[R] newbie:looking for an efficient way to compute distance vector

[papagenic]

hello,

I tried your suggestion ,but the first line :
z <- outer(x, x, "-")
seems to fail pretty quickly as the length of the x vector increases.  This
is probably because it has to create a matrix of dimension dim(x)*dim(x). I
am wondering if that can be quicker than a building a loop. 

if my vector x  holds a time series, I am trying to find for each element i
of x the nb of steps j so that the value  x[j]  differs from x[i] by more
than a predefined step value. 

Regards


Benilton Carvalho wrote:
> 
> 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
>> -- 
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>>
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> 
> 
> ______________________________________________
> 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.
> 
> 

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