[SOLVED] populating a matrix by rows

populating a matrix by rows

I would like to populate a matrix by row based on a function which gives the seq() of a number seq(1) 1 seq(2) 1 2 and so on

    matrixinp = matrix(data=NA, nrow=6, ncol=6) 

> print(matrixinp) 
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]   NA   NA   NA   NA   NA   NA
[2,]   NA   NA   NA   NA   NA   NA
[3,]   NA   NA   NA   NA   NA   NA
[4,]   NA   NA   NA   NA   NA   NA
[5,]   NA   NA   NA   NA   NA   NA
[6,]   NA   NA   NA   NA   NA   NA
    
    # display matrix 
    print(matrixinp) 
    
    # fill the elements with some  
    # 90 in a matrix 
    for (i in 1:6){
      aaa<-seq(i)
      print(aaa)
      for(j in 1:6){ 
        matrixinp[j,] = aaa
      } 
    }

This gave me this:

> print(matrixinp)
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    2    3    1    2    3
[2,]    1    2    3    1    2    3
[3,]    1    2    3    1    2    3
[4,]    1    2    3    1    2    3
[5,]    1    2    3    1    2    3
[6,]    1    2    3    1    2    3

but I would like this:

> print(matrixinp)
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    NA   NA   NA   NA   NA
[2,]    1    2    NA   NA   NA   NA
[3,]    1    2    3    NA   NA   NA
[4,]    1    2    3    4    NA   NA
[5,]    1    2    3    4    5    NA
[6,]    1    2    3    4    5    6

Solution

I bet a one-liner can solve this, R is a vectorized language.

matrixinp <- matrix(data=NA, nrow=6, ncol=6)
matrixinp[lower.tri(matrixinp, diag = TRUE)] <- rep(1:6, 6:1)

matrixinp
#>      [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,]    1   NA   NA   NA   NA   NA
#> [2,]    1    2   NA   NA   NA   NA
#> [3,]    1    2    3   NA   NA   NA
#> [4,]    1    2    3    4   NA   NA
#> [5,]    1    2    3    4    5   NA
#> [6,]    1    2    3    4    5    6

^{Created on 2024-09-17 with reprex v2.1.0}

Wait, I won the bet.
The trick is to know that R's matrices are column major, therefore with the index matrix lower.tri the values will go the right places.