1
votes

I am trying to fill the lower diagonal of my matrix M with a prefilled vector, V

My original matrix looks like similar to this:

M = matrix(c(.3,.2,.1,0), nrow=4, ncol=5)

M  1  2  3  4 5
1 .3 .3 .3 .3 .3
2 .2 .2 .2 .2 .3
3 .1 .1 .1 .1 .1
4 0  0  0  0   0

I have a vector similar to this:

 V
 .4
 .3
 .25
 .1

Now I want to fill the lower triangle with this vector, to get:

0  1  2  3  4  5
1 .3 .3 .3 .3 .1
2 .2 .2 .2 .25 .25
3 .1 .1 .3 .3 .3 
4 0  .4 .4 .4 .4

If I use the lower.tri function it gives out an error so I built a loop which only should fill the columns from the buttom up:

o <- 5
c <- 2
s <- 1
for(s in (1:o)){  
 for(c in (2:o)){
   M[((o-s):o),c] <- V[1:c]}}

My idea was to move upwards like I manually wrote:

M[(5-1):5,2] <- V[1:2]
M[(5-2):5,3] <- V[1:3] 

What's the best way?

2

2 Answers

2
votes

The part of the matrix you want to fill is a reflection of the lower triangular matrix of the original one with respect to row direction, so you can apply a rev function to the lower.tri() result to reverse every row to get the index for replacing the elements, and then it would be straightforward:

Assuming you have matrix M and vector v:

M
#   X1  X2  X3  X4  X5
#1 0.3 0.3 0.3 0.3 0.3
#2 0.2 0.2 0.2 0.2 0.3
#3 0.1 0.1 0.1 0.1 0.1
#4 0.0 0.0 0.0 0.0 0.0

v
# [1] 0.40 0.30 0.25 0.10

lowerIndex = t(apply(lower.tri(M, diag = TRUE), 1, rev))
M[lowIndex] <- (lowIndex * rev(v))[lowIndex]
M

#   X1  X2  X3   X4   X5
#1 0.3 0.3 0.3 0.30 0.10
#2 0.2 0.2 0.2 0.25 0.25
#3 0.1 0.1 0.3 0.30 0.30
#4 0.0 0.4 0.4 0.40 0.40
1
votes

We can do it this way:

Define a full matrix with the values of v to be inserted

N = matrix(rev(v), 4, 5)

Now we can replace elements in M whose column is greater than the reverse of the row with corresponding value in the replacement matrix

R = rev(row(M))
C = col(M)
M[C>R] = N[C>R]

     # [,1] [,2] [,3] [,4] [,5]
# [1,]  0.3  0.3  0.3 0.30 0.10
# [2,]  0.2  0.2  0.2 0.25 0.25
# [3,]  0.1  0.1  0.3 0.30 0.30
# [4,]  0.0  0.4  0.4 0.40 0.40