r - 在R中，当设计矩阵为奇数时，如何估计回归？(In R, how to estimate regression when design matrix is singular?)

Question

n = 50
set.seed(100)
x = matrix(runif(n, -2, 2), nrow=n)
y = 2 + 0.75*sin(x) - 0.75*cos(x) + rnorm(n, 0, 0.2)


nknots = 12
mat.tpb = function(x, nknots){
    n = length(x)
    knots = seq(-2, 2, len=nknots)
    zb = cbind(1, x ,x^2, x^3)
    zt = matrix(NA, nrow=n, ncol=nknots)
    for(i in 1:nknots){zt[,i] = ifelse(x>knots[i], (x-knots[i]) ^3, 0)}
    cbind(zb, zt)}

z = mat.tpb(x, nknots)

I wanted to do this:

(我想这样做：)

> summary(lm(y~z))

Call:
lm(formula = y ~ z)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.35945 -0.06861  0.00875  0.08329  0.39890 

Coefficients: (3 not defined because of singularities)
              Estimate Std. Error t value Pr(>|t|)   
(Intercept) 30636.8811 12890.7059   2.377  0.02291 * 
z1                  NA         NA      NA       NA   
z2          55942.6648 23543.5397   2.376  0.02294 * 
z3          34019.1008 14319.8448   2.376  0.02296 * 
z4           6887.7980  2899.9304   2.375  0.02299 * 
z5                  NA         NA      NA       NA   
z6          -7099.3431  2983.7929  -2.379  0.02277 * 
z7            243.7723    94.1488   2.589  0.01379 * 
z8            -46.0888    15.2702  -3.018  0.00465 **
z9             22.0511     9.1779   2.403  0.02156 * 
z10           -16.0403     8.5338  -1.880  0.06827 . 
z11            13.6275     8.5711   1.590  0.12059   
z12            -6.5516     9.1134  -0.719  0.47685   
z13            -3.5033    10.1582  -0.345  0.73220   
z14            10.1460    12.0295   0.843  0.40456   
z15             0.9644    27.5610   0.035  0.97228   
z16                 NA         NA      NA       NA   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1923 on 36 degrees of freedom
Multiple R-squared:  0.951, Adjusted R-squared:  0.9332 
F-statistic: 53.69 on 13 and 36 DF,  p-value: < 2.2e-16

But it seems the design matrix z is singular,

(但是设计矩阵z似乎是奇异的，)

so I used Moore-Penrose generalized inverse matrix to do this:

(所以我用Moore-Penrose广义逆矩阵来做到这一点：)

> MASS::ginv(t(z)%*%z)%*%t(z)%*%y
             [,1]
 [1,]  0.43424264
 [2,] -0.45336692
 [3,]  0.28938730
 [4,]  0.17283360
 [5,] -0.05730462
 [6,] -0.11531967
 [7,]  1.12054017
 [8,]  0.30900705
 [9,] -3.79858631
[10,]  1.52160075
[11,]  3.67809815
[12,] -4.15539087
[13,] -0.64600815
[14,]  5.77855917
[15,]  1.53625681
[16,]  0.00000000

Please tell me in R, what function or package I can use to do this when the design matrix is singular.

(请在R中告诉我，当设计矩阵为奇数时，我可以使用哪些函数或程序包执行此操作。)

Or the regression coefficient I did by Moore-Penrose generalized inverse matrix is already correct.

(或者我用Moore-Penrose广义逆矩阵所做的回归系数已经是正确的。)

Thank you very much.

(非常感谢你。)

  ask by abba translate from so

Answer 1

等待大神答复