I have some data
data <- diag(5)
I want to use the [<- operator to change a line.
The result should be:
data[1,] <- 2
> data
[,1] [,2] [,3] [,4] [,5]
[1,] 2 2 2 2 2
[2,] 0 1 0 0 0
[3,] 0 0 1 0 0
[4,] 0 0 0 1 0
[5,] 0 0 0 0 1
I know I can do e.g.
`[<-`(data, i=1, j=3, 2)
which gives
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 8 0 0
[2,] 0 1 0 0 0
[3,] 0 0 1 0 0
[4,] 0 0 0 1 0
[5,] 0 0 0 0 1
but how can I operate on line (or column, same issue)?
I tried j=NULL, j=integer(0), it doesn't work. I could do j=1:5 and get what I want but I am wondering how to mimic data[1,] <- 2 and not data[1,1:5] <- 2.
> `[<-`(data, 1, , 2) # blank 2nd argument
[,1] [,2] [,3] [,4] [,5]
[1,] 2 2 2 2 2
[2,] 0 1 0 0 0
[3,] 0 0 1 0 0
[4,] 0 0 0 1 0
[5,] 0 0 0 0 1
You can use ncol to ensure that all columns are set:
`[<-`(data, i = 1, j = 1:ncol(data), 2)
Related
Suppose I have a matrix, mat. Suppose further that the sum of one row of this matrix is equal to zero. Then, I need to set all the coming rows (the rows after the zero row) to zero. For example,
mat <- c(1,2,0,0,0,
3,4,0,2,1,
0,0,0,1,0,
1,2,0,0,0,
0,1,0,1,0)
mat <- matrix(mat,5,5)
mat
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 1 0
[2,] 2 4 0 2 1
[3,] 0 0 0 0 0
[4,] 0 2 1 0 1
[5,] 0 1 0 0 0
All the entries of row 3 are zero. Hence, I want rows 4, and 5 to become zeros as well. I have a list of matrices and would like to apply the same to all the matrices using the lapply function. For simplicity, I make a list of 3 matrices similar to the mat.
mat <- c(1,2,0,0,0,
3,3,0,2,1,
0,0,0,4,0,
1,3,0,0,0,
0,1,0,1,0)
mat <- matrix(mat,5,5)
mat1 <- c(1,2,0,0,0,
3,4,0,2,1,
0,0,0,1,0,
1,2,0,0,0,
0,1,0,1,0)
mat1 <- matrix(mat1,5,5)
mat2 <- c(1,2,0,0,0,
3,4,0,2,1,
0,0,0,2,0,
1,2,0,0,0,
0,2,0,3,0)
mat2 <- matrix(mat2,5,5)
Mat <- list(mat1, mat2, mat3)
You did not actually post mat3 in your data so I just used mat3 <- matrix(1, 5, 5), i.e. a 5x5 matrix of ones. This was to ensure it could handle cases where there is no row where all values are zero.
This will return a list of matrices where all rows are zero after the first row of zeroes:
lapply(Mat, \(mat) {
first_zero_row <- which(rowSums(mat)==0)[1]
if(!is.na(first_zero_row)) {
mat[first_zero_row:nrow(mat),] <- 0
}
mat
})
Output:
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 1 0
[2,] 2 4 0 2 1
[3,] 0 0 0 0 0
[4,] 0 0 0 0 0
[5,] 0 0 0 0 0
[[2]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 1 0
[2,] 2 4 0 2 2
[3,] 0 0 0 0 0
[4,] 0 0 0 0 0
[5,] 0 0 0 0 0
[[3]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 1 1
[2,] 1 1 1 1 1
[3,] 1 1 1 1 1
[4,] 1 1 1 1 1
[5,] 1 1 1 1 1
Another option could be:
lapply(Mat, function(x) {x[cumsum(rowSums(x != 0) == 0) != 0, ] <- 0; x})
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 1 0
[2,] 2 3 0 3 1
[3,] 0 0 0 0 0
[4,] 0 0 0 0 0
[5,] 0 0 0 0 0
[[2]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 1 0
[2,] 2 4 0 2 1
[3,] 0 0 0 0 0
[4,] 0 0 0 0 0
[5,] 0 0 0 0 0
[[3]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 3 0 1 0
[2,] 2 4 0 2 2
[3,] 0 0 0 1 0
[4,] 0 2 2 0 3
[5,] 0 1 0 0 0
I have a column vector with dimensions 4000x1, and I need to make a matrix with that vector, but the matrix needs to have the column vector as a diagonal and the other numbers as zero. Something like this:
Column Vector
> vector <- matrix(c(1:5), ncol=1, nrow=5)
> vector
[,1]
[1,] 1
[2,] 2
[3,] 3
[4,] 4
Matrix
[,1] [,2] [,3] [,4]
a 1 0 0 0
b 0 2 0 0
c 0 0 3 0
How can I produce this output?
This sounds like the diag() function, e.g.,
> my_vect <- 1:5
> diag(my_vect)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 2 0 0 0
[3,] 0 0 3 0 0
[4,] 0 0 0 4 0
[5,] 0 0 0 0 5
By the way, as you have written it vector is actually a 5x1 matrix, so you would need to convert it to, well, a vector:
> diag(as.vector(vector))
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 2 0 0 0
[3,] 0 0 3 0 0
[4,] 0 0 0 4 0
[5,] 0 0 0 0 5
I am not sure if this is possible, but I thought I might post it anyway. Currently I am trying to write a code that will run a nested sort. The problem is, I don't know how many nested loops will need to run, it can vary from 2-7 sorting criteria.
Each loop of the function creates a data set that the lower loops will then use. I know this needs to be done using recursion, but I am having an extremely hard time getting this to work. Any help would be appreciated!
library(abind)
re <- c(5,5,5)
answer = matrix(0,5,5)
for(a in 1:4){
answer <- abind(answer,matrix(0,5,5),along=3)
}
for( i in 1:re[1]){
first <- c(1:re[1])
for(j in 1:re[2]){
if(j %in% first == 1){
second = j
}
print(second)
for(k in 1:re[3]){
if(k == second){
answer[k,j,i] <- k
}
}
}
}
answer
Output
answer
, , 1
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 2 0 0 0
[3,] 0 0 3 0 0
[4,] 0 0 0 4 0
[5,] 0 0 0 0 5
, , 2
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 2 0 0 0
[3,] 0 0 3 0 0
[4,] 0 0 0 4 0
[5,] 0 0 0 0 5
, , 3
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 2 0 0 0
[3,] 0 0 3 0 0
[4,] 0 0 0 4 0
[5,] 0 0 0 0 5
, , 4
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 2 0 0 0
[3,] 0 0 3 0 0
[4,] 0 0 0 4 0
[5,] 0 0 0 0 5
, , 5
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 2 0 0 0
[3,] 0 0 3 0 0
[4,] 0 0 0 4 0
[5,] 0 0 0 0 5
I'm looking to generate all possible 4x4 matrices, where each element can either be a 0 or a 1.
Is there a function in R to do this?
Here is a function that would create such matrices for indices from 0 to 2^16-1:
num2mat = function(num){ matrix(as.integer(intToBits(num)),4,4) }
Here is what it produces:
> num2mat(0)
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 0 0 0 0
[4,] 0 0 0 0
> num2mat(2^15+2^13+2^10+2^8+2^7+2^5+2^2+1)
[,1] [,2] [,3] [,4]
[1,] 1 0 1 0
[2,] 0 1 0 1
[3,] 1 0 1 0
[4,] 0 1 0 1
> num2mat(2^16-1)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 1
[2,] 1 1 1 1
[3,] 1 1 1 1
[4,] 1 1 1 1
I have number of strings in an idiosyncratic format, representing sets. In R, I'd like to convert them into a similarity matrix.
For example, a string showing that 1+2 comprise a set, 3 is alone in a set, and 4,5, and 6 comprise a set is:
"1+2,3,4+5+6"
For the example above, I'd like to be able to produce
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 0 0 0 0
[2,] 1 1 0 0 0 0
[3,] 0 0 1 0 0 0
[4,] 0 0 0 1 1 1
[5,] 0 0 0 1 1 1
[6,] 0 0 0 1 1 1
It seems like this should be a painfully simple task. How would I go about it?
Here's an approach:
out <- lapply(unlist(strsplit("1+2,3,4+5+6", ",")), function(x) {
as.numeric(unlist(strsplit(x, "\\+")))
})
x <- table(unlist(out), rep(seq_along(out), sapply(out, length)))
matrix(x %*% t(x), nrow(x))
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 1 0 0 0 0
## [2,] 1 1 0 0 0 0
## [3,] 0 0 1 0 0 0
## [4,] 0 0 0 1 1 1
## [5,] 0 0 0 1 1 1
## [6,] 0 0 0 1 1 1
Pseudocode:
Split at , to get an array of strings, each describing a set.
For each element of the array:
Split at + to get an array of set members
Mark every possible pairing of members of this set on the matrix
You can create a matrix in R with:
m = mat.or.vec(6, 6)
By default, the matrix should initialize with all entries 0. You can assign new values with:
m[2,3] = 1
Here's another approach:
# write a simple function
similarity <- function(string){
sets <- gsub("\\+", ":", strsplit(string, ",")[[1]])
n <- as.numeric(tail(strsplit(gsub("[[:punct:]]", "", string), "")[[1]], 1))
mat <- mat.or.vec(n, n)
ind <- suppressWarnings(lapply(sets, function(x) eval(parse(text=x))))
for(i in 1:length(ind)){
mat[ind[[i]], ind[[i]]] <- 1
}
return(mat)
}
# Use that function
> similarity("1+2,3,4+5+6")
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 0 0 0 0
[2,] 1 1 0 0 0 0
[3,] 0 0 1 0 0 0
[4,] 0 0 0 1 1 1
[5,] 0 0 0 1 1 1
[6,] 0 0 0 1 1 1
# Using other string
> similarity("1+2,3,5+6+7, 8")
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 1 0 0 0 0 0 0
[2,] 1 1 0 0 0 0 0 0
[3,] 0 0 1 0 0 0 0 0
[4,] 0 0 0 0 0 0 0 0
[5,] 0 0 0 0 1 1 1 0
[6,] 0 0 0 0 1 1 1 0
[7,] 0 0 0 0 1 1 1 0
[8,] 0 0 0 0 0 0 0 1