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Given an n dimensional array X, a d by d-1 dimensional matrix V and two specified dimensions (p1, p2) <= (n, n); I would like a function that preforms matrix multiplication of V along the dimensions (p1, p2) of X.
That is given X:
library(abind)
set.seed(4)
X <- matrix(runif(4), 2, 2)
X <- abind(x, x+5, along = 3)
> a
, , 1
[,1] [,2]
[1,] 1 3
[2,] 2 4
, , 2
[,1] [,2]
[1,] 6 8
[2,] 7 9
and given a matrix V
V <- matrix(c(1, 2))
[,1]
[1,] 1
[2,] 2
For example, if p1=2 and p2=1 I would like to remove the following for loop
p1 <- 1
p2 <- 2
a.out <- array(0, c(2, 1, 2))
for (i in 1:dim(a)[2]){
a.out[,,i] <- a[,,i]%*%V # note indexed along other dimension
}
> a.out
, , 1
[,1]
[1,] 7
[2,] 10
, , 2
[,1]
[1,] 22
[2,] 25
The hard part here is that I want to allow for arbitrary dimensional arrays (i.e., n could be greater than 3).
1st Edit:
This problem is not the same as Indexing slice from 3D Rcpp NumericVector as I am discussing arbitrary number of dimensions >=2 and the question is not only about indexing.
2nd Edit:
Just to be a little more clear here is another example of what I am trying to do. Here the dimension of X is 4, p1 = 2, p3=3, and the dimension of X along the p1 dimension is 12. The following code computes the desired result as X.out for random X and V.
X <- array(rnorm(672), c(4, 7, 12, 2))
V <- matrix(rnorm(132), 12, 11) # p1 = 2, p2 = 3, V is of dimension D x D-1
d <- dim(X)
X.out <- array(0, dim=c(d[1:2], d[3]-1, d[4]))
for(i in 1:d[1]){
for (j in 1:d[4]){
X.out[i,,,j] <- X[i,,,j]%*%V # p1 = 2, p2 = 3
}
}
I have two matrices, call them A (n x 2) and B (q x 2). I'd like to get an n x q x 2 array C, such that C[1,5,] represents the difference between the first row of A and the fifth row of B, taking the subtraction of the first element in the first row of A with the first element in the fifth row of B and the second element similarly subtracted.
I'm trying to perform this function via the outer function, but it also gives me the "non-diagonal" subtractions; i.e. it will also subtract A[1,1] - B[5,2] and A[1,2] - B[5,1] which I am not interested in. Does anyone have a fast, easy way to do this?
Current code
>diffs <- outer(A,B,FUN ='-')
>diffs[1,,5,]
[,1] [,2]
[1,] **-0.3808701** 0.7591052
[2,] 0.2629293 **1.4029046**
I've added the stars to indicate what I actually want.
Thanks for any help in advance
(EDIT)
Here's a simpler case for illustrative purposes
> A <- matrix(1:10, nrow = 5, ncol = 2)
> B <- matrix(4:9, nrow = 3, ncol = 2)
> A
[,1] [,2]
[1,] 1 6
[2,] 2 7
[3,] 3 8
[4,] 4 9
[5,] 5 10
> B
[,1] [,2]
[1,] 4 7
[2,] 5 8
[3,] 6 9
>diffs <- outer(A,B,FUN ='-')
>diffs[1,,3,] == (A[1,] - B[3,])
[,1] [,2]
[1,] TRUE FALSE
[2,] FALSE TRUE
>diffs[1,,3,]
[,1] [,2]
[1,] -5 -8
[2,] 0 -3
Before worrying about the shape of the output I think we should make sure we're getting the correct values.
A <- matrix(1:10, nrow=5, ncol=2)
B <- matrix(4:9, nrow=3, ncol=2)
# long-winded method
dia_long <- c(
c(A[1,] - B[1,]),
c(A[1,] - B[2,]),
c(A[1,] - B[3,]),
c(A[2,] - B[1,]),
c(A[2,] - B[2,]),
c(A[2,] - B[3,]),
c(A[3,] - B[1,]),
c(A[3,] - B[2,]),
c(A[3,] - B[3,]),
c(A[4,] - B[1,]),
c(A[4,] - B[2,]),
c(A[4,] - B[3,]),
c(A[5,] - B[1,]),
c(A[5,] - B[2,]),
c(A[5,] - B[3,]))
# loop method
comb <- expand.grid(1:nrow(A), 1:nrow(B))
dia_loop <- list()
for (i in 1:nrow(comb)) {
dia_loop[[i]] <- A[comb[i, 1], ] - B[comb[i, 2], ]
}
dia_loop <- unlist(dia_loop)
# outer/apply method
dia_outer <- apply(outer(A, B, FUN='-'), c(3, 1), diag)
# they all return the same values
all.identical <- function(l) {
all(sapply(2:length(l), FUN=function(x) identical(l[1], l[x])))
}
all.identical(lapply(list(dia_long, dia_loop, dia_outer), sort))
# TRUE
table(dia_long)
# dia_long
# -5 -4 -3 -2 -1 0 1 2 3
# 1 2 4 5 6 5 4 2 1
Are these the values you are looking for?
My solution: use nested lapply and sapply functions to extract the diagonals. I then needed to do some post-processing (not related to this specific problem), before I then turned it into an array. Should be noted that this is a q x 2 x n array, which turned out to be better for my purposes - this could be permuted with aperm from here though to solve the original question.
A <- matrix(1:10, nrow = 5, ncol = 2)
B <- matrix(4:9, nrow = 3, ncol = 2)
diffs <- outer(A,B, FUN = '-')
diffs <- lapply(X = 1:nrow(A),FUN = function(y){
t(sapply(1:ncol(B), FUN = function(x) diag(diffs[y,,x,])))})
diffs <- array(unlist(lapply(diffs, FUN = t)), dim = c(nrow(B),2,nrow(A)))
is it possible to have a matrix of matrices in R? if yes, how should I define such matrix?
for example to have a 10 x 10 matrix, and each element of this matrix contains a matrix itself.
1) list/matrix Yes, create a list and give it dimensions using matrix:
m <- matrix(1:4, 2)
M <- matrix(list(m, 2*m, 3*m, 4*m), 2)
so element 1,1 of M is m:
> M[[1,1]]
[,1] [,2]
[1,] 1 3
[2,] 2 4
2) list/dim<- This also works:
M <- list(m, 2*m, 3*m, 4*m)
dim(M) <- c(2, 2)
3) array This is not quite what you asked for but depending on your purpose it might satisfy your need:
A <- array(c(m, 2*m, 3*m, 4*m), c(2, 2, 2, 2)) # 2x2x2x2 array
so element 1,1 is:
> A[1,1,,]
[,1] [,2]
[1,] 1 3
[2,] 2 4
I am trying to get a function that is the opposite of diff()
I want to add the values of adjacent columns in a matrix for each column in the matrix.
I do NOT need the sum of the entire column or row.
For example:
If I had:
[ 1 2 4;
3 5 8 ]
I would end up with:
[ 3 6;
8 13 ]
Of course for just one or two columns this is simple as I can just do x[,1]+x[,2], but these matrices are quite large.
I'm surprised that I cannot seem to find an efficient way to do this.
m <- matrix(c(1,3,2,5,4,8), nrow=2)
m[,-1] + m[,-ncol(m)]
[,1] [,2]
[1,] 3 6
[2,] 8 13
Or, just for the fun of it:
n <- ncol(m)
x <- suppressWarnings(matrix(c(1, 1, rep(0, n-1)),
nrow = n, ncol = n-1))
m %*% x
[,1] [,2]
[1,] 3 6
[2,] 8 13
Dummy data
mat <- matrix(sample(0:9, 100, replace = TRUE), nrow = 10)
Solution:
sum.mat <- lapply(1:(ncol(mat)-1), function(i) mat[,i] + mat[,i+1])
sum.mat <- matrix(unlist(sum.mat), byrow = FALSE, nrow = nrow(mat))
You could use:
m <- matrix(c(1,2,4,3,5,8), nrow=2, byrow=T)
sapply(2:ncol(m), function(x) m[,x] + m[,(x-1)])
I have 2 matrices.
The first one:
[1,2,3]
and the second one:
[3,1,2
2,1,3
3,2,1]
I'm looking for a way to multiply them.
The result is supposed to be: [11, 13, 10]
In R, mat1%*%mat2 don't work.
You need the transpose of the second matrix to get the result you wanted:
> v1 <- c(1,2,3)
> v2 <- matrix(c(3,1,2,2,1,3,3,2,1), ncol = 3, byrow = TRUE)
> v1 %*% t(v2)
[,1] [,2] [,3]
[1,] 11 13 10
Or potentially quicker (see ?crossprod) if the real problem is larger:
> tcrossprod(v1, v2)
[,1] [,2] [,3]
[1,] 11 13 10
mat1%%mat2 Actuall y works , this gives [ 16 9 11 ]
but you want mat1 %% t(mat2). This means transpose of second matrix, then u can get [11 13 10 ]
Rcode:
mat1 = matrix(c(1,2,3),nrow=1,ncol=3,byrow=TRUE)
mat2 = matrix(c(3,1,2,2,1,3,3,2,1), nrow=3,ncol=3,byrow=TRUE)
print(mat1)
print(mat2 )
#matrix Multiplication
print(mat1 %*% mat2 )
# matrix multiply with second matrix with transpose
# Note of using function t()
print(mat1 %*% t(mat2 ))
It's difficult to say what the best answer here is because the notation in the question isn't in R, it's in matlab. It's hard to tell if the questioner wants to multiple a vector, 1 row matrix, or 1 column matrix given the mixed notation.
An alternate answer to this question is simply switch the order of the multiplication.
v1 <- c(1,2,3)
v2 <- matrix(c(3,1,2,2,1,3,3,2,1), ncol = 3, byrow = TRUE)
v2 %*% v1
This yields an answer that's a single column rather than a single row matrix.
try this one
x<-c()
y<-c()
for(i in 1:9)
{
x[i]<-as.integer(readline("Enter number for 1st matrix"))
}
for(i in 1:9)
{
y[i]<-as.integer(readline("Enter number for 2nd matrix"))
}
M1 <- matrix(x, nrow=3,ncol = 3, byrow=TRUE)
M2 <- matrix(y, nrow=3,ncol = 3, byrow=TRUE)
print(M1%*%M2)