我正在编写c代码，以便使用R中的lapack lib获取矩阵A的QR分解。 我的结果与使用R语言命令的结果不同。
这是一个lapack的东西，还是我代码中的错误？
表示矩阵（行专业）：
1, 19.5, 43.1, 29.1,
1, 24.7, 49.8, 28.2,
1, 30.7, 51.9, 37.0,
1, 29.8, 54.3, 31.1,
1, 19.1, 42.2, 30.9,
1, 25.6, 53.9, 23.7,
1, 31.4, 58.5, 27.6,
1, 27.9, 52.1, 30.6,
1, 22.1, 49.9, 23.2,
1, 25.5, 53.5, 24.8,
1, 31.1, 56.6, 30.0,
1, 30.4, 56.7, 28.3,
1, 18.7, 46.5, 23.0,
1, 19.7, 44.2, 28.6,
1, 14.6, 42.7, 21.3,
1, 29.5, 54.4, 30.1,
1, 27.7, 55.3, 25.7,
1, 30.2, 58.6, 24.6,
1, 22.7, 48.2, 27.1,
1, 25.2, 51.0, 27.5
结果是：
V1 V2 V3 V4
[1,] -4.4721360 -113.16740034 -2.288392e+02 -123.52039508
[2,] 0.2236068 -21.89587861 -2.107945e+01 -7.27753395
[3,] 0.2236068 0.29484219 8.733781e+00 -14.04825478
[4,] 0.2236068 0.25373857 7.566965e-02 -1.55436071
[5,] 0.2236068 -0.23493787 2.999600e-01 0.26555995
[6,] 0.2236068 0.06192165 -3.343037e-01 0.12188660
[7,] 0.2236068 0.32681168 -2.315941e-01 0.40765540
[8,] 0.2236068 0.16696425 1.213823e-01 -0.18580207
[9,] 0.2236068 -0.09792578 -2.561224e-01 -0.16369010
[10,] 0.2236068 0.05735458 -2.993562e-01 0.52588892
[11,] 0.2236068 0.31311047 -4.660317e-02 0.29409317
[12,] 0.2236068 0.28114099 -1.340150e-01 0.16746961
[13,] 0.2236068 -0.25320615 -2.357881e-01 0.26072358
[14,] 0.2236068 -0.20753545 1.360745e-01 0.18135493
[15,] 0.2236068 -0.44045600 -2.456167e-01 0.15393180
[16,] 0.2236068 0.24003736 3.166468e-02 -0.02119950
[17,] 0.2236068 0.15783011 -2.667146e-01 0.32042553
[18,] 0.2236068 0.27200685 -3.732646e-01 0.05926317
[19,] 0.2236068 -0.07052337 3.634501e-03 -0.20518296
[20,] 0.2236068 0.04365337 -4.566657e-02 -0.03457051
我的结果：
-4.4721, -113.1674, -228.8392, -123.5204,
0.1827, -21.8959, -21.0794, -7.2775,
0.1827, 0.2888, 8.7338, -14.0483,
0.1827, 0.2486, 0.0523, -1.5544,
0.1827, -0.2301, 0.2071, 0.2316,
0.1827, 0.0607, -0.2309, 0.1063,
0.1827, 0.3201, -0.1599, 0.3555,
0.1827, 0.1636, 0.0838, -0.1620,
0.1827, -0.0959, -0.1769, -0.1427,
0.1827, 0.0562, -0.2067, 0.4586,
0.1827, 0.3067, -0.0322, 0.2565,
0.1827, 0.2754, -0.0925, 0.1460,
0.1827, -0.2480, -0.1628, 0.2274,
0.1827, -0.2033, 0.0940, 0.1581,
0.1827, -0.4315, -0.1696, 0.1342,
0.1827, 0.2351, 0.0219, -0.0185,
0.1827, 0.1546, -0.1842, 0.2794,
0.1827, 0.2665, -0.2578, 0.0517,
0.1827, -0.0691, 0.0025, -0.1789,
0.1827, 0.0428, -0.0315, -0.0301,
#include <stdio.h>
#include <R.h>
#include <R_ext/BLAS.h>
#include <R_ext/Lapack.h>
int min(int x, int y) {
if (x <= y)
return x;
else
return y;
}
int max(int x, int y) {
if (x >= y)
return x;
else
return y;
}
void transpose(int* nrow, int* ncol, double* a) {
int i, j, index, k = 0;
double* atransp = malloc(*nrow**ncol * sizeof (double));
//compute transpose
for (i = 0; i<*ncol; i++) {
index = i;
atransp[k] = a[index];
k++;
for (j = 0; j<*nrow - 1; j++) {
index += *ncol;
atransp[k] = a[index];
k++;
}
}
//copy transpose in array a
for (i = 0; i<*nrow**ncol; i++)
a[i] = atransp[i];
//free memory
free(atransp);
}
void getQR(int* rowX, int* colX, double* X, double* Tau) {
const int m = *rowX;
const int n = *colX;
double* a = X;
const int lda = max(1, m);
double* tau = malloc(min(m, n) * sizeof (double));
const int lwork = max(1, n);
double* work = malloc(max(1, lwork) * sizeof (double));
int info;
F77_NAME(dgeqrf)(&m, &n, a, &lda, tau, work, &lwork, &info);
printf("\n dgeqrf() ended with : %d\n",info);
copyTo(min(m, n), tau, Tau);
free(work);
free(tau);
}
int main() {
int rX = 20, cX = 4;
double X[] = {
1, 19.5, 43.1, 29.1,
1, 24.7, 49.8, 28.2,
1, 30.7, 51.9, 37.0,
1, 29.8, 54.3, 31.1,
1, 19.1, 42.2, 30.9,
1, 25.6, 53.9, 23.7,
1, 31.4, 58.5, 27.6,
1, 27.9, 52.1, 30.6,
1, 22.1, 49.9, 23.2,
1, 25.5, 53.5, 24.8,
1, 31.1, 56.6, 30.0,
1, 30.4, 56.7, 28.3,
1, 18.7, 46.5, 23.0,
1, 19.7, 44.2, 28.6,
1, 14.6, 42.7, 21.3,
1, 29.5, 54.4, 30.1,
1, 27.7, 55.3, 25.7,
1, 30.2, 58.6, 24.6,
1, 22.7, 48.2, 27.1,
1, 25.2, 51.0, 27.5
};
//column major
transpose(&rX,&cX,X);
//tau is needed to extract Q later
double* tau = malloc(min(rX, cX) * sizeof (double));
double* QR = malloc(rX*cX*sizeof(double));
copyTo(rX*cX,X,QR);
getQR(&rX, &cX, QR, tau);
//printmat(cX,rX,QR,"QR");
return 0;
}
答案 0 :(得分：4)
首先，R使用LINPACK例程DQRDC2作为默认值。您可以使用带有qr()选项的R中的LAPACK=TRUE命令来使用LAPACK例程：
> QR <- qr(Mat,LAPACK=TRUE)
> QR
$qr
V3 V4 V2 V1
[1,] -229.9739116 -123.04447843 -114.61600066 -4.45006998
[2,] 0.1823682 -19.23736803 -1.81750327 -0.25177355
[3,] 0.1900584 0.41052447 -12.08962672 0.36451274
[4,] 0.1988473 0.04298840 0.18492760 0.02485389
[5,] 0.1545369 0.37519893 -0.14576039 0.48992952
[6,] 0.1973825 -0.32149888 -0.01277324 -0.02631498
[7,] 0.2142277 -0.25359559 0.19391630 0.15368409
[8,] 0.1907908 0.07984478 0.13051129 -0.21692785
[9,] 0.1827344 -0.23371181 -0.11707414 -0.19513972
[10,] 0.1959177 -0.25431801 -0.01531797 0.38150533
[11,] 0.2072699 -0.07795264 0.21041668 0.11596815
[12,] 0.2076361 -0.16711575 0.17604756 -0.02208373
[13,] 0.1702836 -0.14766629 -0.23298857 0.30390902
[14,] 0.1618609 0.20180495 -0.14729488 0.33577876
[15,] 0.1563679 -0.12647957 -0.37138938 0.29164143
[16,] 0.1992135 -0.01062557 0.17041958 -0.12582222
[17,] 0.2025093 -0.25954266 0.06531149 0.14217017
[18,] 0.2145939 -0.40877853 0.13742162 -0.20783552
[19,] 0.1765090 0.01244890 -0.06067115 -0.15702073
[20,] 0.1867626 -0.04646282 0.01506042 -0.06657167
但是，您应该知道在这种情况下函数qr()使用LAPACK例程DGEQP3。与您正在使用的DGEQRF例程相反，DGEQP3使用列旋转计算矩阵的QR分解。
非常正常，你会得到不同的结果，因为你没有使用相同的方法。你应该记住，QR分解不是一个独特的解决方案。要知道您的QR分解是否正确，您只需检查Q和R矩阵是否满足要求。例如在R：
> Q <- qr.Q(QR)
> round( t(Q) %*% Q , 10 )
[,1] [,2] [,3] [,4]
[1,] 1 0 0 0
[2,] 0 1 0 0
[3,] 0 0 1 0
[4,] 0 0 0 1
> all.equal(Q %*% qr.R(QR),Mat)
[1] TRUE
另见?qr。