Related
Suppose that we have a (4,4) matrix. My goal is to change iteratively that cells (1,1),(2,1),(3,1),(1,2),(2,2),(1,3)
I wrote the following
for(i in 1:3){
for(j in 1:3){
if(i>j){
A[i,j] = A[i,j] + sample(c(-1,1),prob=c(0.5,0.5))
}
}
However, it doesn't change the correct cells and misses cells that have to be changed.
The matrix A can be of the form
A = matrix(c(1,1,1,1,1,1,1,0,1,1,0,0,1,0,0,0),4,4,byrow=T)
I think that the following chunk of code might be the solution, at least it gives the correct answer for a few runs that I did.
A = matrix(c(1,1,1,0,1,1,0,0,1,0,0,0,0,0,0,0),4,4,byrow=T)
k = 0
for(i in 1:3){
for(j in 1:(3-k)){
A[i,j] = A[i,j] + sample(c(-1,1),prob=c(0.5,0.5), size = 1)
}
k = k + 1
}
I think you simple forgot to set the size= parameter of sample to get one draw of the Rademacher universe.
set.seed(42)
for (i in 1:3) {
for (j in 1:3) {
if (i > j) {
A[i, j] <- A[i, j] + sample(c(-1, 1), size=1, prob=c(0.5, 0.5))
}
}
}
A
# [,1] [,2] [,3] [,4]
# [1,] 1 1 1 1
# [2,] 0 1 1 0
# [3,] 0 2 0 0
# [4,] 1 0 0 0
Another idea is to use a permutation matrix, which you may subset to your needs, and over which you may loop.
id <- RcppAlgos::permuteGeneral(ncol(B) - 1, ncol(B) - 2, repetition=T)
(id <- id[c(1, 4, 7, 2, 5, 3), ])
# [,1] [,2]
# [1,] 1 1
# [2,] 2 1
# [3,] 3 1
# [4,] 1 2
# [5,] 2 2
# [6,] 1 3
set.seed(42)
for (i in 1:nrow(id)) {
A[id[i, 1], id[i, 2]] <- A[id[i, 1], id[i, 2]] +
sample(c(-1, 1), size=1, prob=c(0.5, 0.5))
}
A
# [,1] [,2] [,3] [,4]
# [1,] 0 0 0 1
# [2,] 0 0 1 0
# [3,] 2 1 0 0
# [4,] 1 0 0 0
We can create a row/column index (vectorized approach) by cbinding the vector of index. Use the index to subset the cells of the matrix and assign (<-) after adding the sample output to those elements
n <- 3
j1 <- rep(seq_len(n), rev(seq_len(n)))
i1 <- ave(j1, j1, FUN = seq_along)
ind <- cbind(i1, j1)
ind
# i1 j1
#[1,] 1 1
#[2,] 2 1
#[3,] 3 1
#[4,] 1 2
#[5,] 2 2
#[6,] 1 3
A[ind] <- A[ind] + sample(c(-1,1),prob=c(0.5,0.5),
size = nrow(ind), replace= TRUE)
I am trying to create the following matrix A for n rows and n+1 columns. n will likely be around 20 or 30, but for the purpose of the question I put it at 4 and 5.
Here is what I have so far:
N <- 5 # n+1
n <- 4 # n
columns <- list()
# first column:
columns[1] <- c(-1, 1, rep(0, N-2))
# all other columns:
for(i in N:2) {
columns[i] <- c((rep(0, N-i), 1, -2, 1, rep(0, i-3)))
}
# combine into matrix:
A <- cbind(columns)
I keep getting the following error msg:
In columns[1] <- c(-1, 1, rep(0, N - 2)) :
number of items to replace is not a multiple of replacement length
And later
"for(i in N:2) {
columns[i] <- c((rep(0, N-i),"
}
Error: unexpected '}' in "}"
I guess you can try the for loop below to create your matrix A:
N <- 5
n <- 4
A <- matrix(0,n,N)
for (i in 1:nrow(A)) {
if (i == 1) {
A[i,1:2] <- c(-1,1)
} else {
A[i,i+(-1:1)] <- c(1,-2,1)
}
}
such that
> A
[,1] [,2] [,3] [,4] [,5]
[1,] -1 1 0 0 0
[2,] 1 -2 1 0 0
[3,] 0 1 -2 1 0
[4,] 0 0 1 -2 1
Another solution is to use outer, and this method would be faster and looks more compact than the for loop approach, i.e.,
A <- `diag<-`(replace(z<-abs(outer(1:n,1:N,"-")),!z %in% c(0,1),0),
c(-1,rep(-2,length(diag(z))-1)))
I thought this would be fast compared to the loop, but when I tested on a 5000x5001 example, the loop in ThomasIsCoding's answer was about 5x faster. Go with that one!
N = 5
n = N - 1
A = matrix(0, nrow = n, ncol = N)
delta = row(A) - col(A)
diag(A) = -2
A[delta %in% c(1, -1)] = 1
A[1, 1] = -1
A
# [,1] [,2] [,3] [,4] [,5]
# [1,] -1 1 0 0 0
# [2,] 1 -2 1 0 0
# [3,] 0 1 -2 1 0
# [4,] 0 0 1 -2 1
You could use data.table::shift to shift the vector c(1, -2, 1, 0) by all increments from -1 (backwards shift / lead by 1) to n - 1 (forward shift / lagged by n - 1) and then cbind all the shifted outputs together. The first-row first-column element doesn't follow this pattern so that's fixed at the end.
library(data.table)
out <- do.call(cbind, shift(c(1, -2, 1, 0), seq(-1, n - 1), fill = 0))
out[1, 1] <- -1
out
# [,1] [,2] [,3] [,4] [,5]
# [1,] -1 1 0 0 0
# [2,] 1 -2 1 0 0
# [3,] 0 1 -2 1 0
# [4,] 0 0 1 -2 1
I would like to write one function whose input is a square matrix, and it returns a square matrix whose numbers from the upper right corner down to lower left corner are preserved and other numbers are zero.
For example
suppose A is a 4*4 matrix in the following.(sorry I do not know how to type the matrix expression)
[1,2,3,4]
[5,6,7,8]
[9,10,11,12]
[13,14,15,16]
How can I write a function in R without any loops to transform the matrix into this?
[0,0,0,4]
[0,0,7,0]
[0,10,0,0]
[13,0,0,0]
This feels like a gymnastics exercise...
xy <- matrix(1:16, ncol = 4, byrow = TRUE)
xy <- apply(xy, MARGIN = 1, rev)
xy[lower.tri(xy)] <- 0
xy[upper.tri(xy)] <- 0
t(apply(xy, MARGIN = 1, rev))
[,1] [,2] [,3] [,4]
[1,] 0 0 0 4
[2,] 0 0 7 0
[3,] 0 10 0 0
[4,] 13 0 0 0
Here is another option.
mat <- matrix(1:16, 4, byrow = TRUE)
idx <- cbind(seq_len(nrow(mat)),
ncol(mat):1)
values <- mat[idx]
mat <- matrix(0, nrow = dim(mat)[1], ncol = dim(mat)[2])
mat[idx] <- values
mat
# [,1] [,2] [,3] [,4]
#[1,] 0 0 0 4
#[2,] 0 0 7 0
#[3,] 0 10 0 0
#[4,] 13 0 0 0
A non-apply solution using some maths to generate the indices stealing xy from #Roman
xy <- matrix(1:16, ncol = 4, byrow = TRUE)
ind <- nrow(xy)
xy[setdiff(1:length(xy), seq(ind, by = ind -1, length.out = ind))] <- 0
xy
# [,1] [,2] [,3] [,4]
#[1,] 0 0 0 4
#[2,] 0 0 7 0
#[3,] 0 10 0 0
#[4,] 13 0 0 0
Trying it on 5 X 5 matrix
xy <- matrix(1:25, 5, byrow = TRUE)
ind <- nrow(xy)
xy[setdiff(1:length(xy), seq(ind, by = ind -1, length.out = ind))] <- 0
xy
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0 0 0 0 5
#[2,] 0 0 0 9 0
#[3,] 0 0 13 0 0
#[4,] 0 17 0 0 0
#[5,] 21 0 0 0 0
This answer takes a slightly different approach than the other answers. Instead of trying to zero out everything except for the diagonal, we can just build the diagonal by itself:
m <- matrix(rep(0,16), nrow = 4, byrow = TRUE)
for (i in 0:15) {
row <- floor(i / 4)
col <- i %% 4
if (i == 3 + (row*3)) {
m[row+1, col+1] <- i+1
}
}
m
[,1] [,2] [,3] [,4]
[1,] 0 0 0 4
[2,] 0 0 7 0
[3,] 0 10 0 0
[4,] 13 0 0 0
I just thought about a way to reverse the original diag function from base R.
You can see it by just typing diag in the console.
Here the highlighted change I made in my diag_reverse:
y <- x[((m - 1L):0L * (dim(x)[1L])) + (1L:m)] # m is min(dim(x))
And here's the complete function (I kept all the code except that one line):
diag_reverse <- function (x = 1, nrow, ncol, names = TRUE)
{
if (is.matrix(x)) {
if (nargs() > 1L && (nargs() > 2L || any(names(match.call()) %in%
c("nrow", "ncol"))))
stop("'nrow' or 'ncol' cannot be specified when 'x' is a matrix")
if ((m <- min(dim(x))) == 0L)
return(vector(typeof(x), 0L))
y <- x[((m - 1L):0L * (dim(x)[1L])) + (1L:m)] # HERE I made the change
if (names) {
nms <- dimnames(x)
if (is.list(nms) && !any(vapply(nms, is.null, NA)) &&
identical((nm <- nms[[1L]][seq_len(m)]), nms[[2L]][seq_len(m)]))
names(y) <- nm
}
return(y)
}
if (is.array(x) && length(dim(x)) != 1L)
stop("'x' is an array, but not one-dimensional.")
if (missing(x))
n <- nrow
else if (length(x) == 1L && nargs() == 1L) {
n <- as.integer(x)
x <- 1
}
else n <- length(x)
if (!missing(nrow))
n <- nrow
if (missing(ncol))
ncol <- n
.Internal(diag(x, n, ncol))
}
Then we can call it:
m <- matrix(1:16,nrow=4,ncol=4,byrow = T)
diag_reverse(m)
#[1] 4 7 10 13
I'll test it on other matrices to see if it gives always the correct answer.
The apply family are really just loops with a bow tie.
Here is a way to do it without apply. With some input checking and should work on any size matrix.
off_diag = function(X)
{
if(!is.matrix(X)) stop('Argument is not a matrix')
n <- nrow(X)
if(ncol(X) != n) stop('Matrix is not square')
if(n<2) return(X)
Y <- X * c(0,rep(rep(c(0,1),c(n-2,1)),n),rep(0,n-1))
return(Y)
}
Now it can handle numeric vectors, character vectors and NAs.
mat <- matrix(1:16, 4, byrow = TRUE)
off_diag(mat)
# [,1] [,2] [,3] [,4]
# [1,] 0 0 0 4
# [2,] 0 0 7 0
# [3,] 0 10 0 0
# [4,] 13 0 0 0
Edit: improvement
I realised my function will fail if there are NAs since NA*0 is NA, additionally it will not work on characters, but doesn't check the matrix has mode as numeric. So instead I use the same setup to make a logical vector
minor_diag = function(X)
{
if(!is.matrix(X)) stop('Argument is not a matrix')
n <- nrow(X)
if(ncol(X) != n) stop('Matrix is not square')
if(n<2) return(X)
index = c(TRUE,rep(rep(c(TRUE,FALSE),c(n-2,1)),n),rep(TRUE,n-1))
X[index]=0
return(X)
}
mat <- matrix(letters[1:16], 4, byrow = TRUE)
minor_diag(mat)
## [,1] [,2] [,3] [,4]
## [1,] "0" "0" "0" "d"
## [2,] "0" "0" "g" "0"
## [3,] "0" "j" "0" "0"
## [4,] "m" "0" "0" "0"
minor_diag(matrix(NA,2,2))
## [,1] [,2]
## [1,] 0 NA
## [2,] NA 0
A one liner without loops
#setup
n <- 5
A <- matrix(1:(n^2), n)
#solution
diag(diag(A[n:1,]))[n:1,]
I would like to transform an object like this
df <- data.frame(ROW = c(1,3),COLUMN =c(2,3),VALUE = c(10,20))
df
ROW COLUMN VALUE
1 2 10
3 3 20
to a matrix like this
m <-matrix(c(0,10,0,0,0,0,0,0,20),ncol = 3,nrow = 3)
m
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 10 0 0
[3,] 0 0 20
I know that the data frame represents a sparse matrix but I did not find any other question that relates exactly to what I am looking for.
We can use sparseMatrix
library(Matrix)
as.matrix( sparseMatrix(i = df$COLUMN, j= df$ROW, x= df$VALUE))
# [,1] [,2] [,3]
#[1,] 0 0 0
#[2,] 10 0 0
#[3,] 0 0 20
Or create a matrix of 0's and then assign
m1 <- matrix(0, 3, 3)
m1[as.matrix(df[2:1])] <- df$VALUE
Note: Based on the output, the ROW/COLUMN index seems to be reversed
We can iterate over the rows of df and fill a matrix according to the row and column indexes included in df:
# initialize
new_mat <- matrix(0, ncol = max(df$ROW), nrow = max(df$COLUMN))
for(i in 1:nrow(df)){
new_mat[df[i,]$COLUMN, df[i,]$ROW] <- df[i,]$VALUE
}
# [,1] [,2] [,3]
# [1,] 0 0 0
# [2,] 10 0 0
# [3,] 0 0 20
as #akrun has noted, it seems as though the row and column indexes are flipped.
Here is a solution with apply
mat <- matrix( 0, ncol = max(df$ROW), nrow = max(df$COLUMN) )
f <- function( x ) { mat[x[1],x[2]] <<- x[3] }
apply( df, 1, f } )
The <<- operator applies the value to the externally defined mat.
May need to use as.numeric etc. for data type conversions.
Apologize for the title. Here is my problem.
I need to generate a L-length binary code permutation.
For L = 2, its like:
00
01
10
11
My idea is to divide the problem into two part:
write a function f that transform an integer to its binary form
loop 1:2^L
Here is my function for step 1:
kbitpermute <- function(input = 2^16 - 1, k = 16){
ret = matrix(0, ncol = k, nrow = 1)
for(i in 1:k){
ret[1,k-i+1] = input & 1
input = bitwShiftR(input, 1)
}
return(ret)
}
And then I use lapply() to obtain the permutation, like:
temp3 = lapply(1:2^2 - 1, kbitpermute, k = 2)
temp3
[[1]]
[,1] [,2]
[1,] 0 0
[[2]]
[,1] [,2]
[1,] 0 1
[[3]]
[,1] [,2]
[1,] 1 1
[[4]]
[,1] [,2]
[1,] 1 1
It seems works, but how to transform the output of lapply to its matrix form? And, is there any easy way to do the whole task?
=== update ===
I find a way to use unlist to do the transformation, like
temp3 = t(matrix(unlist(lapply(1:2^2 - 1, kbitpermute, k = 2)), nrow = 2, ncol=2^2)).
But is there any easy way?
=== update ===
right version of kbitpermute, and special thanks to #Batanichek
kbitpermute <- function(input = 2^16 - 1, k = 16){
ret = matrix(0, ncol = k, nrow = 1)
for(i in 1:k){
ret[1,k-i+1] = bitwAnd(input, 1)
input = bitwShiftR(input, 1)
}
return(ret)
}
> t(sapply(1:2^2 - 1, kbitpermute, k = 2))
[,1] [,2]
[1,] 0 0
[2,] 0 1
[3,] 1 0
[4,] 1 1
Try something like ( from here)
test=function(x,k){
a=rev(as.integer(intToBits(x))[1:k])
return(a)
}
x=2
t(sapply(1:2^x - 1, test,k=x))
# [,1] [,2]
#[1,] 0 0
#[2,] 0 1
#[3,] 1 0
#[4,] 1 1
Time comparison
x=15
system.time(t(sapply(1:2^x - 1, test,k=x)))
#пользователь система прошло
# 0.62 0.21 1.35
system.time(t(sapply(1:2^x - 1, kbitpermute,k=x)))
#пользователь система прошло
# 1.84 0.35 2.48
(P.S.have not english local )