not so sure this helps much, but...
theDist <- function(x, step){
ij <- combn(1:length(x), 2)
mapply(function(i, j, step)
(abs(d <- x[j]-x[i]) > step)*sign(d)*(j-i),
ij[1,], ij[2,], step)
}
set.seed(123)
y <- rnorm(10^3)
system.time(tmp <- theDist(y, 1))
gives me:
user system elapsed
20.287 0.038 20.347
this will be expensive anyways, because you don't say how big is big
when you mention "as the length increases"... at some point, you want
to trade the nested for()'s by vectorized versions, which may cost you
memory to compute things like combn() or outer().
b
On Jan 23, 2008, at 4:39 PM, papagenic wrote:
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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