## Permute the position of a subset of a vector - r

I want to permute a subset of a vector.
For example, say I have a vector (x) and I select a random subset of the vector (e.g., 40% of its values).
What I want to do is output a new vector (x2) that is identical to (x) except the positions of the values within the random subset are randomly swapped.
For example:
x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
random subset = 1, 4, 5, 8
x2 could be = 4, 2, 3, 8, 1, 6, 7, 5, 9, 10
Here's an an example vector (x) and how I'd select the indices of a random subset of 40% of its values. Any help making (x2) would be appreciated!
x <- seq(1,10,1)
which(x%in%sample(x)[seq_len(length(x)*0.40)])

First draw a sample of proportion p from the indices, then sample and re-assign elements with that indices.
f <- \(x, p=0.4) {
r <- sample(seq_along(x), length(x)*p)
x[r] <- sample(x[r])
`attr<-`(x, 'subs', r) ## add attribute w/ indices that were sampled
}
set.seed(42)
f(x)
# [1] 8 2 3 4 1 5 7 10 6 9
# attr(,"subs")
# [1] 1 5 10 8
Data:
x <- 1:10

For sure there is a faster code to do what you are asking, but, a solution would be:
x <- seq(1,10,1)
y <- which(x%in%sample(x)[seq_len(length(x)*0.40)]) # Defined as "y" the vector of the random subset
# required libraries
library(combinat)
permutation <- permn(y) # permn() function in R generates a list of all permutations of the elements of x.
# https://www.geeksforgeeks.org/calculate-combinations-and-permutations-in-r/
permutation_sampled <- sample(permutation,1) # Sample one of the permutations.
x[y] <- permutation_sampled[[1]] # Substitute the selected permutation in x using y as the index of the elements that should be substituted.

## Related

### finding values in a range in r and sum the number of values

I have a question I have the following data c(1, 2, 4, 5, 1, 8, 9) I set a l = 2 and an u = 6 I want to find all the values in the range (3,7) How can I do this?

In base R we can use comparison operators to create a logical vector and use that for subsetting the original vector x[x > 2 & x <= 6] #[1] 3 5 6 Or using a for loop, initialize an empty vector, loop through the elements of 'x', if the value is between 2 and 6, then concatenate that value to the empty vector v1 <- c() for(i in x) { if(i > 2 & i <= 6) v1 <- c(v1, i) } v1 #[1] 3 5 6 data x <- c(3, 5, 6, 8, 1, 2, 1)

### How to make a generalized function update the value of a vector?

I have been trying to write a generalized function that multiplies each value in each row of a matrix by the corresponding value of a vector in terms of their position (i.e. matrix[1,1]*vector[1], matrix[1,2]*vector[2], etc) and then sum them together. It is important to note that the lengths of the vector and the rows of the matrix are always the same, which means that in each row the first value of the vector is multiplied with the first value of the matrix row. Also important to note, I think, is that the rows and columns of the matrix are of equal length. The end sum for each row should be assigned to different existing vector, the length of which is equal to the number of rows. This is the matrix and vector: a <- c(4, -9, 2, -1) b <- c(-1, 3, -8, 2) c <- c(5, 2, 6, 3) d <- c(7, 9, -2, 5) matrix <- cbind(a,b,c,d) a b c d [1,] 4 -1 5 7 [2,] -9 3 2 9 [3,] 2 -8 6 -2 [4,] -1 2 3 5 vector <- c(1, 2, 3, 4) These are the basic functions that I have to generalize for the rows and columns of matrix and a vector of lenghts "n": f.1 <- function() { (matrix[1,1]*vector[1] + matrix[1,2]*vector[2] + matrix[1,3]*vector[3] + matrix[1,4]*vector[4]) } f.2 <- function() { (matrix[2,1]*vector[1] + matrix[2,2]*vector[2] + matrix[2,3]*vector[3] + matrix[2,4]*vector[4]) } and so on... This is the function I have written: ncells = 4 f = function(x) { i = x result = 0 for(j in 1:ncells) { result = result + vector[j] * matrix[i][j] } return(result) } Calling the function: result.cell = function() { for(i in 1:ncells) { new.vector[i] = f(i) } } The vector to which this result should be assigned (i.e. new.vector) has been defined beforehand: new.vector <- c() I expected that the end sum for each row will be assigned to the vector in a corresponding manner (e.g. if the sums for all rows were 1, 2, 3, 4, etc. then new.vector(1, 2, 3, 4, etc) but it did not happen. (Edit) When I do this with the basic functions, the assignment works: new.vector[1] <- f.1() new.vector[2] <- f.2() This does not however work with the generalized function: new.vector[1:ncells] <- result cell[1:ncells] (End Edit) I have also tried setting the length for the the new.vector to be equal to ncells but I don't think it did any good: length(new.vector) = ncells My question is how can I make the new vector take the resulting sums of the multiplied elements of a row of a matrix by the corresponding value of a vector. I hope I have been clear and thanks in advance!

There is no need for a loop here, we can use R's power of matrix multiplication and then sum the rows with rowSums. Note that m and v are used as names for matrix and vector to avoid conflict with those function names. nr <- nrow(m) rowSums(m * matrix(rep(v, nr), nr, byrow = TRUE)) # [1] 45 39 -4 32 However, if the vector v is always going to be the column number, we can simply use the col function as our multiplier. rowSums(m * col(m)) # [1] 45 39 -4 32 Data: a <- c(4, -9, 2, -1) b <- c(-1, 3, -8, 2) c <- c(5, 2, 6, 3) d <- c(7, 9, -2, 5) m <- cbind(a, b, c, d) v <- 1:4

### Finding n tuples from a list whose aggregation satisfies a condition

I have a list of two-element vectors. From this list, I'd like to find n vectors (not necessarily distinct) (x,y), such that the sum of the ys of these vectors is larger or equal to a number k. If multiple vectors satisfy this condition, select the one where the sum of the xs is the smallest. For example, I'd like to find n=2 vectors (x1,y1) and (x2,y2) such that y1+y2 >= k. If there are more than just one which satisfies this condition, select the one where x1+x2 is the smallest. I've so far only managed to set-up the following code: X <- c(3, 2, 3, 8, 7, 7, 13, 11, 12, 12) Y <- c(2, 1, 3, 6, 5, 6, 8, 9, 10, 9) df <- data.frame(A, B) l <- list() for (i in seq(1:nrow(df))){ n <- as.numeric(df[i,]) l[[i]] <- n } Using the values above, let's say n=1, k=9, then I'll pick the tuple (x,y)=(11,9) because even though (12,9) also matches the condition that y=k, the x is smaller. If n=2, k=6, then I'll pick (x1,y1)=(3,3) and (x2,y2)=(3,3) because it's the smallest x1+x2 that satisfies y1+y2 >= 6. If n=2, k=8, then I'll pick (x1,y1)=(3,3) and (x2,y2)=(7,5) because y1+y2>=8 and in the next alternative tuples (3,3) and (8,6), 3+8=11 is larger than 3+7. I feel like a brute-force solution would be possible: all possible n-sized combinations of each vector with the rest, for each permutation calculate yTotal=y1+y2+y3... find all yTotal combinations that satisfy yTotal>=k and of those, pick the one where xTotal=x1+x2+x3... is minimal. I definitely struggle putting this into R-code and wonder if it's even the right choice. Thank you for your help!

First, it seems from your question that you allow to select from Y with replacement. The code basically does your brute-force approach: use the permutations in the gtools library to generate the permutations. Then basically do the filtering for sum(Y)>=k, and ordering first by smallest sum(Y) and then sum(X). X <- c(3, 2, 3, 8, 7, 7, 13, 11, 12, 12) Y <- c(2, 1, 3, 6, 5, 6, 8, 9, 10, 9) n<-1 perm<-gtools::permutations(n=length(Y),r=n, repeats.allowed=T) result<-apply(perm,1,function(x){ c(sum(Y[x]),sum(X[x])) }) dim(result) # 2 10 k=9 ## Case of n=1, k=9 keep<-which(result[1,]>=k) result[,keep[order(result[1,keep],result[2,keep])[1]]] # 9 and 11 ##### n=2 cases ########## n<-2 perm<-gtools::permutations(n=length(Y),r=n, repeats.allowed=T) result<-apply(perm,1,function(x){ c(sum(Y[x]),sum(X[x])) }) dim(result) # 2 100 ## n=2, k=6 keep<-which(result[1,]>=6) keep[order(result[1,keep],result[2,keep])[1]] # the 23 permutation perm[23,] # 3 3 is (Y1,Y2) result[,keep[order(result[1,keep],result[2,keep])[1]]] # sum(Y)=6 and sum(X)=6 ## n=2, k=8 keep<-which(result[1,]>=8) keep[order(result[1,keep],result[2,keep])[1]] # the 6 permutation perm[6,] # 1 6 is (Y1,Y2) result[,keep[order(result[1,keep],result[2,keep])[1]]] # sum(Y)=8 and sum(X)=10

### Variable sample upper value in R

I have the following matrix m <- matrix(c(2, 4, 3, 5, 1, 5, 7, 9, 3, 7), nrow=5, ncol=2,) colnames(x) = c("Y","Z") m <-data.frame(m) I am trying to create a random number in each row where the upper limit is a number based on a variable value (in this case 1*Y based on each row's value for for Z) I currently have: samp<-function(x){ sample(0:1,1,replace = TRUE)} x$randoms <- apply(m,1,samp) which work works well applying the sample function independently to each row, but I always get an error when I try to alter the x in sample. I thought I could do something like this: samp<-function(x){ sample(0:m$Z,1,replace = TRUE)} x$randoms <- apply(m,1,samp) but I guess that was wishful thinking. Ultimately I want the result: Y Z randoms 2 5 4 4 7 7 3 9 3 5 3 1 1 7 6 Any ideas?

The following will sample from 0 to x$Y for each row, and store the result in randoms: x$randoms <- sapply(x$Y + 1, sample, 1) - 1 Explanation: The sapply takes each value in x$Y separately (let's call this y), and calls sample(y + 1, 1) on it. Note that (e.g.) sample(y+1, 1) will sample 1 random integer from the range 1:(y+1). Since you want a number from 0 to y rather than 1 to y + 1, we subtract 1 at the end. Also, just pointing out - no need for replace=T here because you are only sampling one value anyway, so it doesn't matter whether it gets replaced or not.

Based on #mathematical.coffee suggestion and my edited example this is the slick final result: m <- matrix(c(2, 4, 3, 5, 1, 5, 7, 9, 3, 7), nrow=5, ncol=2,) colnames(m) = c("Y","Z") m <-data.frame(m) samp<-function(x){ sample(Z + 1, 1)} m$randoms <- sapply(m$Z + 1, sample, 1) - 1

### Complement of empty index vector is empty index vector

I am removing values from a vector by using - (minus sign) in front of the index vector. Like this: scores <- scores[-indexes.to.delete] Sometimes indexes.to.delete vector is empty, that is N/A. So the scores vector should then remain unchanged. However, I am getting empty scores vector when indexes.to.delete is empty. Example: x <- c(1, 2, 3); y <- c(4, 5, 6); indexes.to.delete <- which(y < x); # will return empty vector y <- y[-indexes.to.delete]; # returns empty y vector, but I want y stay untouched I could code an if statement checking whether indexes.to.delete is empty, but I am wondering if there is a simpler way?

Maybe use; x <- c(1, 2, 3) y <- c(4, 5, 6) y[!y<x] > y[!y<x] [1] 4 5 6 x <- c(1, 2, 3) y <- c(4, 1, 6) > y[!y<x] [1] 4 6 >