R help - Jan 2008 - newbie:looking for an efficient way to compute distance vector

dear experts,

I am new to R and am trying to compute a vector y from  a vector x where :
y[i] = sign(x[j]-x[i])*(j-i)  with j the first index after i where
abs(x[j]-x[i]) > to a given step
y[i] is 0 if there is no such j

I can write this in R as follows
for(i in 1:length(x)) {
   y[i]=0
   for(j in i:length(x)) {
      if (abs(x[j]-x[i]) > step) {
         y[i]=sign(x[j]-x[i])*(j-i)
         break;
      }
   }
}

but I wonder if there is a more efficient way to write this. I understand
explicit looping can often be avoided in R using vector notation.

Thanks for your help
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i'm not so sure i understood, but you might want something in the  
lines of:

z <- outer(x, x, "-")
(abs(z)>step)*outer(1:length(x), 1:length(x))*z

(not tested)

b


On Jan 23, 2008, at 11:32 AM, papagenic wrote:
>
> dear experts,
>
> I am new to R and am trying to compute a vector y from  a vector x  
> where :
> y[i] = sign(x[j]-x[i])*(j-i)  with j the first index after i where
> abs(x[j]-x[i]) > to a given step
> y[i] is 0 if there is no such j
>
> I can write this in R as follows
> for(i in 1:length(x)) {
>   y[i]=0
>   for(j in i:length(x)) {
>      if (abs(x[j]-x[i]) > step) {
>         y[i]=sign(x[j]-x[i])*(j-i)
>         break;
>      }
>   }
> }
>
> but I wonder if there is a more efficient way to write this. I  
> understand
> explicit looping can often be avoided in R using vector notation.
>
> Thanks for your help
> -- 
> View this message in context:
http://www.nabble.com/newbie%3Alooking-for-an-efficient-way-to-compute-distance-vector-tp15045583p15045583.html
> Sent from the R help mailing list archive at Nabble.com.
>
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> PLEASE do read the posting guide
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