Suppose I have a matrix m and a positive integer vector v, what I want to do is get a new matrix m_new and each row of m (say m[i, ]) are replicated by v[i] times in m_new. For example:
m = matrix(1:6, nrow = 3)
## [,1] [,2]
## [1,] 1 4
## [2,] 2 5
## [3,] 3 6
v = c(3, 1, 2)
And m_new should be:
[,1] [,2]
[1,] 1 4 # m[1, ] is replicated by
[2,] 1 4 # v[1] = 3
[3,] 1 4 # times
[4,] 2 5
[5,] 3 6
[6,] 3 6
A for loop will make it for the small case:
m_new = matrix(0, sum(v), ncol(m))
k = 1
for(i in 1:nrow(m)){
for(j in k:(k+v[i]-1)){
m_new[j, ] = m[i, ]
}
k = k + v[i]
}
, but the row number of m in real world is usually big. Is there any effient way to do this?
m[rep(1:nrow(m), times = v), ]
# [,1] [,2]
# [1,] 1 4
# [2,] 1 4
# [3,] 1 4
# [4,] 2 5
# [5,] 3 6
# [6,] 3 6
> m <- matrix(1:25, ncol=5)
> m
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
[5,] 5 10 15 20 25
> apply(m, 2, function(c) rep(c,v))
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 2 7 12 17 22
[4,] 3 8 13 18 23
[5,] 3 8 13 18 23
[6,] 3 8 13 18 23
[7,] 4 9 14 19 24
[8,] 4 9 14 19 24
[9,] 4 9 14 19 24
[10,] 4 9 14 19 24
[11,] 5 10 15 20 25
[12,] 5 10 15 20 25
[13,] 5 10 15 20 25
[14,] 5 10 15 20 25
[15,] 5 10 15 20 25
Related
n <- 5
a <- matrix(c(1:n**2),nrow = n, byrow = T)
output is
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[2,] 6 7 8 9 10
[3,] 11 12 13 14 15
[4,] 16 17 18 19 20
[5,] 21 22 23 24 25
how do I shift the '1' to the current position of '25' to look like this:
[,1] [,2] [,3] [,4] [,5]
[1,] 2 3 4 5 6
[2,] 7 8 9 10 11
[3,] 12 13 14 15 16
[4,] 17 18 19 20 21
[5,] 22 23 24 25 1
a <- t(a); a[] <- c(a[-1], a[1]); a <- t(a)
a
# [,1] [,2] [,3] [,4] [,5]
# [1,] 2 3 4 5 6
# [2,] 7 8 9 10 11
# [3,] 12 13 14 15 16
# [4,] 17 18 19 20 21
# [5,] 22 23 24 25 1
c(a) unwinds or unlists the matrix into a vector. It does this column-first, so c(a) results in [1] 1 6 11 16 21 2 .... We want it to be row-first, though, so
t(a) transposes it, so that what was a row-first is now column-first, allowing c(a) and such to work.
c(a[-1], a[1]) is just "concatenate all except the first with the first", the classic way to put the first element of a vector at the end.
a[] <- is a way to do calcs on its values where the calcs do not preserve the "dimensionality" of the object.
After we've rearranged, we then transpose back to the original shape and row/column-order.
Here is a base R one-liner
> t(`dim<-`(t(a)[seq_along(a)%%length(a)+1],rev(dim(a))))
[,1] [,2] [,3] [,4] [,5]
[1,] 2 3 4 5 6
[2,] 7 8 9 10 11
[3,] 12 13 14 15 16
[4,] 17 18 19 20 21
[5,] 22 23 24 25 1
I am a newbie in R, I now have a matrix of 3 columns and 8, 000 rows, with groups of 500 rows, which means 16 sets of 500*3 matrices stacked on top of each other in rows. Now I want to take the first 300 rows of each group of matrices, put 16 groups of 300 by 3 into a new matrix, what do I do?
Two 6 * 2 matrices on top of each other:
m <- matrix(1:24, ncol = 2)
# [,1] [,2]
# [1,] 1 13
# [2,] 2 14
# [3,] 3 15
# [4,] 4 16
# [5,] 5 17
# [6,] 6 18
# [7,] 7 19
# [8,] 8 20
# [9,] 9 21
#[10,] 10 22
#[11,] 11 23
#[12,] 12 24
Make it an array:
a <- array(m, c(6, 2, 2))
a <- aperm(a, c(1, 3, 2))
First three rows of each matrix:
a[1:3,,]
#, , 1
#
# [,1] [,2]
#[1,] 1 13
#[2,] 2 14
#[3,] 3 15
#
#, , 2
#
# [,1] [,2]
#[1,] 7 19
#[2,] 8 20
#[3,] 9 21
Use this if you need a matrix:
matrix(aperm(a[1:3,,], c(1, 3, 2)), ncol = 2)
# [,1] [,2]
#[1,] 1 13
#[2,] 2 14
#[3,] 3 15
#[4,] 7 19
#[5,] 8 20
#[6,] 9 21
You need to generate the sequence 1:300, 501:800, ... etc, then subset out these rows. If your matrix is called mat you can do that like this:
new_mat <- mat[as.numeric(sapply((0:15 * 500), "+", 1:300)),]
If you're looking for just a way to select the first 300 rows from your matrix for each group, this could be a solution.
Given m your matrix of 8000x3 composed by 16 groups on top of each other, then:
r <- 500 # rows for each group
g <- 16 # number of groups
n <- 300 # first n rows to select
new_m <- m[rep(rep(c(T,F), c(n,r-n)), g), ]
dim(new_m)
#> [1] 4800 3
new_m is now a matrix 4800x3
In case you are working with keras or reticulate, you could use array_reshape.
#### 0. parameters
nrows <- 4 # 500 in your example # rows for each group
ncols <- 3 # 3 in your example
ngrps <- 2 # 16 in your example # number of groups
nslct <- 3 # 300 in your example # first n rows to select
#### 1. create an example matrix
m <- matrix(1:24, nrows*ngrps, ncols)
m
#> [,1] [,2] [,3]
#> [1,] 1 9 17
#> [2,] 2 10 18
#> [3,] 3 11 19
#> [4,] 4 12 20
#> [5,] 5 13 21
#> [6,] 6 14 22
#> [7,] 7 15 23
#> [8,] 8 16 24
dim(m)
#> [1] 8 3
#--> c(ngrps * nrows, ncols)
#### 2. reshape in groups
m <- reticulate::array_reshape(m, c(ngrps,nrows,ncols))
dim(m)
#> [1] 2 4 3
# --> c(n_groups, n_rows, n_cols)
m[1,,]
#> [,1] [,2] [,3]
#> [1,] 1 9 17
#> [2,] 2 10 18
#> [3,] 3 11 19
#> [4,] 4 12 20
m[2,,]
#> [,1] [,2] [,3]
#> [1,] 5 13 21
#> [2,] 6 14 22
#> [3,] 7 15 23
#> [4,] 8 16 24
#### 3. select first nslct rows for each group
new_m <- m[,seq_len(nslct),]
# that's the result for each group
new_m[1,,]
#> [,1] [,2] [,3]
#> [1,] 1 9 17
#> [2,] 2 10 18
#> [3,] 3 11 19
new_m[2,,]
#> [,1] [,2] [,3]
#> [1,] 5 13 21
#> [2,] 6 14 22
#> [3,] 7 15 23
#### 4. recreate one matrix
reticulate::array_reshape(new_m, c(nslct*ngrps,ncols))
#> [,1] [,2] [,3]
#> [1,] 1 9 17
#> [2,] 2 10 18
#> [3,] 3 11 19
#> [4,] 5 13 21
#> [5,] 6 14 22
#> [6,] 7 15 23
Created on 2020-11-23 by the reprex package (v0.3.0)
A simple way to generate an array that mimics the fill of your matrix is to use the transpose of the matrix as the input for the array function. Here is a simple example:
n <- rep(1:3, each = 4)
m1 <- matrix(n, ncol = 2, byrow = TRUE)
> m1
[,1] [,2]
[1,] 1 1
[2,] 1 1
[3,] 2 2
[4,] 2 2
[5,] 3 3
[6,] 3 3
m2 <- t(m1)
a1 <- array(m2, c(2, 2, 3))
> a1
, , 1
[,1] [,2]
[1,] 1 1
[2,] 1 1
, , 2
[,1] [,2]
[1,] 2 2
[2,] 2 2
, , 3
[,1] [,2]
[1,] 3 3
[2,] 3 3
I´m trying convert a vector to multiple matrices and save them in a list.
#Create list to save matrix
BSEPRA=vector("list", 420)
#Vector size 6720
temporalRA
I need to build 4*4 size matrices with the first 16 elements then with the next ones (17:32) and so on up to 6705:6720 to have 420 and save them in the list. But this doesn't work:
for (i in 1:length(temporalRA)){
temp2<-matrix(temporalRA[seq(1,6720, 16), ],nrow = 4,ncol = 4, )
BSEPRA[[i]]=temp2
}
vec <- 1:32
split(vec, (seq_along(vec) - 1) %/% (4*4))
# $`0`
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
# $`1`
# [1] 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
lapply(split(vec, (seq_along(vec) - 1) %/% 16), matrix, nrow = 4, ncol = 4)
# $`0`
# [,1] [,2] [,3] [,4]
# [1,] 1 5 9 13
# [2,] 2 6 10 14
# [3,] 3 7 11 15
# [4,] 4 8 12 16
# $`1`
# [,1] [,2] [,3] [,4]
# [1,] 17 21 25 29
# [2,] 18 22 26 30
# [3,] 19 23 27 31
# [4,] 20 24 28 32
If there's any concern that you will not have a perfect multiple of 16, I encourage you to take one extra step.
The problem:
vec <- 1:30
vecs <- split(vec, (seq_along(vec) - 1) %/% (4*4))
vecs
# $`0`
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
# $`1`
# [1] 17 18 19 20 21 22 23 24 25 26 27 28 29 30
lapply(vecs, matrix, nrow = 4)
# Warning in FUN(X[[i]], ...) :
# data length [14] is not a sub-multiple or multiple of the number of rows [4]
# $`0`
# [,1] [,2] [,3] [,4]
# [1,] 1 5 9 13
# [2,] 2 6 10 14
# [3,] 3 7 11 15
# [4,] 4 8 12 16
# $`1`
# [,1] [,2] [,3] [,4]
# [1,] 17 21 25 29
# [2,] 18 22 26 30
# [3,] 19 23 27 17
# [4,] 20 24 28 18
(Notice how 17:18 are "recycled".)
The fix:
vec <- 1:30
vecs <- split(vec, (seq_along(vec) - 1) %/% (4*4))
vecs <- lapply(vecs, `length<-`, max(lengths(vecs)))
vecs
# $`0`
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
# $`1`
# [1] 17 18 19 20 21 22 23 24 25 26 27 28 29 30 NA NA
lapply(vecs, matrix, nrow = 4)
# $`0`
# [,1] [,2] [,3] [,4]
# [1,] 1 5 9 13
# [2,] 2 6 10 14
# [3,] 3 7 11 15
# [4,] 4 8 12 16
# $`1`
# [,1] [,2] [,3] [,4]
# [1,] 17 21 25 29
# [2,] 18 22 26 30
# [3,] 19 23 27 NA
# [4,] 20 24 28 NA
Another approach that I think will be more efficient than splitting the vector and iterating over the list to make matrices is to make an array and split it:
vec <- 1:32
mdim <- 4
mdimsq <- mdim^2
asplit(array(vec, dim = c(mdim, mdim, length(vec) / mdimsq)), 3)
[[1]]
[,1] [,2] [,3] [,4]
[1,] 1 5 9 13
[2,] 2 6 10 14
[3,] 3 7 11 15
[4,] 4 8 12 16
[[2]]
[,1] [,2] [,3] [,4]
[1,] 17 21 25 29
[2,] 18 22 26 30
[3,] 19 23 27 31
[4,] 20 24 28 32
Likewise, if the vector is not a perfect multiple it can be padded with NA:
vec <- 1:30
vln <- ceiling(length(vec)/mdimsq) * mdimsq
if (length(vec) < vln) vec[vln] <- NA
asplit(array(vec, dim = c(mdim, mdim, length(vec) / mdimsq)), 3)
[[1]]
[,1] [,2] [,3] [,4]
[1,] 1 5 9 13
[2,] 2 6 10 14
[3,] 3 7 11 15
[4,] 4 8 12 16
[[2]]
[,1] [,2] [,3] [,4]
[1,] 17 21 25 29
[2,] 18 22 26 30
[3,] 19 23 27 NA
[4,] 20 24 28 NA
I want to iterate the following matrix and print sets of 2 cell values. Is there a way to do this without a for-loop?
Input:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 4 7 10 13 16
[2,] 2 5 8 11 14 17
[3,] 3 6 9 12 15 18
Expected Output:
[,1] [,2]
[1,] 1 4
[2,] 7 10
[3,] 13 16
[4,] 2 5
[5,] 8 11
[6,] 14 17
[7,] 3 6
[8,] 9 12
[9,] 15 18
This my code:
mat<-matrix(data=seq(1:18), nrow=3,ncol=6)
r <- rep(seq(1,3),each=2)
c1 <- seq(1,6,2)
c2 <- seq(2,6,2)
m <- mat[r,c(c1:c2)] # This does not work, it only output first two cells
We can get the transpose of the matrix, then convert back to matrix by specifying the ncol
matrix(t(mat), ncol=2, byrow=TRUE)
# [,1] [,2]
# [1,] 1 4
# [2,] 7 10
# [3,] 13 16
# [4,] 2 5
# [5,] 8 11
# [6,] 14 17
# [7,] 3 6
# [8,] 9 12
# [9,] 15 18
I'm a beginner R user and I need to write a function that sums the rows of a data frame over a fixed interval (every 4 rows).
I've tried the following code
camp<-function(X){
i<-1
n<-nrow(X)
xc<-matrix(nrow=36,ncol=m)
for (i in 1:n){
xc<-apply(X[i:(i+4),],2,sum)
rownames(xc[i])<-rownames(X[i])
i<-i+5
}
return(xc)
}
the result is "Error in X[i:(i + 4), ] : index out of range".
How can I solve? Any suggestion?
Thanks.
The zoo package has rollapply which is pretty handy for stuff like this...
# Make some data
set.seed(1)
m <- matrix( sample( 10 , 32 , repl = TRUE ) , 8 )
# [,1] [,2] [,3] [,4]
#[1,] 3 7 8 3
#[2,] 4 1 10 4
#[3,] 6 3 4 1
#[4,] 10 2 8 4
#[5,] 3 7 10 9
#[6,] 9 4 3 4
#[7,] 10 8 7 5
#[8,] 7 5 2 6
# Sum every 4 rows
require( zoo )
tmp <- rollapply( m , width = 4 , by = 4 , align = "left" , FUN = sum )
# [,1] [,2] [,3] [,4]
#[1,] 23 13 30 12
#[2,] 29 24 22 24
You can also use rowSums() on the result if you actually wanted to aggregate the columns into a single value for each of the 4 rows...
rowSums( tmp )
#[1] 78 99
Here is a way to do it :
## Sample data
m <- matrix(1:36, nrow=12)
## Create a "group" index
fac <- (seq_len(nrow(m))-1) %/% 4
## Apply sum
apply(m, 2, function(v) tapply(v, fac, sum))
Sample data :
[,1] [,2] [,3]
[1,] 1 13 25
[2,] 2 14 26
[3,] 3 15 27
[4,] 4 16 28
[5,] 5 17 29
[6,] 6 18 30
[7,] 7 19 31
[8,] 8 20 32
[9,] 9 21 33
[10,] 10 22 34
[11,] 11 23 35
[12,] 12 24 36
Result :
[,1] [,2] [,3]
0 10 58 106
1 26 74 122
2 42 90 138