R help - Oct 2010 - Question about assigning values in a matrix, conditional on column first row; how to do the loop.

Hello, I?m new at programming and I will greatly appreciate if you can help me
with this.


I have a very large matrix (hundreds of rows and columns), with the first raw
filled with different numbers (between 0 and 1). The rest of the matrix is
filled with values 0, 1, 2. What I need is to replace the values in the matrix
(except the first row will has to remain intact). In each column I have to
replace value 2 for the value of the first row. Also, I want to replace the
values 0 and 1 by -999. I was able to write the loop for a vector but I do not
know how to expand the loop to the matrix.

Please see these examples

For a vector I did this and it works
> V = c(0.1,1,0,2)
> V
[1] 0.1 1.0 0.0 2.0
> for (i in 2:length(V)){if(V[i]==2)V[i]=V[1]
+                                 else (V[i]=-999)}
> V
[1]    0.1 -999.0 -999.0    0.1

For a matrix I could not perform the task. Here it is an example of a small
matrix and the expected result, but the original has more rows and columns.
> matr <- matrix(c(0.1,1,0,2,0.5,1,2,2), nrow=4, ncol=2)
> matr
     [,1] [,2]
[1,]  0.1  0.5
[2,]  1.0  1.0
[3,]  0.0  2.0
[4,]  2.0  2.0
> matr.expectedresult= matrix(c(0.1,-999,-999,0.1,0.5,-999,0.5,0.5), nrow=4,
ncol=2)
> matr.expectedresult
       [,1]   [,2]
[1,]    0.1    0.5
[2,] -999.0 -999.0
[3,] -999.0    0.5
[4,]    0.1    0.5

Thank you very much

Paula

try this:
> matr
     [,1] [,2]
[1,]  0.1  0.5
[2,]  1.0  1.0
[3,]  0.0  2.0
[4,]  2.0  2.0
> result <- apply(matr[-1,], 1, function(.row){
+     ifelse(.row == 2, matr[1,], -999)
+ })
> # data back together
> rbind(matr[1,], t(result))
       [,1]   [,2]
[1,]    0.1    0.5
[2,] -999.0 -999.0
[3,] -999.0    0.5
[4,]    0.1    0.5
>

On Tue, Oct 5, 2010 at 11:54 AM, Paula Fergnani Salvia
<paulafergnani at yahoo.com.ar> wrote:
> Hello, I?m new at programming and I will greatly appreciate if you can help
me with this.
>
>
> I have a very large matrix (hundreds of rows and columns), with the first
raw filled with different numbers (between 0 and 1). The rest of the matrix is
filled with values 0, 1, 2. What I need is to replace the values in the matrix
(except the first row will has to remain intact). In each column I have to
replace value 2 for the value of the first row. Also, I want to replace the
values 0 and 1 by -999. I was able to write the loop for a vector but I do not
know how to expand the loop to the matrix.
>
> Please see these examples
>
> For a vector I did this and it works
>
>> V = c(0.1,1,0,2)
>> V
> [1] 0.1 1.0 0.0 2.0
>
>> for (i in 2:length(V)){if(V[i]==2)V[i]=V[1]
> + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? else (V[i]=-999)}
>
>> V
> [1] ? ?0.1 -999.0 -999.0 ? ?0.1
>
> For a matrix I could not perform the task. Here it is an example of a small
matrix and the expected result, but the original has more rows and columns.
>
>> matr <- matrix(c(0.1,1,0,2,0.5,1,2,2), nrow=4, ncol=2)
>> matr
> ? ? [,1] [,2]
> [1,] ?0.1 ?0.5
> [2,] ?1.0 ?1.0
> [3,] ?0.0 ?2.0
> [4,] ?2.0 ?2.0
>
>> matr.expectedresult= matrix(c(0.1,-999,-999,0.1,0.5,-999,0.5,0.5),
nrow=4, ncol=2)
>> matr.expectedresult
> ? ? ? [,1] ? [,2]
> [1,] ? ?0.1 ? ?0.5
> [2,] -999.0 -999.0
> [3,] -999.0 ? ?0.5
> [4,] ? ?0.1 ? ?0.5
>
> Thank you very much
>
> Paula
>
>
>
>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

Paula-

Look at this example.  R in this case is your data matrix.

-tgs

R<-matrix(sample(0:2,100,replace=T),nrow=10)
R[1,]<-round(runif(10),3)

f<-function(x){
  show(x)
  y<-x[-1]
  y[y!=2]<--999
  y[y==2]<-x[1]
  c(x[1],y)
  }

apply(R,2,f)

On Tue, Oct 5, 2010 at 11:54 AM, Paula Fergnani Salvia <
paulafergnani@yahoo.com.ar> wrote:
> Hello, I’m new at programming and I will greatly appreciate if you can help
> me with this.
>
>
> I have a very large matrix (hundreds of rows and columns), with the first
> raw filled with different numbers (between 0 and 1). The rest of the matrix
> is filled with values 0, 1, 2. What I need is to replace the values in the
> matrix (except the first row will has to remain intact). In each column I
> have to replace value 2 for the value of the first row. Also, I want to
> replace the values 0 and 1 by -999. I was able to write the loop for a
> vector but I do not know how to expand the loop to the matrix.
>
> Please see these examples
>
> For a vector I did this and it works
>
> > V = c(0.1,1,0,2)
> > V
> [1] 0.1 1.0 0.0 2.0
>
> > for (i in 2:length(V)){if(V[i]==2)V[i]=V[1]
> +                                 else (V[i]=-999)}
>
> > V
> [1]    0.1 -999.0 -999.0    0.1
>
> For a matrix I could not perform the task. Here it is an example of a small
> matrix and the expected result, but the original has more rows and columns.
>
> > matr <- matrix(c(0.1,1,0,2,0.5,1,2,2), nrow=4, ncol=2)
> > matr
>     [,1] [,2]
> [1,]  0.1  0.5
> [2,]  1.0  1.0
> [3,]  0.0  2.0
> [4,]  2.0  2.0
>
> > matr.expectedresult= matrix(c(0.1,-999,-999,0.1,0.5,-999,0.5,0.5),
> nrow=4, ncol=2)
> > matr.expectedresult
>       [,1]   [,2]
> [1,]    0.1    0.5
> [2,] -999.0 -999.0
> [3,] -999.0    0.5
> [4,]    0.1    0.5
>
> Thank you very much
>
> Paula
>
>
>
>
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
	[[alternative HTML version deleted]]

Paula,

Just for fun, another way:

## Create 10 by 10 example matrix
set.seed(100)
a <- matrix(sample(c(0,1,2),100, replace=T), nrow=10, ncol=10)
a[1,] <- sample(c(0.25,0.5,0.75),ncol(a), replace=T)

## Use apply to substitiute matrix values by row
b <- apply(a, 1, function(x)
{
  row <- x
  row[row==0 | row==1] <- -999
  row[which(row==2)] <- a[1,which(x==2)]
  return(row)
})

## Transpose result matrix
b <- t(b)



Cheers,
Jeremy

Jeremy Hetzel
Boston University
-- 
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