Combine matrices row by row - r

I have two matrices of arbitrary sizes, e.g. matrix 1 (n * m) and matrix 2 (k * l). Is there a (convenient) way in R to cbind them row-by-row, to form a (n * k) * (m + l) matrix where each row of matrix 1 has a chance to be cbinded to each row of matrix 2? It is a complete row-by-row combination so the order does not matter.
For example, is there a function f such that:
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Thanks!

In the future, please include example data that is copy/pasteable, not just a picture.
m1 = matrix(1:6, ncol = 2)
m2 = matrix(7:12, ncol = 3)
combos = expand.grid(1:nrow(m1), 1:nrow(m2))
cbind(m1[combos$Var1, ], m2[combos$Var2, ])
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 4 7 9 11
# [2,] 2 5 7 9 11
# [3,] 3 6 7 9 11
# [4,] 1 4 8 10 12
# [5,] 2 5 8 10 12
# [6,] 3 6 8 10 12

There might be a better solution, but this works for smaller size problems:
A <- matrix(c(1,2,3,2,3,4), nrow = 3)
B <- matrix(c(5,6,6,7,7,8), nrow = 2)
temp <- lapply(1:nrow(A), function(x){
C = apply(B, 1, function(y){
c(A[x,],y)
})
return(t(C))
})
output <- do.call(rbind, temp)

cbind(A[rep(1:(x<-nrow(A)),each=y<-nrow(B)),],B[rep(1:y,x),])
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 5 6 7
[2,] 1 2 6 7 8
[3,] 2 3 5 6 7
[4,] 2 3 6 7 8
[5,] 3 4 5 6 7
[6,] 3 4 6 7 8
Break down:
x=nrow(A)
y=nrow(B)
cbind(A[rep(1:x,each=y),],B[rep(1:y,x),])

Related

How can I create this exact 5x5 matrix using only for loops?

0 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
So I was given this matrix and was told to create it using only for loops. What i have done so far is using cbind(0:4,1+(0:4),2+(0:4),3+(0:4),4+(0:4)) but i cant figure out a way to do so with the for function.
You were on the right track. If you rewrite your current
cbind(0:4,1+(0:4),2+(0:4),3+(0:4),4+(0:4))
as
cbind(0+(0:4),1+(0:4),2+(0:4),3+(0:4),4+(0:4))
you might notice that the thing that you are adding to 0:4 is implicitly a loop index.
Make it explicit:
m = c()
for(i in 0:4){
m = cbind(m,i+(0:4))
}
print(m)
Output:
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 2 3 4
[2,] 1 2 3 4 5
[3,] 2 3 4 5 6
[4,] 3 4 5 6 7
[5,] 4 5 6 7 8
One way:
mat <- matrix(0L, nrow=5, ncol=5)
for (i in 0:4) {
for (j in 0:4) {
mat[i + 1, j + 1] <- i + j
}
}
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 1 2 3 4
# [2,] 1 2 3 4 5
# [3,] 2 3 4 5 6
# [4,] 3 4 5 6 7
# [5,] 4 5 6 7 8
And technically *apply functions are loops as well:
sapply(0:4, \(x) 0:4 + x)
You can just create an empty matrix first and then fill it with two for-loops iterating over rows and columns. Playing a little bit around with the variable to write into the matrix (count) I figured out that this is a suitable solution.
matrix2fill <- matrix(NA, 5,5)
count = 0
for (i in 1:5){
for (j in 1:5){
matrix2fill[j,i] = count
count = count + 1
}
count = i
}
matrix2fill
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 2 3 4
[2,] 1 2 3 4 5
[3,] 2 3 4 5 6
[4,] 3 4 5 6 7
[5,] 4 5 6 7 8
Yet another way:
mymat <- matrix(NA, nrow = 5, ncol = 5)
i_mat <- 1
for (i in 0:4) {
mymat[seq(i_mat, i_mat + 4)] <- seq(i, i + 4)
i_mat <- i_mat + 5
}
mymat
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 2 3 4
[2,] 1 2 3 4 5
[3,] 2 3 4 5 6
[4,] 3 4 5 6 7
[5,] 4 5 6 7 8
Try this:
a = matrix(1:25, nrow=5, ncol=5)
for (i in 1:5) {
for (j in 1:5) {
a[i][j] = (i-1) + (j-1)
}
}
print(a)

Minimum of cells in two matrices within a moving kernel

I have two matrices m1 and m2.
m1 <- matrix(1:16, ncol = 4)
m2 <- matrix(16:1, ncol = 4)
# > m1
# [,1] [,2] [,3] [,4]
# [1,] 1 5 9 13
# [2,] 2 6 10 14
# [3,] 3 7 11 15
# [4,] 4 8 12 16
# > m2
# [,1] [,2] [,3] [,4]
# [1,] 16 12 8 4
# [2,] 15 11 7 3
# [3,] 14 10 6 2
# [4,] 13 9 5 1
I want to find the minimum between the two matrices for each cell within a moving kernel of 3x3. The outer margines should be ignored, i.e. they can be filled with NAs and the min function should then have na.rm = TRUE. The result should look like this:
# > m3
# [,1] [,2] [,3] [,4]
# [1,] 1 1 3 3
# [2,] 1 1 2 2
# [3,] 2 2 1 1
# [4,] 3 3 1 1
I have already tried a combination of pmin{base} and runmin{caTools} like this:
pmin(runmin(m1, 3, endrule = "keep"),
runmin(m2, 3, endrule = "keep"))
However, this did not work. Probably due to the fact that
"If x is a matrix than each column will be processed separately."
(from ?runmin)
Is there any package, that performs such operations, or is it possible to apply?
Here is a base R approach:
m = pmin(m1, m2)
grid = expand.grid(seq(nrow(m)), seq(ncol(m)))
x = apply(grid, 1, function(u) {
min(m[max(1,u[1]-1):min(nrow(m), u[1]+1), max(1,u[2]-1):min(ncol(m), u[2]+1)])
})
dim(x) = dim(m)
#> x
# [,1] [,2] [,3] [,4]
#[1,] 1 1 3 3
#[2,] 1 1 2 2
#[3,] 2 2 1 1
#[4,] 3 3 1 1

Creating matrix from random block rows of another matrix without loop (in R)?

I am trying to create a matrix by drawing random block rows from another matrix. I have managed to do so with a loop.
set.seed(1)
a_matrix <- matrix(1:10,10,5) # the matrix with original sample
b_matrix <- matrix(NA,10, 5) # a matrix to store the bootstrap sample
S2<- seq(from =1 , to = 10, by =2) #[1] 1 3 5 7 9
m <- 2 # block size of m
for (r in S2){ start_point<-sample(1:(nrow(a_matrix)-1), 1, replace=T)
#randomly choose a number 1 to length of a_matrix -1
b_block <- a_matrix[start_point:(start_point+(m-1)), 1:ncol(a_matrix)]
# randomly select blocks from matrix a
b_matrix[r,]<-as.matrix((b_block)[1,])
b_matrix[(r+1),]<-as.matrix((b_block)[2,]) # put the blocks into matrix b
}
b_matrix
#we now have a b_matrix that is made of random blocks (size m=2)
#of the original a_matrix
The loop method works but it is clearly not very efficient and it is not possible to extend it to other block size (for e.g. having a blocksize of 3) .What is a cleaner and expandable approach ? Thanks in advance
Here I tried to clean it up a bit and generalize the use of m:
random_block_sample <- function(a_matrix, m = 2L) {
N <- nrow(a_matrix)
stopifnot(m <= N)
n <- ceiling(N / m)
s <- sample(N - m + 1L, n, TRUE) # start_point
i <- unlist(lapply(s, seq, length.out = m))
b_matrix <- a_matrix[i, , drop = FALSE]
head(b_matrix, N)
}
set.seed(1L)
random_block_sample(a_matrix, m = 2L)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 3 3 3 3 3
# [2,] 4 4 4 4 4
# [3,] 4 4 4 4 4
# [4,] 5 5 5 5 5
# [5,] 6 6 6 6 6
# [6,] 7 7 7 7 7
# [7,] 9 9 9 9 9
# [8,] 10 10 10 10 10
# [9,] 2 2 2 2 2
# [10,] 3 3 3 3 3
set.seed(1L)
random_block_sample(a_matrix, m = 5L)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 2 2 2 2 2
# [2,] 3 3 3 3 3
# [3,] 4 4 4 4 4
# [4,] 5 5 5 5 5
# [5,] 6 6 6 6 6
# [6,] 3 3 3 3 3
# [7,] 4 4 4 4 4
# [8,] 5 5 5 5 5
# [9,] 6 6 6 6 6
# [10,] 7 7 7 7 7

Repeatedly Complementary subsets

I want to repeatedly divide a set into two complementary subsets with known size and keep them as the columns of two matrix. For example assume the main set is {1, 2, ..., 10}, the size of first sample is 8 and I want to repeat sampling 3 times. I want to have:
[,1] [,2] [,3]
[1,] 10 9 1
[2,] 8 1 10
[3,] 3 7 5
[4,] 4 2 3
[5,] 1 8 8
[6,] 6 4 2
[7,] 9 5 7
[8,] 5 10 6
and
[,1] [,2] [,3]
[1,] 2 3 4
[2,] 7 6 9
Any idea how to implement it in R avoiding for loops?
I would use replicate + sample, like this:
set.seed(1) # Just so you can replicate my results
A <- replicate(3, sample(10, 8, FALSE)) # Change 3 to the number of replications
A
# [,1] [,2] [,3]
# [1,] 3 7 8
# [2,] 4 1 9
# [3,] 5 2 4
# [4,] 7 8 6
# [5,] 2 5 7
# [6,] 8 10 2
# [7,] 9 4 3
# [8,] 6 6 1
For the other set, I would use apply + setdiff, like this:
B <- apply(A, 2, function(x) setdiff(1:10, x))
B
# [,1] [,2] [,3]
# [1,] 1 3 5
# [2,] 10 9 10
Another option as suggested by #thelatemail (which would be more efficient) is to just create use replicate to create your original matrix, and use basic subsetting to create your separate matrices.
A <- replicate(3, sample(10))
B <- A[-(seq_len(8)), ]
A <- A[seq_len(8), ]

In R, using `unique()` with extra conditions to extract submatrices: easy solution without plyr

In R, let M be the matrix
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 1 3 3
[3,] 2 4 5
[4,] 6 7 8
I would like to select the submatrix m
[,1] [,2] [,3]
[1,] 1 3 3
[2,] 2 4 5
[3,] 6 7 8
using unique on M[,1], specifying to keep the row with the maximal value in the second columnM.
At the end, the algorithm should keep row [2,] from the set \{[1,], [2,]\}. Unfortunately unique() returns me a vector with actual values, and not row numbers, after elimination of duplicates.
Is there a way to get the asnwer without the package plyr?
Thanks a lot,
Avitus
Here's how:
is.first.max <- function(x) seq_along(x) == which.max(x)
M[as.logical(ave(M[, 2], M[, 1], FUN = is.first.max)), ]
# [,1] [,2] [,3]
# [1,] 1 3 3
# [2,] 2 4 5
# [3,] 6 7 8
You're looking for duplicated.
m <- as.matrix(read.table(text="1 2 3
1 3 3
2 4 5
6 7 8"))
m <- m[order(m[,2], decreasing=TRUE), ]
m[!duplicated(m[,1]),]
# V1 V2 V3
# [1,] 6 7 8
# [2,] 2 4 5
# [3,] 1 3 3
Not the most efficient:
M <- matrix(c(1,1,2,6,2,3,4,7,3,3,5,8),4)
t(sapply(unique(M[,1]),function(i) {temp <- M[M[,1]==i,,drop=FALSE]
temp[which.max(temp[,2]),]
}))
# [,1] [,2] [,3]
#[1,] 1 3 3
#[2,] 2 4 5
#[3,] 6 7 8

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