Question

我想在r中生成30000 x 30000的矩阵，通过其转置乘以30000个元素的向量，然后获得该矩阵的SVD，但程序告诉我r无法找到大小为900000000的向量。我，我该怎么办？

y <- read.csv("C:\\Users\\jmarescr\\Desktop\\BigLetra50.csv",header=TRUE)

x <- matrix(y[1:30000,1],30000,1)
tx <- as.matrix(t(x))

mat <- x %*% tx

Error: can not allocate vector of length 900000000

s <- svd(mat)

Error in svd (x): object 'mat' not found

Answer 1

SVD的一部分优点是你不需要采用x的交叉产品来获得交叉产品的SVD。

您可以直接从x%*%t(x)的SVD元素中获取tcrossprod(x)（又名x）的SVD。具体而言（直到U列的符号）SVD（x％*％t（x））= U D ^ 2 t（U），其中U和D取自x的SVD。（供参考，see here。）

要查看它的实际效果，请尝试一个较小的示例：

set.seed(1)
x <- matrix(rnorm(15), ncol=5)


svd(x)$d
# [1] 3.046842 1.837539 1.411585
sqrt(svd(tcrossprod(x))$d)
# [1] 3.046842 1.837539 1.411585

svd(x)$u
#            [,1]       [,2]      [,3]
# [1,] -0.3424029  0.7635281 0.5475264
# [2,] -0.8746155 -0.4719093 0.1111273
# [3,]  0.3432316 -0.4408248 0.8293766
svd(tcrossprod(x))$u
#            [,1]       [,2]      [,3]
# [1,] -0.3424029  0.7635281 0.5475264
# [2,] -0.8746155 -0.4719093 0.1111273
# [3,]  0.3432316 -0.4408248 0.8293766
svd(tcrossprod(x))$v
#            [,1]       [,2]      [,3]
# [1,] -0.3424029  0.7635281 0.5475264
# [2,] -0.8746155 -0.4719093 0.1111273
# [3,]  0.3432316 -0.4408248 0.8293766

另一种看法：

sss <- svd(x)

with(sss, u %*% diag(d)^2 %*% t(u))
#           [,1]      [,2]      [,3]
# [1,]  3.654154  1.684675 -1.322649
# [2,]  1.684675  7.877802 -1.900721
# [3,] -1.322649 -1.900721  3.120415

tcrossprod(x)
#           [,1]      [,2]      [,3]
# [1,]  3.654154  1.684675 -1.322649
# [2,]  1.684675  7.877802 -1.900721
# [3,] -1.322649 -1.900721  3.120415

R中非常大的矩阵的SVD

1 个答案: