On Fri, Aug 24, 2012 at 10:34:14AM +0200, Petr Savicky wrote:
> On Thu, Aug 23, 2012 at 09:49:33PM -0700, Gopi Goteti wrote:
> > I would like to know whether there is a faster way to do the below
> > operation (updating vec1).
> >
> > My objective is to update the elements of a vector (vec1), where a
> > particular element i is dependent on the previous one. I need to do this on
> > vectors that are 1 million or longer and need to repeat that process
> > several hundred times. The for loop works but is slow. If there is a faster
> > way, please let me know.
> >
> > probs <- c(.1, .3, .2, .4, .7, .9, .3, .4, .5, .6)
> > p10 <- 0.6
> > p00 <- 0.4
> > vec1 <- rep(0, 10)
> > for (i in 2:10) {
> > vec1[i] <- ifelse(vec1[i-1] == 0,
> > ifelse(probs[i]<p10, 0, 1),
> > ifelse(probs[i]<p00, 0, 1))
> > }
>
> Hi.
>
> If p10 is always more than p00, then try the following.
[...]
> # modification
> a10 <- ifelse(probs<p10, 0, 1)
> a00 <- ifelse(probs<p00, 0, 1)
> vec2 <- ifelse(a10 == a00, a10, NA)
> vec2[1] <- 0
> n <- length(vec2)
> while (any(is.na(vec2))) {
> shift <- c(NA, vec2[-n])
> vec2 <- ifelse(is.na(vec2), shift, vec2)
> }
Hi.
Let me suggest a variant of this, which can be more efficient
in some cases.
probs <- c(.1, .3, .2, .4, .7, .9, .3, .4, .5, .6)
p10 <- 0.6
p00 <- 0.4
# original code
vec1 <- rep(0, 10)
for (i in 2:10) {
vec1[i] <- ifelse(vec1[i-1] == 0,
ifelse(probs[i]<p10, 0, 1),
ifelse(probs[i]<p00, 0, 1))
}
# modification
a10 <- ifelse(probs<p10, 0, 1)
a00 <- ifelse(probs<p00, 0, 1)
vec2 <- ifelse(a10 == a00, a10, NA)
vec2[1] <- 0
while (1) {
i <- which(is.na(vec2))
if (length(i) == 0) break
vec2[i] <- vec2[i-1]
}
all(vec1 == vec2)
[1] TRUE
Hope this helps.
Petr Savicky.
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