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

[David Huffer]

You could try either of the two examples below. They both assume that min and max are invariant: 

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
)

##  this way gives same random values for
##  each integer between 0 and
##  max (mymat):

milton <- function ( x , min = 0.4 , max = 0.7 ) {
  .rep <- runif ( n = max ( x ) , min = min , max = max )
  for ( i in 1:max ( x ) ) {
  x[x == i] <- .rep[i]
  }
  x
}
milton ( x = mymat )

##  this way gives different random values
##  for each integer between
##  0 and max (mymat):

milton <- function ( x , min = 0.4 , max = 0.7 ) {
  for ( i in 1:max ( x ) ) {
    x [ x == i ] <- runif (
      sum ( x == i )
      , min = min
      , max = max
    )
  }
  x
}
milton ( x = mymat )
 
--
 David
 
 -----------------------------------------------------
 David Huffer, Ph.D.               Senior Statistician
 CSOSA/Washington, DC           david.huffer at csosa.gov
 -----------------------------------------------------


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of milton ruser
Sent: Wednesday, August 26, 2009 1:18 PM
To: David Winsemius
Cc: r-help at r-project.org
Subject: Re: [R] changing equal values on matrix by same random number

Hi David,
Thanks for the reply. This is what I need:

> 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

But as my real landscapes have values from 1 to large number (~10,0000),
so I think that if I put this on a for() looping it will be very time
expensive,
and as I have a lot of landscapes, I need to speed up it.

Any suggestion?

bests

milton



On Wed, Aug 26, 2009 at 1:12 PM, David Winsemius <dwinsemius at comcast.net>wrote:

>
> 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
>
>

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