More efficient function and for loop - r

Question

I am trying to do a more efficient for loop. I know the existence of sapply, laaply, etc. but I don't know how to implement it in my code.
I have my function which I don't know if it is very efficient. I think I should improve this but I don't know how.
myfun <- function(a, b, c) {
sum <- 0
iter <- 0
while (sum < c) {
nr <- runif(1, a, b)
sum <- sum + nr
iter <- iter + 1
}
return(iter)
}
And here is the part which I would like to use an sapply or something similar.
a <- 0
b <- 1
c <- 2
x <- 0
for (i in 1:10^9) {
x <- x + myfun(a, b, c)
}
Also, I need to make a sapply similar to this
sapply(1:10^9, functie(a ,b ,c))
But the sapply uses 1:10^9 as parameters, instead of a, b, c.

Answer 1

I think replicate() is what you may be looking for (I changed your n to something smaller).
set.seed(1234)
n <- 10^2
y <- replicate(n, myfun(a,b,c))
sum(y)
# [1] 462
This matches your prior result.
set.seed(1234)
a <-0
b <-1
c <-2
x <-0
for (i in 1:n){
x <- x + myfun(a,b,c)
}
x
# [1] 462

Answer 2

Recursion
Here is a recursive function f() that does the same job as myfun().
f <- function(s=0) {
if (s[length(s)] >= 2) {
return(length(s) - 1L)
} else {
f(c(s, s[length(s)] + runif(1, 0L, 1L)))
}
}
set.seed(42)
f()
# [1] 3
replicate(8, f())
# [1] 4 5 4 4 3 5 3 5
stopifnot(all.equal({set.seed(42);f()}, {set.seed(42);myfun(0, 1, 2)}))
However (and most likely for that reason), it's just cooler, not faster.
Rcpp
Learning from that, we may define the while loop in Rcpp.
library(Rcpp)
cppFunction('
double myfun_cpp() {
double s = 0;
int i = 0;
while (s < 2) {
s = s + R::runif(0, 1);
i++;
}
return i;
}
')
set.seed(42)
myfun_cpp()
# [1] 3
replicate(8, myfun_cpp())
# [1] 4 5 4 4 3 5 3 5
stopifnot(all.equal({set.seed(42);myfun_cpp()}, {set.seed(42);myfun(0, 1, 2)}))
Now it's lightning fast:
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# f 22.244076 22.639439 25.345203 22.927777 24.089196 35.82683 100 c
# myfun_cpp 3.204448 3.260542 3.843632 3.294618 3.347971 13.71213 100 a
# myfun 16.823981 17.125346 20.605663 17.516248 27.385791 28.63267 100 b
set.seed(42); R <- 1e3
microbenchmark::microbenchmark(
f=replicate(R, f()),
myfun_cpp=replicate(R, myfun_cpp()),
myfun=replicate(R, myfun(0, 1, 2)), times=1e2L,
control=list(warmup=1e1L))

Answer 3

Here are some options
A base R recursion method
f_TIC <- function(x, y, z) ifelse(z <= 0, 0, f_TIC(x, y, z - runif(1, x, y)) + 1)
Rcpp implementation of f_TIC
library(Rcpp)
cppFunction("
int f_TIC_cpp(double x, double y, double z) {
if (z <= 0) {
return 0;
} else {
return f_TIC_cpp(x, y, z- R::runif(0,1))+1;
}
}
")
Benchmarking
library(Rcpp)
f <- function(s = 0) {
if (s[length(s)] >= 2) {
return(length(s) - 1L)
} else {
f(c(s, s[length(s)] + runif(1, 0L, 1L)))
}
}
f_TIC <- function(x, y, z) ifelse(z <= 0, 0, f_TIC(x, y, z - runif(1, x, y)) + 1)
cppFunction("
double myfun_cpp() {
double s = 0;
int i = 0;
while (s < 2) {
s = s + R::runif(0, 1);
i++;
}
return i;
}
")
cppFunction("
int f_TIC_cpp(double x, double y, double z) {
if (z <= 0) {
return 0;
} else {
return f_TIC_cpp(x, y, z- R::runif(0,1))+1;
}
}
")
myfun <- function(a, b, c) {
sum <- 0
iter <- 0
while (sum < c) {
nr <- runif(1, a, b)
sum <- sum + nr
iter <- iter + 1
}
return(iter)
}
set.seed(42)
R <- 1e3
microbenchmark::microbenchmark(
f = replicate(R, f()),
f_TIC = replicate(R, f_TIC(0, 1, 2)),
f_TIC_cpp = replicate(R, f_TIC_cpp(0,1,2)),
myfun_cpp = replicate(R, myfun_cpp()),
myfun = replicate(R, myfun(0, 1, 2)),
times = 1e2L,
control = list(warmup = 1e1L)
)
and we will see
Unit: milliseconds
expr min lq mean median uq max neval
f 11.9342 12.50330 14.161982 13.02100 14.96575 22.7116 100
f_TIC 20.1925 21.69420 23.678240 22.28255 24.86350 34.1577 100
f_TIC_cpp 2.0293 2.10080 2.639625 2.17505 2.36190 7.9715 100
myfun_cpp 1.7351 1.79415 2.094577 1.83810 2.00495 6.7481 100
myfun 9.1408 9.45240 11.783504 10.32355 14.68815 19.5400 100

Answer 4

I would probably solve this using purrr::map(). E.g. like this:
c(1:1e9) %>%
purrr::map_dbl(
~ myfun(a, b, c)
) %>%
sum()
This first calls myfun() the same number of times as the length of c(1:1e9), and stores the results in a numeric vector, then it uses sum() to add the results together.
My tests shows it's a bit faster than using replicate().

Answer 5

You're doing it right, in my honest opinion.
Since you don't need to return a vectorized or multi-dimensional result but instead update an existing object at each iteration, the for loop you're suggesting is more than adequate.
If you want to take a look at some great discussion about this topic I suggest you to look at this link: https://r4ds.had.co.nz/iteration.html
Edit: just to address the speed argument
start <- Sys.time()
purrr::map_dbl(1:1000, function(x) y + myfun(a, b, c)) %>% sum
end <- Sys.time()
end - start
# Time difference of 0.02593184 secs
start <- Sys.time()
y <- replicate(1000, myfun(a,b,c))
cumsum(y)[1000]
end <- Sys.time()
end - start
# Time difference of 0.01755929 secs
y <- 0
start <- Sys.time()
for(i in 1:1000){
y<- y + myfun(a,b,c)
}
end <- Sys.time()
end - start
# Time difference of 0.01459098 secs