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)
Related
I am thinking about problem. How to count in R
(A-square matrix,k-any natural number) WITHOUT "for"?
If I've interpreted your notation correctly, perhaps something like this in base R...
A <- matrix(c(1,2,3,4), nrow = 2) #example matrix
k <- 10
B <- Reduce(`%*%`, (rep(list(A), k)), accumulate = TRUE) #list of A^(1:k)
BB <- lapply(1:k, function(k) B[[k]]/k) #list of A^(1:k)/k
Reduce(`+`, BB) #sum of series BB
[,1] [,2]
[1,] 603684.8 1319741
[2,] 879827.1 1923425
I have a 4x100 matrix where I would like to multiply column 1 with row 1 in its transpose etc and store these matrices somewhere to be able to take the sum of these new matrices lateron.
I really don't know where to start due to the fact that I get 4x4 matrices after the column-row-multiplication. Due to this fact I cannot store them in a matrix
data:
mm num[1:4,1:100]
mm_t num[1:100,1:4]
I'm thinking of creating a list in some way
list1=list()
for(i in 1:100){
list1[i] <- mm[,i]%*%mm_t[i,]
}
but I need some more indices i think because this just leaves me with a number in each argument..
First, your call for data is not clear. Second, are you tryign to multiply each value by itself, or do matrix multiplication
We create a 4x100 matrix and its transpose:
mm <- matrix(1:400, nrow = 4, ncol = 100)
mm.t <- t(mm)
Then we can do the matrix multiplication (which is what you did, and you get a 4 x 4 matrix from the definition of matrix multiplication https://www.wikiwand.com/en/Matrix_multiplication)
If we want to multiply each index by itself (so mm[1,1] by mm [1,1]) then:
mm * mm
This will result in 4x100 matrix where each value is the square of the original value.
If we want the matrix multiplication of each column with itself, then:
sapply(1:100, function(x) {
mm[, x] %*% mm[, x]
})
This results in 100 values: each one is the matrix product of a 4x1 vector with itself.
Let's start with some sample data. Please get in the habit of including things like this in your question:
nr = 4
nc = 100
set.seed(47)
mm = matrix(runif(nr * nc), nrow = nr)
Here's a working answer, very similar to your attempt:
result = list()
for (i in 1:ncol(mm)) result[[i]] = mm[, i] %*% t(mm[, i])
result[1:2]
# [[1]]
# [,1] [,2] [,3] [,4]
# [1,] 0.9544547 0.3653018 0.7439585 0.8035430
# [2,] 0.3653018 0.1398132 0.2847378 0.3075428
# [3,] 0.7439585 0.2847378 0.5798853 0.6263290
# [4,] 0.8035430 0.3075428 0.6263290 0.6764924
#
# [[2]]
# [,1] [,2] [,3] [,4]
# [1,] 0.3289532 0.3965557 0.2231443 0.2689613
# [2,] 0.3965557 0.4780511 0.2690022 0.3242351
# [3,] 0.2231443 0.2690022 0.1513691 0.1824490
# [4,] 0.2689613 0.3242351 0.1824490 0.2199103
As to why yours didn't work, we can experiment and see that indeed we get a number rather than a matrix. The reason is that when you subset a single row or column of a matrix, the dimensions are "dropped" and it is coerced to a plain vector. And when you matrix multiply two vectors, you get their dot product.
mmt = t(mm)
mm[, 1] %*% mmt[1, ]
# [,1]
# [1,] 2.350646
dim(mm[, 1])
# NULL
dim(mmt[1, ])
# NULL
We can avoid this by specifying drop = FALSE in the subset code
dim(mmt[1, , drop = FALSE])
# [1] 1 4
And thus slightly modify your attempt, just adding drop = FALSE will make it work.
res2 = list()
for (i in 1:ncol(mm)) res2[[i]] = mm[, i] %*% mmt[i, , drop = FALSE]
identical(result, res2)
# [1] TRUE
I would like to add each coefficient of a vector to each different column of a matrix. For example, if I have a vector and a matrix:
x <- c(1,2,3)
M <- matrix(c(5,6,7), nrow = 3, ncol = 3)
I would like to in my new matrix M1 1+5 in the first column, 2+6 in the second and 3+7 in the last one.
Is there any function in R that does this task?
try this:
M + rep(x, each = nrow(M))
or this:
apply(M, 1, `+`, x)
result:
[,1] [,2] [,3]
[1,] 6 7 8
[2,] 7 8 9
[3,] 8 9 10
EDIT:
akrun commented on two other great solutions:
M + x[col(M)]
and
sweep(M, 2, x, "+")
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've got a matrix (mat1), say 100 rows and 100 columns; I want to create another matrix where every row is the same as the 1st row in mat1 (except that I want to keep the 1st col as the original values)
I've managed to do this using a loop:
mat2 <- mat1
for(i in 1:nrow(mat1))
{
mat2[i,2:ncol(mat2)] <- mat1[1,2:ncol(mat1)]
}
this works and produces the result I expect; however, I'd have thought there should be a way to do it without a loop; I've tried:
mat2 <- mat1
mat2[c(2:100),2:ncol(mat2)] <- mat1[1,2:ncol(mat1)]
Can someone point out my error?!
Thanks,
Chris
The problem is the way R fills matrices, by columns. Here is a simple example that illustrates this:
mat1 <- matrix(1:9, ncol = 3)
mat2 <- matrix(1:9, ncol = 3)
mat2[-1, -1] <- mat1[1, -1]
mat2
> mat2
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 4 4
[3,] 3 7 7
mat1[1, -1] is the vector 4,7, which you can see that R has used to fill the bit of mat2 column-wise. You wanted a row-wise operation.
One solution is to replicate the replacement vector as many times as is required:
> mat2[-1, -1] <- rep(mat1[1, -1], each = nrow(mat1)-1)
> mat2
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 4 7
[3,] 3 4 7
This works because the rep() call replicates each value in the vector when we use the "each" argument, instead of replicating (repeating) the vector:
> rep(mat1[1, -1], each = nrow(mat1)-1)
[1] 4 4 7 7
The default behaviour would also give the wrong answer:
> rep(mat1[1, -1], nrow(mat1)-1)
[1] 4 7 4 7
In part, the problem you are seeing is also the way R extends arguments to the appropriate length for the replacement. R actually, and silently, extended the replacement vector exactly in the way rep(mat1[1, -1], nrow(mat1)-1) does, which when coupled with the fill-by-column principle gave the behaviour you saw.
Try
mat2[c(2:nrow(mat2)), 2:ncol(mat2)] <- mat1[rep.int(1,nrow(mat1)-1),2:ncol(mat1)]
Another option...
n = 5
mat1 = matrix(sample(n^2, n^2), n, n)
# use matrix with byrow to copy 1st row n times
mat2 = matrix(rep(mat1[1, ], n), n, n, byrow = TRUE)
# copy 1st column
mat2[ , 1] = mat1[ , 1]
mat1
mat2