Suppose I have a list of nine, 2 x 2 matrices as defined by:
mat_list <- list(matrix(1, 2, 2), matrix(2, 2, 2), matrix(3, 2, 2),
matrix(4, 2, 2), matrix(5, 2, 2), matrix(6, 2, 2),
matrix(7, 2, 2), matrix(8, 2, 2), matrix(9, 2, 2))
I would like to merge these matrices into a single 6 x 6 matrix. It would look like this:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 4 4 7 7
[2,] 1 1 4 4 7 7
[3,] 2 2 5 5 8 8
[4,] 2 2 5 5 8 8
[5,] 3 3 6 6 9 9
[6,] 3 3 6 6 9 9
I can accomplish this task using the following code:
do.call( cbind, list( do.call( rbind, mat_list[1:3]),
do.call( rbind, mat_list[4:6]),
do.call( rbind, mat_list[7:9])) )
But how can this be generalized for a very large list of matrices? It would be too tedious to write out the list of do.call functions.
Maybe we can do like this
do.call(
cbind,
lapply(
split(mat_list, ceiling(seq_along(mat_list) / 3)),
function(x) do.call(rbind, x)
)
)
which gives
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 4 4 7 7
[2,] 1 1 4 4 7 7
[3,] 2 2 5 5 8 8
[4,] 2 2 5 5 8 8
[5,] 3 3 6 6 9 9
[6,] 3 3 6 6 9 9
or
> do.call(cbind, Map(function(x) do.call(rbind, x), data.frame(matrix(mat_list, 3, 3))))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 4 4 7 7
[2,] 1 1 4 4 7 7
[3,] 2 2 5 5 8 8
[4,] 2 2 5 5 8 8
[5,] 3 3 6 6 9 9
[6,] 3 3 6 6 9 9
Related
I have a vector comprised of a set of numbers, for example:
vec <- c(1, 2, 3, 4, 5)
I wish to produce a matrix that contains the sum of each pairwise element within that vector - in this case:
[,1] [,2] [,3] [,4] [,5]
[1,] 2 3 4 5 6
[2,] 3 4 5 6 7
[3,] 4 5 6 7 8
[4,] 5 6 7 8 9
[5,] 6 7 8 9 10
You can use outer()
outer(vec,vec,"+")
Output:
[,1] [,2] [,3] [,4] [,5]
[1,] 2 3 4 5 6
[2,] 3 4 5 6 7
[3,] 4 5 6 7 8
[4,] 5 6 7 8 9
[5,] 6 7 8 9 10
Note: this may also be written as:
outer(vec,vec,`+`)
Here's one way.
vec <- c(1, 2, 3, 4, 5)
matrix(rowSums(expand.grid(vec, vec)), ncol = length(vec))
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 2 3 4 5 6
#> [2,] 3 4 5 6 7
#> [3,] 4 5 6 7 8
#> [4,] 5 6 7 8 9
#> [5,] 6 7 8 9 10
sapply can do but maybe not as efficient as outer
> sapply(vec,`+`,vec)
[,1] [,2] [,3] [,4] [,5]
[1,] 2 3 4 5 6
[2,] 3 4 5 6 7
[3,] 4 5 6 7 8
[4,] 5 6 7 8 9
[5,] 6 7 8 9 10
Goal
I want to use a long vector of numbers, to create a matrix where each column is a successive offset (lag or lead) of the original vector. If n is the maximum offset, the matrix will have dimensions [length(vector), n * 2 + 1] (because we want offsets in both directions, and include the 0 offset, i.e. the original vector).
Example
To illustrate, consider the following vector:
test <- c(2, 8, 1, 10, 7, 5, 9, 3, 4, 6)
[1] 2 8 1 10 7 5 9 3 4 6
Expected output
Now we create offsets of values, let's say for n == 3:
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] NA NA NA 2 8 1 10
[2,] NA NA 2 8 1 10 7
[3,] NA 2 8 1 10 7 5
[4,] 2 8 1 10 7 5 9
[5,] 8 1 10 7 5 9 3
[6,] 1 10 7 5 9 3 4
[7,] 10 7 5 9 3 4 6
[8,] 7 5 9 3 4 6 NA
[9,] 5 9 3 4 6 NA NA
[10,] 9 3 4 6 NA NA NA
I am looking for an efficient solution. data.table or tidyverse solutions more than welcome.
Returning only the rows that have no NA's (i.e. rows 4 to 7) is also ok.
Current solution
lags <- lapply(3:1, function(x) dplyr::lag(test, x))
leads <- lapply(1:3, function(x) dplyr::lead(test, x))
l <- c(lags, test, leads)
matrix(unlist(l), nrow = length(test))
In base R, you can use embed to get rows 4 through 7. You have to reverse the column order, however.
embed(test, 7)[, 7:1]
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 2 8 1 10 7 5 9
[2,] 8 1 10 7 5 9 3
[3,] 1 10 7 5 9 3 4
[4,] 10 7 5 9 3 4 6
data
test <- c(2, 8, 1, 10, 7, 5, 9, 3, 4, 6)
This will produce what you need...
n <- 3
t(embed(c(rep(NA,n), test, rep(NA,n)), length(test)))[length(test):1,]
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] NA NA NA 2 8 1 10
[2,] NA NA 2 8 1 10 7
[3,] NA 2 8 1 10 7 5
[4,] 2 8 1 10 7 5 9
[5,] 8 1 10 7 5 9 3
[6,] 1 10 7 5 9 3 4
[7,] 10 7 5 9 3 4 6
[8,] 7 5 9 3 4 6 NA
[9,] 5 9 3 4 6 NA NA
[10,] 9 3 4 6 NA NA NA
This can be solved by constructing the matrix from a long vector and returning only the wanted columns and rows:
test <- c(2, 8, 1, 10, 7, 5, 9, 3, 4, 6)
n_offs <- 3L
n_row <- length(test) + n_offs + 1L
matrix(rep(c(rep(NA, n_offs), test), n_row), nrow = n_row)[1:length(test), 1:(n_offs * 2L + 1L)]
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] NA NA NA 2 8 1 10
[2,] NA NA 2 8 1 10 7
[3,] NA 2 8 1 10 7 5
[4,] 2 8 1 10 7 5 9
[5,] 8 1 10 7 5 9 3
[6,] 1 10 7 5 9 3 4
[7,] 10 7 5 9 3 4 6
[8,] 7 5 9 3 4 6 NA
[9,] 5 9 3 4 6 NA NA
[10,] 9 3 4 6 NA NA NA
A variant which just returns the same result as embed(test, 7)[, 7:1] is:
matrix(rep(test, length(test) + 1L), nrow = length(test) + 1L)[
seq_len(length(test) - 2L * n_offs), seq_len(n_offs * 2L + 1L)]
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 2 8 1 10 7 5 9
[2,] 8 1 10 7 5 9 3
[3,] 1 10 7 5 9 3 4
[4,] 10 7 5 9 3 4 6
I have two vectors and I want a new vector which elements are the sum of an element of vector 1 and an element of vector 2.
v1<-c(1,2,3,4,5,6)
v2<-c(0,1,1,2,2,1)
for(i in 1:length(v1)){
for(j in 1:length(v2)){
n<-vector()
n<-v1[i]+v2[j]
}
m<-NULL
m[n]<-m
}
When I run the loop, I get m=NULL and n is numeric class with NA. Any idea?
Perhaps we need
tapply(c(v1, v2), c(v1, v2), FUN = sum)
Or just
v1 + v2
Or could be outer
outer(v1, v2, FUN = "+")
If you want to correct your code, you can try something like this:
v1<-c(1,2,3,4,5,6)
v2<-c(0,1,1,2,2,1)
m<-matrix(rep(0,length(v1)*length(v2)), nrow=length(v1))
for(i in 1:length(v1)){
for(j in 1:length(v2)){
m[i,j] <- v1[i]+v2[j]
}
}
m
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 2 2 3 3 2
[2,] 2 3 3 4 4 3
[3,] 3 4 4 5 5 4
[4,] 4 5 5 6 6 5
[5,] 5 6 6 7 7 6
[6,] 6 7 7 8 8 7
This can also be done this way
outer(v1, v2, FUN='+')
or in this way
matrix(apply(expand.grid(1:length(v1), 1:length(v2))[2:1], 1,
function(x)v1[x[1]]+v2[x[2]]), nrow=length(v1), byrow=TRUE)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 2 2 3 3 2
[2,] 2 3 3 4 4 3
[3,] 3 4 4 5 5 4
[4,] 4 5 5 6 6 5
[5,] 5 6 6 7 7 6
[6,] 6 7 7 8 8 7
Is it possible in R to say - I want all indices from position i to the end of vector/matrix?
Say I want a submatrix from 3rd column onwards. I currently only know this way:
A = matrix(rep(1:8, each = 5), nrow = 5) # just generate some example matrix...
A[,3:ncol(A)] # get submatrix from 3rd column onwards
But do I really need to write ncol(A)? Isn't there any elegant way how to say "from the 3rd column onwards"? Something like A[,3:]? (or A[,3:...])?
Sometimes it's easier to tell R what you don't want. In other words, exclude columns from the matrix using negative indexing:
Here are two alternative ways that both produce the same results:
A[, -(1:2)]
A[, -seq_len(2)]
Results:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
But to answer your question as asked: Use ncol to find the number of columns. (Similarly there is nrow to find the number of rows.)
A[, 3:ncol(A)]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
For rows (not columns as per your example) then head() and tail() could be utilised.
A <- matrix(rep(1:8, each = 5), nrow = 5)
tail(A, 3)
is almost the same as
A[3:dim(A)[1],]
(the rownames/indices printed are different is all).
Those work for vectors and data frames too:
> tail(1:10, 4)
[1] 7 8 9 10
> tail(data.frame(A = 1:5, B = 1:5), 3)
A B
3 3 3
4 4 4
5 5 5
For the column versions, you could adapt tail(), but it is a bit trickier. I wonder if NROW() and NCOL() might be useful here, rather than dim()?:
> A[, 3:NCOL(A)]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
Or flip this on its head and instead of asking R for things, ask it to drop things instead. Here is a function that encapsulates this:
give <- function(x, i, dimen = 1L) {
ind <- seq_len(i-1)
if(isTRUE(all.equal(dimen, 1L))) { ## rows
out <- x[-ind, ]
} else if(isTRUE(all.equal(dimen, 2L))) { ## cols
out <- x[, -ind]
} else {
stop("Only for 2d objects")
}
out
}
> give(A, 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 2 3 4 5 6 7 8
[2,] 1 2 3 4 5 6 7 8
[3,] 1 2 3 4 5 6 7 8
> give(A, 3, dimen = 2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
You can use the following instruction:
A[, 3:length(A[, 1])]
A dplyr readable renewed approach for the same thing:
A %>% as_tibble() %>%
select(-c(V1,V2))
A %>% as_tibble() %>%
select(V3:ncol(A))
Is it possible in R to say - I want all indices from position i to the end of vector/matrix?
Say I want a submatrix from 3rd column onwards. I currently only know this way:
A = matrix(rep(1:8, each = 5), nrow = 5) # just generate some example matrix...
A[,3:ncol(A)] # get submatrix from 3rd column onwards
But do I really need to write ncol(A)? Isn't there any elegant way how to say "from the 3rd column onwards"? Something like A[,3:]? (or A[,3:...])?
Sometimes it's easier to tell R what you don't want. In other words, exclude columns from the matrix using negative indexing:
Here are two alternative ways that both produce the same results:
A[, -(1:2)]
A[, -seq_len(2)]
Results:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
But to answer your question as asked: Use ncol to find the number of columns. (Similarly there is nrow to find the number of rows.)
A[, 3:ncol(A)]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
For rows (not columns as per your example) then head() and tail() could be utilised.
A <- matrix(rep(1:8, each = 5), nrow = 5)
tail(A, 3)
is almost the same as
A[3:dim(A)[1],]
(the rownames/indices printed are different is all).
Those work for vectors and data frames too:
> tail(1:10, 4)
[1] 7 8 9 10
> tail(data.frame(A = 1:5, B = 1:5), 3)
A B
3 3 3
4 4 4
5 5 5
For the column versions, you could adapt tail(), but it is a bit trickier. I wonder if NROW() and NCOL() might be useful here, rather than dim()?:
> A[, 3:NCOL(A)]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
Or flip this on its head and instead of asking R for things, ask it to drop things instead. Here is a function that encapsulates this:
give <- function(x, i, dimen = 1L) {
ind <- seq_len(i-1)
if(isTRUE(all.equal(dimen, 1L))) { ## rows
out <- x[-ind, ]
} else if(isTRUE(all.equal(dimen, 2L))) { ## cols
out <- x[, -ind]
} else {
stop("Only for 2d objects")
}
out
}
> give(A, 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 2 3 4 5 6 7 8
[2,] 1 2 3 4 5 6 7 8
[3,] 1 2 3 4 5 6 7 8
> give(A, 3, dimen = 2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 3 4 5 6 7 8
[2,] 3 4 5 6 7 8
[3,] 3 4 5 6 7 8
[4,] 3 4 5 6 7 8
[5,] 3 4 5 6 7 8
You can use the following instruction:
A[, 3:length(A[, 1])]
A dplyr readable renewed approach for the same thing:
A %>% as_tibble() %>%
select(-c(V1,V2))
A %>% as_tibble() %>%
select(V3:ncol(A))