[R] for loop optimization help

[Jinsong Zhao]

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