I would like to apply on a matrix a function of both the value, the row index and the column index for every value in the matrix and get the transformed matrix.
For example
mat<-matrix(c(1,2,3,4),2,2)
mat
[,1] [,2]
[1,] 1 3
[2,] 2 4
f<-function(x,i,j){x+i+j}
mat2 <- my.apply(f,mat)
mat2
[,1] [,2]
[1,] 3 6
[2,] 5 8
The example above is for illustration purposes, f can be much more complex.
apply does not do the job, because of the way the extra arguments are handled.
apply(mat,1:2,f,seq_along(mat[,1]),seq_along(mat[1,]))
, , 1
[,1] [,2]
[1,] 3 4
[2,] 5 6
, , 2
[,1] [,2]
[1,] 5 6
[2,] 7 8
I can not find either a way with the lapply family. A for loop can do the job but it won't be efficient nor elegant.
Any suggestions?
Thanks
Try mapply
mat <- matrix(c(1, 2, 3, 4), 2, 2)
mat
## [,1] [,2]
## [1,] 1 3
## [2,] 2 4
matrix(mapply(function(x, i, j) x + i + j, mat, row(mat), col(mat)), nrow = nrow(mat))
## [,1] [,2]
## [1,] 3 6
## [2,] 5 8
Here is an ugly use of apply, just for some quick and dirty job. The trick is adding an additional column (or row) for row (or column) indices.
mat <- matrix(c(1, 2, 3, 4), 2, 2)
t(apply(cbind(mat, 1:nrow(mat)), 1, function(x){x[1:ncol(mat)] + 1:ncol(mat) + x[ncol(mat)+1]}))
## [,1] [,2]
##[1,] 3 5
##[2,] 6 8
If you have a function f(x, i, j) already, you can also try:
apply(cbind(mat, 1:nrow(mat)), 1, function(x){a = numeric(); for(j in 1:ncol(mat)){a[j] = f(x[j], x[ncol(mat)+1], j)}; a})
Related
I have a list containing matrices of the same size in R. I would like to apply a function over the same element of all matrices. Example:
> a <- matrix(1:4, ncol = 2)
> b <- matrix(5:8, ncol = 2)
> c <- list(a,b)
> c
[[1]]
[,1] [,2]
[1,] 1 3
[2,] 2 4
[[2]]
[,1] [,2]
[1,] 5 7
[2,] 6 8
Now I want to apply the mean function and would like to get a matrix like that:
[,1] [,2]
[1,] 3 5
[2,] 4 6
One conceptual way to do this would be to sum up the matrices and then take the average value of each entry. Try using Reduce:
Reduce('+', c) / length(c)
Output:
[,1] [,2]
[1,] 3 5
[2,] 4 6
Demo here:
Rextester
Another option is to construct an array and then use apply.
step 1: constructing the array.
Using the abind library and do.call, you can do this:
library(abind)
myArray <- do.call(function(...) abind(..., along=3), c)
Using base R, you can strip out the structure and then rebuild it like this:
myArray <- array(unlist(c), dim=c(dim(a), length(c)))
In both instances, these return the desired array
, , 1
[,1] [,2]
[1,] 1 3
[2,] 2 4
, , 2
[,1] [,2]
[1,] 5 7
[2,] 6 8
step 2: use apply to calculate the mean along the first and second dimensions.
apply(myArray, 1:2, mean)
[,1] [,2]
[1,] 3 5
[2,] 4 6
This will be more flexible than Reduce, since you can swap out many more functions, but it will be slower for this particular application.
I want to apply a function over one margin (column in my example) of a matrix. The problem is that the function returns matrix and apply converts it to vector so that it returns a matrix. My goal is to get three-dimensional array. Here is the example (note that matrix() is not the function of interest, just an example):
x <- matrix(1:12, 4, 3)
apply(x, 2, matrix, nrow = 2, ncol = 2)
The output is exactly the same as the input. I have pretty dull solution to this:
library(abind)
abind2 <- function (x, ...)
abind(x, ..., along = dim(x) + 1)
apply(x, 2, list) %>%
lapply(unlist) %>%
lapply(matrix, nrow = 2, ncol = 2) %>%
do.call(what = 'abind2')
I believe there must exist something better than this. Something that does not include list()ing and unlist()ing columns.
Edit:
Also, the solution should be ready to be easily applicable to any-dimensional array with any choice of MARGIN which my solution is not.
This, for example, I want to return 4-dimensional array.
x <- array(1:24, c(4,3,2))
apply(x, 2:3, list) %>%
lapply(unlist) %>%
lapply(matrix, nrow = 2, ncol = 2) %>%
do.call(what = 'abind2')
Not that complicated at all. Simply use
array(x, dim = c(2, 2, ncol(x)))
Matrix and general arrays are stored by column into a 1D long array in physical address. You can just reallocate dimension.
OK, here is possibly what you want to do in general:
tapply(x, col(x), FUN = matrix, nrow = 2, ncol = 2)
#$`1`
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
#
#$`2`
# [,1] [,2]
#[1,] 5 7
#[2,] 6 8
#
#$`3`
# [,1] [,2]
#[1,] 9 11
#[2,] 10 12
You can try to convert your matrix into a data.frame and use lapply to apply your function on the columns (as a data.frame is a list), it will return a list, where each element represents the function result for a column:
lapply(as.data.frame(x), matrix, nrow = 2, ncol = 2)
# $V1
# [,1] [,2]
# [1,] 1 3
# [2,] 2 4
# $V2
# [,1] [,2]
# [1,] 5 7
# [2,] 6 8
# $V3
# [,1] [,2]
# [1,] 9 11
# [2,] 10 12
EDIT with the second definition of x:
x <- array(1:24, c(4,3,2))
lapply(as.data.frame(x), matrix, nrow = 2, ncol = 2)
# $V1
# [,1] [,2]
# [1,] 1 3
# [2,] 2 4
# $V2
# [,1] [,2]
# [1,] 5 7
# [2,] 6 8
# $V3
# [,1] [,2]
# [1,] 9 11
# [2,] 10 12
# $V4
# [,1] [,2]
# [1,] 13 15
# [2,] 14 16
# $V5
# [,1] [,2]
# [1,] 17 19
# [2,] 18 20
# $V6
# [,1] [,2]
# [1,] 21 23
# [2,] 22 24
EDIT2: a try to get an arry as result
Based on this similar question, you may try this code:
x <- array(1:24, c(4,3,2))
sapply(1:3,
function(y) sapply(1:ncol(x[, y, ]),
function(z) matrix(x[,y,z], ncol=2, nrow=2),
simplify="array"),
simplify="array")
Dimension of the result is 2 2 2 3.
Actually, the problem here is that it needs two different calls to apply when x is an array of more than 2 dimension. In the last example of the quesion (with x <- array(1:24, c(4,3,2))), we want to apply to each element of third dimension a function that apply to each element of second dimension the matrix function.
How to repeat a matrix A along 3rd dimension n times? array(rep(A,n),c(dim(A),n)) should work but is there a simpler way? I was hoping A[,,rep(1,n)] would do the trick but apparently not.
You forgot to change the dimensions first:
A = matrix(1:4, nrow = 2)
dim(A) = c(dim(A), 1)
A[, , rep(1, 3)]
#, , 1
#
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
#
#, , 2
#
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
#
#, , 3
#
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
I guess the following is a little shorter:
myArray <- replicate(n, A)
Here is an example:
temp <- matrix(1:10, 5)
replicate(3, temp)
I have
mat1 = matrix(c(2, 4, 3, 6, 7, 8), nrow=2, ncol=3)
mat2 = matrix(c(5, 6, 7, 1, 2, 3), nrow=2, ncol=3)
mat3 = matrix(c(8, 5, 8, 6, 7, 9), nrow=2, ncol=3)
which gives me 3 matrices:
[,1] [,2] [,3]
[1,] 2 3 7
[2,] 4 6 8
[,1] [,2] [,3]
[1,] 5 7 2
[2,] 6 1 3
[,1] [,2] [,3]
[1,] 8 8 7
[2,] 5 6 9
What I would like to do is compare the three matrices per row per first column, and select the row of the matrix that has the highest value on the first column.
For example: in row 1 column 1, matrix3 has the highest value (8) compared to matrix1 (2) and matrix2 (5). In row 2 column 1, matrix2 has the highest value (6). I would like to create a new matrix that copies the row of the matrix that has that highest value, resulting in:
[,1] [,2] [,3]
[1,] 8 8 7 <- From mat3
[2,] 6 1 3 <- From mat2
I know how to get a vector with the highest values from column 1, but I cannot get the whole row of the matrix copied into a new matrix. I have:
mat <- (mat1[1,])
which just copies the first row of the first matrix
[1] 2 3 7
I can select which number is the maximum number:
max(mat1[,1],mat2[,1],mat3[,1])
[1] 8
But I cannot seem to combine the two to return a matrix with the whole row.
Getting the code to loop for each row will be no problem, but I cannot seem to get it to work for the first row and as such, I am missing the essential code. Any help would be greatly appreciated. Thank you.
Are you working interactively? Do you manipulate multiple matrices spread in your workspace? A straightforward answer to your problem could be:
#which matrices have the largest element of column 1 in each row?
max.col(cbind(mat1[, 1], mat2[, 1], mat3[, 1]))
#[1] 3 2
rbind(mat3[1, ], mat2[2, ]) #use the above information to get your matrix
# [,1] [,2] [,3]
#[1,] 8 8 7
#[2,] 6 1 3
On a more ganeral use-case, a way could be:
mat_ls = list(mat1, mat2, mat3) #put your matrices in a "list"
which_col = 1 #compare column 1
which_mats = max.col(do.call(cbind, lapply(mat_ls, function(x) x[, which_col])))
do.call(rbind, lapply(seq_along(which_mats),
function(i) mat_ls[[which_mats[i]]][i, ]))
# [,1] [,2] [,3]
#[1,] 8 8 7
#[2,] 6 1 3
Probably not the prettiest solution
temp <- rbind(mat1, mat2, mat3)
rbind(temp[c(T,F),][which.max(temp[c(T,F),][, 1]),],
temp[c(F,T),][which.max(temp[c(F,T),][, 1]),])
## [,1] [,2] [,3]
## [1,] 8 8 7
## [2,] 6 1 3
You may also try:
a2 <- aperm(simplify2array( mget(ls(pattern="mat"))),c(3,2,1)) #gets all matrices with name `mat`
t(sapply(1:(dim(a2)[3]), function(i) {x1 <- a2[,,i]; x1[which.max(x1[,1]),]}))
# [,1] [,2] [,3]
#[1,] 8 8 7
#[2,] 6 1 3
I have two matrices A and B. Matrix B is a two-column matrix, each row containing one index of an element in matrix A. I want to change those elements in matrix A, which are indexed by each row in matrix B, to 0.
Is there a way to avoid using the loop shown below?
> A <- matrix(1:12, 3, 4)
> B <- matrix(c(1, 2, 2, 2, 3, 4), byrow = TRUE, ncol = 2)
> A
[,1] [,2] [,3] [,4]
[1,] 1 4 7 10
[2,] 2 5 8 11
[3,] 3 6 9 12
> B
[,1] [,2]
[1,] 1 2
[2,] 2 2
[3,] 3 4
> for (i in 1:nrow(B)) {
+ A[B[i, 1], B[i, 2]] <- 0
+ }
> A
[,1] [,2] [,3] [,4]
[1,] 1 0 7 10
[2,] 2 0 8 11
[3,] 3 6 9 0
It's done like this
A[B] <- 0;