On 2012-08-27 9:35, David Winsemius wrote:
On Aug 26, 2012, at 5:06 PM, Jinsong Zhao wrote:
Hi there,
In my code, there is a for loop like the following:
pmatrix <- matrix(NA, nrow = 99, ncol = 10000)
qmatrix <- matrix(NA, nrow = 99, ncol = 3)
paf <- seq(0.01, 0.99, 0.01)
for (i in 1:10000) {
p.r.1 <- rnorm(1000, 1, 0.5)
p.r.2 <- rnorm(1000, 2, 1.5)
p.r.3 <- rnorm(1000, 3, 1)
pmatrix[,i] <- quantile(c(p.r.1, p.r.2, p.r.3), paf)
}
for (i in 1:99) {
qmatrix[i,] <- quantile(pmatrix[i,], c(0.05, 0.5, 0.95))
}
Because of the number of loop is very large, e.g., 10000 here, the
code is very slow.
I would think that picking the seq(0.01, 0.99, 0.01) items in the first
case and the 500th, 5000th and the 9500th in the second case, rather
than asking for what `quantile` would calculate, would surely be more
"statistical", in the sense of choose order statistics anyway. Likely
much faster.
Yes, you are right. Following your suggestions, the execution time of
`sort' is much shorter than `quantile' in the following code:
pmatrix <- matrix(NA, nrow = 99, ncol = 10000)
qmatrix <- matrix(NA, nrow = 99, ncol = 3)
paf <- seq(0.01, 0.99, 0.01)
for (i in 1:10000) {
p.r.1 <- rnorm(1000, 1, 0.5)
p.r.2 <- rnorm(1000, 2, 1.5)
p.r.3 <- rnorm(1000, 3, 1)
pmatrix[,i] <- sort(c(p.r.1, p.r.2, p.r.3))[paf*3000]
}
qmatrix <- pmatrix[,c(0.05, 0.5, 0.95)*10000]
Thanks again.
Regards,
Jinsong
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