user00110014 - 7 months ago 72

R Question

I want to calculate the variables

`fn_x`

`Fn_x`

`y <- seq(0,2,0.01)`

z <- sort(rexp(100,1))

U <- round(runif(100), 0)

myfun <- function(x) 0.75 * (1-x^2) * (abs(x)<1)

fn_x <- matrix(0, length(y), 1)

Fn_x <- matrix(0, length(y), 1)

for(j in 1:length(y)){

fn_x[j] <- (1/(100*2)) * sum(myfun((y[j]-z)/2))

Fn_x[j] <- (1/100)*sum(I(z <=y[j] & U==1))

}

My function is using two different matrices with different dimensions for calculating each element, so the function

`apply`

Answer

Since you're already preallocating vectors before executing the loop, you're doing a lot of the heavy lifting needed to speed up calculations. At this point, `data.table`

or pure implementation in C++ using e.g. `Rcpp`

package would boost the speed.

```
library(microbenchmark)
microbenchmark(
original = {
fn_x <- matrix(NA, length(y), 1)
Fn_x <- matrix(NA, length(y), 1)
for(j in 1:length(y)){
fn_x[j] <- (1/(100*2)) * sum(myfun((y[j]-z)/2))
Fn_x[j] <- (1/100)*sum(I(z <=y[j] & U==1))
}
},
new = {
fn_x2 <- sapply(y, FUN = function(x, z) {
(1/(100*2)) * sum(myfun((x-z)/2))
}, z = z)
Fn_x2 <- sapply(y, FUN = function(x, z, U) {
(1/100) * sum(I(z <= x & U == 1))
}, z = z, U = U)
}
)
Unit: milliseconds
expr min lq mean median uq max
original 9.550934 10.407091 12.13302 10.895803 11.95638 22.87758
new 8.734813 9.126127 11.18128 9.264137 10.12684 87.68265
```

Source (Stackoverflow)