How to add a constant in a for loop by keeping the original matrix in each iteration? - r

Question

For example,
x<-matrix(c(1,2,3,4),2,2)
1 2
3 4
I want to add the constant "c" to each element of the matrix separately like this.
Iteration 1
1+c 2
3 4
Iteration 2
1 2+c
3 4
Iteration 3
1 2
3+c 4
Iteration 4
1 2
3 4+c
I have tried the following R code, but it retains the updated value while performing second iteration.
x= matrix of order nxm
for(i in 1:r)
{
for(j in 1:c)
{
x[i,j]=x[i,j]+c
print(x)
}
}
In this code the values getting updated and printing the updated value for each iteration.
Please help me... Thanks in Advance.

Answer 1

R prefers array operations.
Any matrix x is just an array of its entries, laid out column by column. You may successively add the constant c to the first, second, third, ... entry to copies of x, so that the original x remains unchanged. Do this by constructing arrays of the same length as x with all zero entries except for c in the desired location. The code shown at the end of this post does this by concatenating a bunch of zeros, c, and more zeros so that c appears in position i:
c(rep(0,i-1), cnst, rep(0,n-i)
If you loop with i=1, 2, 3, etc, the results will work down through each column of x, moving left to right. To do the operations in the order presented in the question, which works through each row, moving top to bottom, simply apply the procedure to the transpose of x and transpose the outputs.
Even for large matrices, this approach of adding an entire array is at least twice as fast on my system as adding c just to the i position of a copy of x.
Here is R code for the general procedure. It works on any non-empty matrix x. Beware: the output consists of length(x) copies of x and therefore can be quite large. In this example--which takes about a second to run on my system--x has 10,000 entries and therefore the output has 100,000,000 entries. You might want to test it on smaller matrices first!
x <- matrix(1:(100^2), 100) # Any nonempty matrix
cnst <- 1 # Value to add successively to each term in `x`
#
# The algorithm begins here.
#
n <- length(x)
lapply(1:n, function(i) matrix(as.vector(x)+c(rep(0,i-1),cnst,rep(0,n-i)), nrow(x)))

Answer 2

You just need to make a copy of the matrix:
x_safely_stored <- matrix of order nxm
for(i in 1:r) {
for(j in 1:c) {
x <- x_safely_stored
x[i,j]=x[i,j]+c
print(x)
}
}