unavoidable loop? a better way??

Hi all, I have the following problem, best expressed by my present
solution:

# p is a vector
myfunc <- function (p) {
   x[1] <- p[1]
   for (i in c(2:length(p))) {
     x[i] <- 0.8*p[i] + 0.2*p[i-1]
   }
   return (x)
}

myfunc <- function (p) {

myfunc(1:10)

That is, I'm calculating a time-weighted average. Unfortunately the
scale of the problem is big. length(p) in this case is such that each
call takes about 6 seconds, and I have to call it about 2000 times
(~3 hours). And, I'd like to do this each day. Thus, a more efficient
method is desirable.

Of course, this could be done faster by writing it in c, but I want
to avoid doing that if there already exists something internal to do
the operation quickly (because I've never programmed c for use in R).

Can anybody offer a solution?

I apologise if this is a naive question.

James

On Saturday 13 November 2004 00:51, James Muller wrote:

Hi all, I have the following problem, best expressed by my present
solution:

# p is a vector
myfunc <- function (p) {
   x[1] <- p[1]
   for (i in c(2:length(p))) {
     x[i] <- 0.8*p[i] + 0.2*p[i-1]
   }
   return (x)
}

Does this work at all? I get

myfunc <- function (p) {

+    x[1] <- p[1]
+    for (i in c(2:length(p))) {
+      x[i] <- 0.8*p[i] + 0.2*p[i-1]
+    }
+    return (x)
+ }

myfunc(1:10)

Error in myfunc(1:10) : Object "x" not found


Anyway, simple loops are almost always avoidable. e.g.,

myfunc <- function (p) {
   x <- p
   x[-1] <- 0.8 * p[-1] + 0.2 * p[-length(p)]
   x
}

Deepayan

That is, I'm calculating a time-weighted average. Unfortunately the
scale of the problem is big. length(p) in this case is such that each
call takes about 6 seconds, and I have to call it about 2000 times
(~3 hours). And, I'd like to do this each day. Thus, a more efficient
method is desirable.

Of course, this could be done faster by writing it in c, but I want
to avoid doing that if there already exists something internal to do
the operation quickly (because I've never programmed c for use in R).

Can anybody offer a solution?

I apologise if this is a naive question.

James

Take 3:

# p is a vector
myfunc <- function (p) {
  x <- rep(0,length(p))
  x[1] <- p[1]
  for (i in c(2:length(p))) {
    x[i] <- 0.8*p[i] + 0.2*x[i-1]   # note the x in the last term
  }
  return (x)
}

James





On Sat, 13 Nov 2004 01:12:50 -0600, Deepayan Sarkar <deepayan at stat.wisc.edu> 
wrote:

On Saturday 13 November 2004 00:51, James Muller wrote:

Hi all, I have the following problem, best expressed by my present
solution:

# p is a vector
myfunc <- function (p) {
   x[1] <- p[1]
   for (i in c(2:length(p))) {
     x[i] <- 0.8*p[i] + 0.2*p[i-1]
   }
   return (x)
}

Does this work at all? I get

myfunc <- function (p) {

+    x[1] <- p[1]
+    for (i in c(2:length(p))) {
+      x[i] <- 0.8*p[i] + 0.2*p[i-1]
+    }
+    return (x)
+ }

myfunc(1:10)

Error in myfunc(1:10) : Object "x" not found


Anyway, simple loops are almost always avoidable. e.g.,

myfunc <- function (p) {
   x <- p
   x[-1] <- 0.8 * p[-1] + 0.2 * p[-length(p)]
   x
}

Deepayan

That is, I'm calculating a time-weighted average. Unfortunately the
scale of the problem is big. length(p) in this case is such that each
call takes about 6 seconds, and I have to call it about 2000 times
(~3 hours). And, I'd like to do this each day. Thus, a more efficient
method is desirable.

Of course, this could be done faster by writing it in c, but I want
to avoid doing that if there already exists something internal to do
the operation quickly (because I've never programmed c for use in R).

Can anybody offer a solution?

I apologise if this is a naive question.

James