I am starting to learn R and trying to multiply a 5X3 matrix with a 3X1 column vector in R; However while creating a new variable to perform the operation, R throws the error "non-conformable arrays". Can someone please point out my mistake in the code below -
*#5X3 Matrix*
X <- matrix(c(1,25.5,1.23,1,40.8,1.89,1,30.2,1.55,1,4.3,1.18,1,10.7,1.68),nrow=5,ncol=3,byrow=TRUE)
*3X1 Column vector*
b1 <- matrix(c(23,0.1,-8), nrow = 3, ncol = 1, byrow = TRUE)
v1 <- X * b1
v1
Appreciate your help :)
You need the matrix-multiplication operator %*%:
X <- matrix(c(1,25.5,1.23,1,40.8,1.89,1,30.2,1.55,1,4.3,1.18,1,10.7,1.68),nrow=5,ncol=3,byrow=TRUE)
b1 <- matrix(c(23,0.1,-8), nrow = 3, ncol = 1, byrow = TRUE)
v1 <- X %*% b1
v1
#> [,1]
#> [1,] 15.71
#> [2,] 11.96
#> [3,] 13.62
#> [4,] 13.99
#> [5,] 10.63
Normally one would use the first alternative below but the others are possible too. The first four alternatives below give a column vector as the result while the others give a plain vector without dimensions. The first three work even if b1 has more than one column. The remainder assume b1 has one column but could be generalized.
X %*% b1
crossprod(t(X), b1)
library(einsum)
einsum("ij,jk -> ik", X, b1)
out <- matrix(0, nrow(X), ncol(b1))
for(i in 1:nrow(X)) {
for(k in 1:ncol(X)) out[i] <- out[i] + X[i, k] * b1[k, 1]
}
out
colSums(t(X) * c(b1))
apply(X, 1, crossprod, b1)
sapply(1:nrow(X), function(i) sum(X[i, ] * b1))
rowSums(mapply(`*`, as.data.frame(X), b1))
rowSums(sapply(1:ncol(X), function(j) X[, j] * b1[j]))
X[, 1] * b1[1, 1] + X[, 2] * b1[2, 1] + X[, 3] * b1[3, 1]
Note
The input shown in the question is:
X <- matrix(c(1,25.5,1.23,1,40.8,1.89,1,30.2,1.55,1,4.3,1.18,1,10.7,1.68),nrow=5,ncol=3,byrow=TRUE)
b1 <- matrix(c(23,0.1,-8), nrow = 3, ncol = 1, byrow = TRUE)
Related
I try to loop trough a matrix but cant find a easy and elegant way instead of writing many (>10) equations... Can anyone help me please?
My Matrix looks like this:
and I want to calculate the following:
(0 * 0 * 4/24) + (0 * 1 * 6/24) + (0 * 2 * 3/24) + (1 * 0 * 3/24) + (1 * 1 * 4/24) + (1 * 2 * 4/24)
instead of using
__
btw: my code for the matrix
vals<- c(4/24, 6/24, 3/24, 3/24, 4/24, 4/24)
x <- c(0,1)
y <- c(0,1,2)
df <- matrix(vals, byrow = TRUE, nrow = 2, ncol = 3,
dimnames = list(x,y))
instead of calculation each step manually, I think there should be a for-loop method, but cant figure it out..
A possible solution:
c(x %*% df %*% y)
#> [1] 0.5
Another possible solution, based on outer:
sum(outer(x, y, Vectorize(\(x,y) x*y*df[x+1,y+1])))
#> [1] 0.5
x <- c(0, 1)
y <- c(0, 1, 2)
vals<- c(4/24, 6/24, 3/24, 3/24, 4/24, 4/24)
mat <- matrix(vals, byrow = TRUE, nrow = 2, ncol = 3,
dimnames = list(x,y)) ## not a data frame; don't call it "df"
There is even a better way than a for loop:
sum(tcrossprod(x, y) * mat)
#[1] 0.5
sum((x %o% y) * df)
Explanation:
x %o% y gets the outer product of vectors x and y which is:
#> [,1] [,2] [,3]
#> [1,] 0 0 0
#> [2,] 0 1 2
Since that has the same dimensions as df, you can multiply the corresponding elements and get the sum: sum((x %o% y) * df)
If you are new to R (as I am), here is the loop approach.
result = 0
for (i in 1:length(x)) {
for (j in 1:length(y)) {
result = result + x[i] * y[j] * df[i, j]
}
}
result
In an earlier question (R: Logical Conditions Not Being Respected), I learned how to make the following simulation :
Step 1: Keep generating two random numbers "a" and "b" until both "a" and "b" are greater than 12
Step 2: Track how many random numbers had to be generated until it took for Step 1 to be completed
Step 3: Repeat Step 1 and Step 2 100 times
res <- matrix(0, nrow = 0, ncol = 3)
for (j in 1:100){
a <- rnorm(1, 10, 1)
b <- rnorm(1, 10, 1)
i <- 1
while(a < 12 | b < 12) {
a <- rnorm(1, 10, 1)
b <- rnorm(1, 10, 1)
i <- i + 1
}
x <- c(a,b,i)
res <- rbind(res, x)
}
head(res)
[,1] [,2] [,3]
x 12.14232 12.08977 399
x 12.27158 12.01319 1695
x 12.57345 12.42135 302
x 12.07494 12.64841 600
x 12.03210 12.07949 82
x 12.34006 12.00365 782
Question: Now, I am trying to make a slight modification to the above code - Instead of "a" and "b" being produced separately, I want them to be produced "together" (in math terms: "a" and "b" were being produced from two independent univariate normal distributions, now I want them to come from a bivariate normal distribution).
I tried to modify this code myself:
library(MASS)
Sigma = matrix(
c(1,0.5, 0.5, 1), # the data elements
nrow=2, # number of rows
ncol=2, # number of columns
byrow = TRUE) # fill matrix by rows
res <- matrix(0, nrow = 0, ncol = 3)
for (j in 1:100){
e_i = data.frame(mvrnorm(n = 1, c(10,10), Sigma))
e_i$i <- 1
while(e_i$X1 < 12 | e_i$X2 < 12) {
e_i = data.frame(mvrnorm(n = 1, c(10,10), Sigma))
e_i$i <- i + 1
}
x <- c(e_i$X1, e_i$X2 ,i)
res <- rbind(res, x)
}
res = data.frame(res)
But this is producing the following error:
Error in while (e_i$X1 < 12 | e_i$X2 < 12) { : argument is of length
zero
If I understand your code correctly you are trying to see how many samples occur before both values are >=12 and doing that for 100 trials? This is the approach I would take:
library(MASS)
for(i in 1:100){
n <- 1
while(any((x <- mvrnorm(1, mu=c(10,10), Sigma=diag(0.5, nrow=2)+0.5))<12)) n <- n+1
if(i==1) res <- data.frame("a"=x[1], "b"=x[2], n)
else res <- rbind(res, data.frame("a"=x[1], "b"=x[2], n))
}
Here I am assigning the results of a mvrnorm to x within the while() call. In that same call, it evaluates whether either are less than 12 using the any() function. If that evaluates to FALSE, n (the counter) is increased and the process repeated. Once TRUE, the values are appended to your data.frame and it goes back to the start of the for-loop.
Regarding your code, the mvrnorm() function is returning a vector, not a matrix, when n=1 so both values go into a single variable in the data.frame:
data.frame(mvrnorm(n = 1, c(10,10), Sigma))
Returns:
mvrnorm.n...1..c.10..10...Sigma.
1 9.148089
2 10.605546
The matrix() function within your data.frame() calls, along with some tweaks to your use of i, will fix your code:
library(MASS)
Sigma = matrix(
c(1,0.5, 0.5, 1), # the data elements
nrow=2, # number of rows
ncol=2, # number of columns
byrow = TRUE) # fill matrix by rows
res <- matrix(0, nrow = 0, ncol = 3)
for (j in 1:10){
e_i = data.frame(matrix(mvrnorm(n = 1, c(10,10), Sigma), ncol=2))
i <- 1
while(e_i$X1[1] < 12 | e_i$X2[1] < 12) {
e_i = data.frame(matrix(mvrnorm(n = 1, c(10,10), Sigma), ncol=2))
i <- i + 1
}
x <- c(e_i$X1, e_i$X2 ,i)
res <- rbind(res, x)
}
res = data.frame(res)
I would like to do symbolic matrix operations with Ryacas using a function that converts base R matrices to Ryacas format. The result of the function seems to match Ryacas format. But when I attempt to multiply the matrices, the error
# Error in aa %*% aa : requires numeric/complex matrix/vector arguments
throws. The code below is a minimal example that shows the case.
Any suggestion, please?
library(Ryacas)
conv.mat <- function(x) {
conv <- lapply(1:nrow(x), function(i) paste0(x[i, ], collapse = ", "))
conv <- paste0("List(", paste0("List(", unlist(conv), ")", collapse = ", "), ")")
noquote(conv)
}
# Writing a matrix manually for Ryacas format
a <- List(List(1, 2), List(3, 7))
a * a
# expression(list(list(7, 16), list(24, 55)))
# Writing a matrix in R and convert it to Ryacas format by the function conv.mat
aa <- matrix(c(1, 2, 3, 7), 2, byrow = TRUE)
aa <- conv.mat(aa)
# [1] List(List(1, 2), List(3, 7))
aa * aa
# Error in aa * aa : non-numeric argument to binary operator
Firstly, to multiply Ryacas matrices you want aa * aa rather than aa %*% aa. But that alone doesn't help in your case as conv.mat doesn't give exactly what we need (an expression).
We may use, e.g.,
conv.mat <- function(x)
do.call(List, lapply(1:nrow(x), function(r) do.call(List, as.list(x[r, ]))))
Then
M <- matrix(c(1, 2, 3, 7), 2, byrow = TRUE)
M %*% M
# [,1] [,2]
# [1,] 7 16
# [2,] 24 55
M <- conv.mat(M)
M * M
# expression(list(list(7, 16), list(24, 55)))
I have a dataset with very large matrices and vectors. I would like to multiply a matrix with a vector that has one "1" element with the rest being zero. I would like to loop this calculation so that every possible 1 and 0 combination within the vector has been multiplied with the matrix, and the store the results in a vector.
I give an example of what I'm trying to do.
I have two matrices, a and b:
a <- matrix(1:16, nrow = 4, byrow = TRUE)
b <- matrix(17:32, nrow = 4, byrow = TRUE)
and a vector with 1's and 0's. As I don't know how to loop well yet, I write down the code for each combination:
c1 <- rep(0, times = 4)
c1[1] <- 1
c2 <- rep(0, times = 4)
c2[2] <- 1
c3 <- rep(0, times = 4)
c3[3] <- 1
c4 <- rep(0, times = 4)
c4[4] <- 1
I multiply a with each of the vector combinations c, diagonalize it, multiply this with b and sum each row and column. I then store this result in a vector results:
d1 <- sum(colSums(b %*% diag(as.vector(a %*% c1), nrow = 4)))
d2 <- sum(colSums(b %*% diag(as.vector(a %*% c2), nrow = 4)))
d3 <- sum(colSums(b %*% diag(as.vector(a %*% c3), nrow = 4)))
d4 <- sum(colSums(b %*% diag(as.vector(a %*% c4), nrow = 4)))
results <- cbind(d1, d2, d3, d4)
that gives:
d1 d2 d3 d4
[1,] 2824 3216 3608 4000
Is there a good line of code that does this more efficiently than what I did here?
Because of the special structure of your calculation you can shorten it to
a <- matrix(1:16, nrow = 4, byrow = TRUE)
b <- matrix(17:32, nrow = 4, byrow = TRUE)
results <- apply(a, 2, function(ai) sum(b %*% ai))
# [1] 2824 3216 3608 4000
or even shorter
colSums(b %*% a)
# [1] 2824 3216 3608 4000
I am implementing a technique whereby I take two matrices (M1 and M2) and multiply them each by the same "skewer" vector (B), producing results vectors R1 and R2, then taking a correlation of these vectors, as so:
P1 <- data.frame(split(rnorm(5*16, 1, 1), 1:16))
M1 <- matrix(unlist(P1[1,]), nrow = 4)
M1[upper.tri(M1)] <- t(M1)[upper.tri(M1)]
P2 <- data.frame(split(rnorm(5*16, 1, 1), 1:16))
M2 <- matrix(unlist(P2[1,]), nrow = 4)
M2[upper.tri(M2)] <- t(M2)[upper.tri(M2)]
B <- rnorm(4, 0, 1)
R1 <- M1 %*% B
R2 <- M2 %*% B
cor(R1, R2)
However, I need to extend this in two ways: i) I need to do this for n (4000, but showing here for 2) vectors of B, which I have achieved using a function as below, and ii) performing this for each iteration of a posterior distribution of the matrices (1000, using 5 here in the example), which I have achieved using a for loop inside the function. This returns a data frame with one row per iteration, and 1 column per skewer, and each cell giving the correlation of response vectors. While this works, the for loop is slow -
com_rsk_p2 <- function(m1, m2, n = 2){
nitt <- length(m1[,1])
k <- sqrt(length(m1))
B <- split(rnorm(n*k, 0, 1), 1:n)
rv_cor <- split(rep(NA, times = n*nitt), 1:nitt)
for(i in 1:nitt){
R1 <- sapply(B, function(x) x %*% matrix(unlist(m1[i,]), ncol = k))
R2 <- sapply(B, function(x) x %*% matrix(unlist(m2[i,]), ncol = k))
rv_cor[[i]] <- diag(matrix(mapply(cor, list(R1), list(R2)), ncol = n))
}
return(t(data.frame(rv_cor)))
}
I've been working on this for a couple of days, but coming up short - is it possible to use a non-looping/apply approach so that each iteration of M1 and M2 are multiplied by each skewer, storing the result vector correlations for each case? I'm sure there must be some trick that I am missing!
> out <- com_rsk_p2(P1, P2)
> out
[,1] [,2]
X1 0.7622732 0.8156658
X2 0.4414054 0.4266134
X3 0.4388098 -0.1248999
X4 0.5438046 0.7723585
X5 -0.5833943 -0.5294521
Ideally I'd like to have the objects R1 and R2 remain within the function because I will use these for some other things later on when I add to this function (calculating angles between vectors etc.).
Updated 26/04/2018 I have created a list of the matrices, and matrix of the B vectors, and I can multiply a single matrix by each vector of B as below - the key I am looking for is to extend this to an efficient approach that multiplies each matrix in the list by each vector of B:
P1 <- data.frame(split(round(rnorm(5*16, 1, 1),2), 1:16))
P2 <- data.frame(split(round(rnorm(5*16, 1, 1),2), 1:16))
nitt <- length(P1[,1])
k <- sqrt(length(P1))
M1L <- list(rep(NA, times = nitt))
M2L <- list(rep(NA, times = nitt))
for(i in 1:nitt){
M <- matrix(P1[i,], byrow = T, ncol = k)
M[lower.tri(M)] <- t(M)[lower.tri(M)]
M1L[[i]] <- M
M <- matrix(P2[i,], byrow = T, ncol = k)
M[lower.tri(M)] <- t(M)[lower.tri(M)]
M2L[[i]] <- M
}
B <- matrix(round(rnorm(2*4, 0, 1),2), ncol = 2)
matrix(unlist(M2L[[1]]), ncol = 4) %*% B
> matrix(unlist(M2L[[1]]), ncol = 4) %*% B
[,1] [,2]
[1,] 0.1620 -0.3203
[2,] 0.6027 0.8148
[3,] 0.9763 -1.3177
[4,] -0.5369 0.5605