问题描述
我有以下R矩阵，它是2x3和3x3子矩阵的组合，并且可以是两个以上具有不同维度的子矩阵(例如m1xp和m2xp和m3xp，其中m1，m2，m3分别== p)
I have the following R matrix that is a combination of 2x3 and 3x3 submatrices and it can be more than 2 submatrices with different dimension (e.g. m1xp and m2xp and m3xp where each of m1,m2,m3 <= p)
A2 <- list(rbind(c(1,1,1),c(-1,1,-1)),
rbind(c(-1,1,1),c(1,-1,2),c(2,-1,2)))
library(Matrix)
A2 <- as.matrix(Matrix::bdiag(A2))
Rhs <- matrix(c(0,5,0.5,4),nrow = 4)
beta <- c(rep(1.2,3),c(0.5,0.2,0.1))
> A2
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 1 0 0 0
[2,] -1 1 -1 0 0 0
[3,] 0 0 0 -1 1 1
[4,] 0 0 0 1 -1 2
[5,] 0 0 0 2 -1 2
我想获得第一个子矩阵和第二个子矩阵之间的所有行索引组合，以解决线性优化问题.组合必须来自两个子矩阵，然后求解新的beta，然后检查是否满足条件Aq %*% beta == Rhs，然后停止.如果不是，则采取另一种组合.我认为下面是子矩阵之间的所有行组合:
I would like to get all the rows indices combination between the first sub-matrix and the 2nd sub-matrix to solve an linear optimization problem. The combination has to be from both submatrices then solve for new beta and then check if the condition Aq %*% beta == Rhs is satisfy, stop. If not, then take another combination. I think below is all the rows combination between the sub-matrices:
第一个子矩阵和第二个子矩阵的组合
A combination as one from the first sub-matrix and one from the second sub-matrix
Aq <- A2[c(1,3),]
Aq <- A2[c(1,4),]
Aq <- A2[c(1,5),]
Aq <- A2[c(2,3),]
Aq <- A2[c(2,4),]
Aq <- A2[c(2,5),]
然后，将第一个矩阵中的一个与第二个矩阵中的2组合为一个
Then, a combination as one from the first and 2 from the second matrix
Aq <- A2[c(1,3,4),]
Aq <- A2[c(1,3,5),]
Aq <- A2[c(1,4,5),]
Aq <- A2[c(2,3,4),]
Aq <- A2[c(2,3,5),]
Aq <- A2[c(2,4,5),]
然后，将第一个矩阵中的一个与第二个矩阵中的3组合为一个
Then, a combination as one from the first and 3 from the second matrix
Aq <- A2[c(1,3,4,5),]
Aq <- A2[c(2,3,4,5),]
然后，第一个矩阵的组合为2，第二个矩阵的组合为
Then, a combination as 2 from the first and one from the second matrix
Aq <- A2[c(1,2,3),]
Aq <- A2[c(1,2,4),]
Aq <- A2[c(1,2,5),]
然后，第一个矩阵的组合为2，第二个矩阵的组合为
Then, a combination as 2 from the first and 2 from the second matrix
Aq <- A2[c(1,2,3,4),]
Aq <- A2[c(1,2,3,5),]
Aq <- A2[c(1,2,4,5),]
然后，第一个矩阵的组合为2，第二个矩阵的组合为
Then, a combination as 2 from the first and 3 from the second matrix
Aq <- A2[c(1,2,3,4,5),]
是否有更好的方法来获取所有组合?然后，我想创建一个循环，一次选择上述组合中的一个，并检查是否
Is there a better way to get all the combinations?Then I would like to create a loop that choice one on the above combination at a time and check if
if (Aq %*% beta == Rhs) {
break
} else {
TAKE ANOTHER COMBINATION Aq
}
请注意，我可能有两个以上的子矩阵来创建块矩阵.然后，我必须创建第一个，第二个和第三个矩阵之间的所有行组合.我希望在R中有一种简单的方法.我尝试了grid.expand函数，但没有给我想要的输出.
Please note I could have more than 2 submatrices that create the block matrix. Then I have to create all row combinations between from the first, 2nd and 3rd matrix. I am hoping there is easy way to do in R. I have tried grid.expand function but it is not giving me the desired output.
推荐答案
一种可能的基本R方法:
A possible base R approach:
indices1 <- 1:2
indices2 <- 3:5
apply(expand.grid(seq_along(indices1), seq_along(indices2)), 1,
function(x) t(apply(
expand.grid(combn(indices1, x[1], simplify=FALSE),
combn(indices2, x[2], simplify=FALSE)),
1, unlist)))
输出:
[[1]]
Var1 Var2
[1,] 1 3
[2,] 2 3
[3,] 1 4
[4,] 2 4
[5,] 1 5
[6,] 2 5
[[2]]
Var11 Var12 Var2
[1,] 1 2 3
[2,] 1 2 4
[3,] 1 2 5
[[3]]
Var1 Var21 Var22
[1,] 1 3 4
[2,] 2 3 4
[3,] 1 3 5
[4,] 2 3 5
[5,] 1 4 5
[6,] 2 4 5
[[4]]
Var11 Var12 Var21 Var22
[1,] 1 2 3 4
[2,] 1 2 3 5
[3,] 1 2 4 5
[[5]]
Var1 Var21 Var22 Var23
[1,] 1 3 4 5
[2,] 2 3 4 5
[[6]]
Var11 Var12 Var21 Var22 Var23
[1,] 1 2 3 4 5
添加一个更通用的版本:
edit: adding a more general version:
#identifying the indices
indices <- split(seq_len(nrow(A2)), max.col(abs(A2) > 0, "first"))
#generating the combinations
apply(expand.grid(lapply(indices, seq_along)), 1L,
function(idx) {
t(apply(
expand.grid(
lapply(seq_along(idx),
function(k) {
combn(indices[[k]], idx[k], simplify=FALSE)
})),
1L, unlist))
})
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