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 <dwinsem...@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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