Processing the matrix data by using which(), and return all the row names-CodePudding

I would like to process the following matrix data by using which(), and return all the row names, for example:

m = matrix(seq(-1,1, 0.5), nrow = 3)

#     [,1] [,2]
#[1,] -1.0  0.5
#[2,] -0.5  1.0
#[3,]  0.0 -1.0

which(m==0.5,arr.ind=TRUE)
#       row col
# [1,]   1   2

How can I get the matrix like this? All row names can be shown in the table, and the missing value in col is NA.

#       row col
# [1,]   1   2
# [2,]   2   NA
# [3,]   3   NA
# [4,]   4   NA

CodePudding user response：

Here is a method that first change the matrix into a dataframe, and use tidyr::complete() to "expand" the dataframe based on the number of rows of m. Finally change it back to a matrix.

library(tidyverse)

as.data.frame(which(m==0.5,arr.ind=TRUE)) %>% 
  complete(row = 1:nrow(m)) %>% 
  as.matrix()

    row col
[1,]   1   2
[2,]   2  NA
[3,]   3  NA

CodePudding user response：

Using only base R, you could build a function. Please find below a reprex:

Reprex

Your data

m = matrix(seq(-1,1, 0.5), nrow = 3)

Code of the function

indexfunct <- function(x, value){
  res <- matrix(NA, nrow = dim(x)[1], ncol = dim(x)[2], dimnames = (list(seq(dim(x)[1]), c("row", "col"))))
  res[,1] <- seq(dim(x)[1])
  res[which(x==value,arr.ind=TRUE)[,1],2] <- which(x==value,arr.ind=TRUE)[,2]
  return(res)
}

Output of the function

indexfunct(m, value = 0.5)
#>   row col
#> 1   1   2
#> 2   2  NA
#> 3   3  NA

^{Created on 2022-03-06 by the reprex package (v2.0.1)}

CodePudding user response：

Another possible base R solution:

m = matrix(seq(-1,1, 0.5), nrow = 3)

rbind(which(m == 0.5, arr.ind = TRUE), 
      cbind(setdiff(1:nrow(m), which(m == 0.5, arr.ind = TRUE)[,"row"]), NA))

#>      row col
#> [1,]   1   2
#> [2,]   2  NA
#> [3,]   3  NA

In case a function is needed:

getindexes <- function(mat, value)
{
  rbind(which(mat == value, arr.ind = TRUE), 
        cbind(setdiff(1:nrow(mat), which(mat == value,arr.ind = TRUE)[,"row"]), NA))
}  

getindexes(m, 0.5)

#>      row col
#> [1,]   1   2
#> [2,]   2  NA
#> [3,]   3  NA

CodePudding user response：

We indeed can try this base R option without which

> v <- rowSums(col(m) * (m == 0.5))

> cbind(row = 1:nrow(m), col = v * NA^(!v))
     row col
[1,]   1   2
[2,]   2  NA
[3,]   3  NA