B1 = 0.20
mycov1 = exp(B0 + -2 * B1) / (1 + exp(B0 + -2 * B1))
mycov2 = exp(B0 + -1 * B1) / (1 + exp(B0 + -1 * B1))
mycov3 = exp(B0 + 0 * B1) / (1 + exp(B0 + 0 * B1))
mycov4 = exp(B0 + 1 * B1) / (1 + exp(B0 + 1 * B1))
trans1 = matrix(c(1 - 0.25 - mycov1, mycov1, 0.25 * 0.80, 0,
0, 1 - 0.50, 0, 0.50 * 0.75,
0, 0, 1, 0,
0, 0, 0, 1),
nrow=4, ncol=4, byrow=TRUE)
trans2 = matrix(c(1 - 0.25 - mycov2, mycov2, 0.25 * 0.80, 0,
0, 1 - 0.50, 0, 0.50 * 0.75,
0, 0, 1, 0,
0, 0, 0, 1),
nrow=4, ncol=4, byrow=TRUE)
trans3 = matrix(c(1 - 0.25 - mycov3, mycov3, 0.25 * 0.80, 0,
0, 1 - 0.50, 0, 0.50 * 0.75,
0, 0, 1, 0,
0, 0, 0, 1),
nrow=4, ncol=4, byrow=TRUE)
trans4 = matrix(c(1 - 0.25 - mycov4, mycov4, 0.25 * 0.80, 0,
0, 1 - 0.50, 0, 0.50 * 0.75,
0, 0, 1, 0,
0, 0, 0, 1),
nrow=4, ncol=4, byrow=TRUE)
trans2b <- trans1 %*% trans2
trans3b <- trans2b %*% trans3
trans4b <- trans3b %*% trans4
trans4b
#
# This is the expected answer
#
# [,1] [,2] [,3] [,4]
# [1,] 0.01819965 0.1399834 0.3349504 0.3173467
# [2,] 0.00000000 0.0625000 0.0000000 0.7031250
# [3,] 0.00000000 0.0000000 1.0000000 0.0000000
# [4,] 0.00000000 0.0000000 0.0000000 1.0000000
#
这是我的C
代码。 C
代码相当长,因为我不清楚C
是否足够有效:
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
char quit;
int main(){
int i, j, k, ii, jj, kk ;
double B0, B1, mycov ;
double trans[4][4] = {0} ;
double prevtrans[4][4] = {{1,0,0,0},
{0,1,0,0},
{0,0,1,0},
{0,0,0,1}};
B0 = -0.40 ;
B1 = 0.20 ;
for (i=1; i <= 4; i++) {
mycov = exp(B0 + B1 * (-2+i-1)) / (1 + exp(B0 + B1 * (-2+i-1))) ;
trans[0][0] = 1 - 0.25 - mycov ;
trans[0][1] = mycov ;
trans[0][2] = 0.25 * 0.80 ;
trans[0][3] = 0 ;
trans[1][0] = 0 ;
trans[1][1] = 1 - 0.50 ;
trans[1][2] = 0 ;
trans[1][3] = 0.50 * 0.75 ;
trans[2][0] = 0 ;
trans[2][1] = 0 ;
trans[2][2] = 1 ;
trans[2][3] = 0 ;
trans[3][0] = 0 ;
trans[3][1] = 0 ;
trans[3][2] = 0 ;
trans[3][3] = 1 ;
for (ii=0; ii<4; ii++){
for(jj=0; jj<4; jj++){
for(kk=0; kk<4; kk++){
trans[ii][jj] += trans[ii][kk] * prevtrans[kk][jj] ;
}
}
}
prevtrans[0][0] = trans[0][0] ;
prevtrans[0][1] = trans[0][1] ;
prevtrans[0][2] = trans[0][2] ;
prevtrans[0][3] = trans[0][3] ;
prevtrans[1][0] = trans[1][0] ;
prevtrans[1][1] = trans[1][1] ;
prevtrans[1][2] = trans[1][2] ;
prevtrans[1][3] = trans[1][3] ;
prevtrans[2][0] = trans[2][0] ;
prevtrans[2][1] = trans[2][1] ;
prevtrans[2][2] = trans[2][2] ;
prevtrans[2][3] = trans[2][3] ;
prevtrans[3][0] = trans[3][0] ;
prevtrans[3][1] = trans[3][1] ;
prevtrans[3][2] = trans[3][2] ;
prevtrans[3][3] = trans[3][3] ;
}
printf("To close this program type 'quit' and hit the return key\n");
printf(" \n");
scanf("%d", &quit);
return 0;
}
以下是上述C
代码返回的最终矩阵:
0.4821 3.5870 11.68 381.22
0 1 0 76.875
0 0 5 0
0 0 0 5
答案 0 :(得分:1)
这一行
trans[ii][jj] += trans[ii][kk] * prevtrans[kk][jj] ;
不对。当您仍在使用它来计算结果矩阵时,您正在修改trans
。您需要另一个矩阵来临时存储乘法结果。然后使用:
// Store the resultant matrix in temp.
for (ii=0; ii<4; ii++){
for(jj=0; jj<4; jj++){
temp[ii][jj] = 0.0;
for(kk=0; kk<4; kk++){
temp[ii][jj] += trans[ii][kk] * prevtrans[kk][jj] ;
}
}
}
// Transfer the data from temp to trans
for (ii=0; ii<4; ii++){
for(jj=0; jj<4; jj++){
trans[ii][jj] = temp[ii][jj];
}
}