使用apply R的递归循环

Question

I have got two loops I want to transform using some apply function in order to reduce the computation time. First one seems easy to do, the problem arises with the second one because the S is updated with its own value in each iteration.

S0   = 100
a    = 0.00016
b    = 0.0126
sim    = 10000 
drifts = 1000
Si = rep(0,sim)

for(i in (1:sim))
{
 S =  S0
 for (j in (1:drifts))
  {
   z = rnorm(1, mean = 0, sd = 1)
   S = S * exp(a + b*z)
  }
  Si[i] =S 
}

Can anyone help?

Answer 1

calc_s <- function(S, i = 1) {
  S <- S * exp(a + b * rnorm(1, mean = 0, sd = 1))
  return (if (i < drifts) calc_s(S, i + 1) else S)}
S2 <- sapply(1:sim, function(x) {
  calc_s(S0)
})

It isn't faster though

Answer 2

How about something like the code below. I have replaced the j-loop with a product, I think the math is correct.

sapply(1:sim,function(x) S0*prod(exp(a + b*rnorm(drifts))))

it is also significantly faster:

> system.time(for(i in (1:sim))
+ {
+   S =  S0
+   for (j in (1:drifts))
+   {
+     z = rnorm(1, mean = 0, sd = 1)
+     S = S * exp(a + b*z)
+   }
+   Si[i] =S 
+ }
+ )
   user  system elapsed 
  23.29    0.02   23.34 
> 
> system.time(Si<-sapply(1:sim,function(x) S0*prod(exp(a + b*rnorm(drifts)))))
   user  system elapsed 
   1.76    0.00    1.76