[R] complex matrix manipulation question

[Berton Gunter]

Mark:

I did not see a reply to your question. Did you get one? If not,here's a
solution using a while() loop which should be fast. One could also use
recursion here in a natural way. This solution assumes that there are no
NA's anywhere -- it's a bit trickier if there are NA's in the x column.
Also, I have omitted matrix notation and just assumed x and y are vectors. I
didn't test this exhaustively, so there might be a few fussy details that
still need debugging. The major problem would be that I did not interpret
your question correctly, but I hope I got it right.

xcsum<-cumsum(x)
i <- 1; k <- 0; n <-length(x); z <- rep(NA,n)
while(i <=  n) {
    xcsum <- xcsum - k
    inew <- which(xcsum > W)[1]
    if(is.na(inew)) break
    else{
        z[inew]<-sum(y[i:inew])
        i<-inew+1
        k<-xcsum[inew]
        }
    }

-- Bert

-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
 
"The business of the statistician is to catalyze the scientific learning
process."  - George E. P. Box
 
 

> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch 
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Mark Leeds
> Sent: Tuesday, January 10, 2006 5:55 PM
> To: R-Stat Help
> Subject: [R] complex matrix manipulation question
> 
> I've done stuff like this before but
> it's been a while and I'm stuck.
>  
> Suppose I have a matrix with one
> column x and another column y
> and both are numeric and let the
> row index of the matrix be i
>  
> Starting at index i ( i would equal on the first iteration )
> when the cumulative sum of x_i+1 - x_i
> is greater than W = some constant, I want to mark that spot in the
> row, call it  i^* and sum all the values in y between  i and  i^* and
> put that value
> a third column z. Otherwise, the values in the indices of z
> between i and  i^*-1  should be NA.
>  
> Then, start at i^*+1 and do the same thing again.
> and keep doing thisn until I get all the way through the rows
> of the matrix.
>  
> I think this is tricky but I used to do it and I forgot how to.
> If it has to be done using loops, that's okay but
> from previous experience, I don't think looping is necessary.
>  
> Thanks.
>  
>                                       Mark
>   
> 
> 
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