[R] changing equal values on matrix by same random number

[David Winsemius]

On Aug 26, 2009, at 12:53 PM, milton ruser wrote:

> Dear all,
>
> I have about 30,000 matrix (512x512), with values from 1 to N.
> Each value on a matrix represent a habitat patch on my
> matrix (i.e. my landscape). Non-habitat are stored as ZERO.
> No I need to change each 1-to-N values for the same random
> number.
>
> Just supose my matrix is:
> mymat<-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
> 0,0,0,0,2,2,2,0,0,0,0,
> 0,0,0,0,2,2,2,0,0,0,0,
> 3,3,0,0,0,0,0,0,0,4,4,
> 3,3,0,0,0,0,0,0,0,0,0), nrow=5)
>
> I would like that all cells with 1 come to be
> runif(1,min=0.4, max=0.7), and cells with 2
> be replace by another runif(...).

First the wrong way and then the right way:

 > mymat[mymat==1] <- runif(1,min=0.4,max=0.7)
 > mymat
           [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.4573161    0    0    2    0    0    0    0    0     3     0
[2,] 0.4573161    0    0    2    0    2    0    0    0     0     0
[3,] 0.4573161    0    0    2    0    2    0    0    4     0     0
[4,] 0.0000000    0    0    0    0    2    3    0    4     0     0
[5,] 0.0000000    0    0    0    0    0    3    0    3     0     0

All the values are the same, clearly not what was desired.

Put it back to your starting point:

 > mymat<-matrix(c(1,1,1,0,0,0,0,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 0,0,0,0,2,2,2,0,0,0,0,
+ 3,3,0,0,0,0,0,0,0,4,4,
+ 3,3,0,0,0,0,0,0,0,0,0), nrow=5)

# So supply the proper number of random realizations:

 > mymat[mymat==1] <- runif(sum(mymat==1),min=0.4,max=0.7)
 > mymat
           [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 0.5745665    0    0    2    0    0    0    0    0     3     0
[2,] 0.6956418    0    0    2    0    2    0    0    0     0     0
[3,] 0.6935466    0    0    2    0    2    0    0    4     0     0
[4,] 0.0000000    0    0    0    0    2    3    0    4     0     0
[5,] 0.0000000    0    0    0    0    0    3    0    3     0     0

If you want to supply a matrix of max and min values for the other  
integers there would probably be an *apply approach that could be used.

>
> I can do it using for(), but it is very time expensive.
> Any help are welcome.
>
> cheers
>

David Winsemius, MD
Heritage Laboratories
West Hartford, CT