Far*_*ghi 6 matlab r matrix-inverse
ginv()MASS与MATLAB pinv()函数相比,R中的函数生成完全不同的值.他们都宣称生产矩阵的Moore-Penrose广义逆.
我试图为R实现设置相同的容差,但差异仍然存在.
max(size(A)) * norm(A) * eps(class(A))sqrt(.Machine$double.eps)再生产:
R:
library(MASS)
A <- matrix(c(47,94032,149, 94032, 217179406,313679,149,313679,499),3,3)
ginv(A)
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输出:
[,1] [,2] [,3]
[1,] 1.675667e-03 -8.735203e-06 5.545605e-03
[2,] -8.735203e-06 5.014084e-08 -2.890907e-05
[3,] 5.545605e-03 -2.890907e-05 1.835313e-02
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svd(A)
输出:
$d
[1] 2.171799e+08 4.992800e+01 2.302544e+00
$u
[,1] [,2] [,3]
[1,] -0.0004329688 0.289245088 -9.572550e-01
[2,] -0.9999988632 -0.001507826 -3.304234e-06
[3,] -0.0014443299 0.957253888 2.892454e-01
$v
[,1] [,2] [,3]
[1,] -0.0004329688 0.289245088 -9.572550e-01
[2,] -0.9999988632 -0.001507826 -3.304234e-06
[3,] -0.0014443299 0.957253888 2.892454e-01
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MATLAB:
A = [47 94032 149; 94032 217179406 313679; 149 313679 499]
pinv(A)
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输出:
ans =
0.3996 -0.0000 -0.1147
-0.0000 0.0000 -0.0000
-0.1147 -0.0000 0.0547
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SVD:
[U, S, V] = svd(A)
U =
-0.0004 0.2892 -0.9573
-1.0000 -0.0015 -0.0000
-0.0014 0.9573 0.2892
S =
1.0e+008 *
2.1718 0 0
0 0.0000 0
0 0 0.0000
V =
-0.0004 0.2892 -0.9573
-1.0000 -0.0015 -0.0000
-0.0014 0.9573 0.2892
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运行debugonce(MASS::ginv),我们看到不同之处在于奇异值分解做了什么。
具体来说,R 检查以下内容:
Xsvd <- svd(A)
Positive <- Xsvd$d > max(tol * Xsvd$d[1L], 0)
Positive
# [1] TRUE TRUE FALSE
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如果第三个元素为真(我们可以通过设置强制tol = 0,正如@nicola 所建议的那样),MASS::ginv将返回:
Xsvd$v %*% (1/Xsvd$d * t(Xsvd$u))
# [,1] [,2] [,3]
# [1,] 3.996430e-01 -7.361507e-06 -1.147047e-01
# [2,] -7.361507e-06 5.014558e-08 -2.932415e-05
# [3,] -1.147047e-01 -2.932415e-05 5.468812e-02
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(即,与 MATLAB 相同)。
相反,它返回:
Xsvd$v[, Positive, drop = FALSE] %*% ((1/Xsvd$d[Positive]) *
t(Xsvd$u[, Positive, drop = FALSE]))
# [,1] [,2] [,3]
# [1,] 1.675667e-03 -8.735203e-06 5.545605e-03
# [2,] -8.735203e-06 5.014084e-08 -2.890907e-05
# [3,] 5.545605e-03 -2.890907e-05 1.835313e-02
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感谢@FaridCher 指出pinv.
我不确定我是否 100% 理解 MATLAB 代码,但我认为这归结为使用方式的不同tol。PositiveR中的MATLAB对应为:
`r = sum(s>tol)`
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tol用户提供的东西在哪里;如果没有提供,我们得到:
m = 0;
% I don't get the point of this for loop -- why not just `m = max(size(A))`?
for i = 1:n
m = max(m,length(A(:,i)));
end
% contrast with simply `tol * Xsvd$d[1L]` in R
% (note: i believe the elements of d are sorted largest to smallest)
tol = m*eps(max(s));
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