shany - 7 months ago 41

R Question

I would like to create two matrices in R such that the elements of matrix x should be random from any distribution and then I calculate the colSums and rowSums of this 2*2 matrix.Then, I would like to generate matrix y from any distribution such that the first subset 2*2 elements are random and then the third row and column are the sum of row and column elements and then the fourth row and column are random in such a way that the rowSums and colSums of 3:4 are equal to the colSums and row Sums of first matrix x. How can I create two random matrices with these properties in R ?

`x=matrix(c(100, 50, 0, 250), nrow=2)`

csum1=colSums(x)

rsum1=rowSums(x)

y=matrix(c(15, 60, 75, 25,60, 25, 85, 215, 75, 85, 160, 240,

75, 165, 240, 0), ncol=4) 1

csum2=colSums(y[3:4,c(-3,-4)])

rsum2=rowSums(y[c(-3,-4),3:4])

> x

[,1] [,2]

[1,] 100 0

[2,] 50 250

> y

[,1] [,2] [,3] [,4]

[1,] 15 60 75 75

[2,] 60 25 85 165

[3,] 75 85 160 240

[4,] 25 215 240 0

Answer

I think you just need to generate two `2x2`

random matrices and then all the other values can be obtained by them :

```
set.seed(1234) # just for reproducibility
# generate the two 2x2 matrices with your desired distribution
# (e.g. here I'm generating integers using the uniform distribution)
x <- matrix(as.integer(runif(n = 4,min=10,max=100)),nrow=2)
y22 <- matrix(as.integer(runif(n = 4,min=10,max=100)),nrow=2)
# compute col and row sums of first matrix
csumX=colSums(x)
rsumX=rowSums(x)
# initialize the 4x4 matrix with NAs and set the upper-left 2x2 corner
y <- matrix(NA,nrow=4,ncol=4)
y[1:2,1:2] <- y22
# define the values of 3rd row and colum as sum of other values
# (exluding elements on the 4th row and column)
y[1:2,3] <- colSums(y[1:2,1:2])
y[3,1:2] <- rowSums(y[1:2,1:2])
y[3,3] <- sum(y[1:2,1:2]) # by definition y[3,3] == y[3,1]+y[3,2] == y[1,3]+y[2,3]
# define first two elements of 4th row and column
y[4,1] <- csumX[1] - y[3,1]
y[4,2] <- csumX[2] - y[3,2]
y[1,4] <- rsumX[1] - y[1,3]
y[2,4] <- rsumX[2] - y[2,3]
# complete 3rd row and column values
y[4,3] <- y[4,1] + y[4,2] # these two will be equal by definition
y[3,4] <- y[1,4] + y[2,4]
# element y[4,4] (always zero ?)
y[4,4] <- 0
```

Result :

```
> x
[,1] [,2]
[1,] 20 64
[2,] 66 66
> y
[,1] [,2] [,3] [,4]
[1,] 87 10 154 -70
[2,] 67 30 40 92
[3,] 97 97 194 22
[4,] -11 33 22 0
```

Source (Stackoverflow)