为什么 R 用基本计算计算 m for 循环要花这么多时间？

Question

Why does the computation of the following code in R take so much time? It takes many minutes, so I have interruped the calculations.

My aim is to adapt my simulated random numbers ( sumzv , dim(sumzv) = 1000000 x 10) to my market model S_t (geometric brownian motion). The vectors m and s describe the drift and the deviation of the GBM and are vectors containing 10 numbers. DEL is the variable for the time steps. S_0 is a vector containing 10 stock prices at time 0.

n <- 1000000
k <- 10

S_t <- data.frame(matrix(0, nrow = n, ncol = k))

i <- 1
j <- 1
t <- 10

for (j in 1:k) {
  
  for (i in 1:n) {
    S_t[i, j] <- S_0[j] * exp(m[j] * t * DEL + s[j] * sqrt(DEL) * sumzv[i, j])
  
  }
  
}

Thank you for your help. Please keep in mind that I'm a beginner:)

Unfortunately, I couldn't find any helpful information so far on the inte.net. Some pages said, vectorization is helpful to speed up an R Code, but this doesn't seem helpful to me. I tried to break down the data frames into vectors but this got very complex.

Answer 1

The following code with vectorized inner loop is equivalent to the posted code.
It also pre-computes some inner loop vectors, fac1 and fac2 .

S_t <- data.frame(matrix(0, nrow = n, ncol = m))
fac1 <- m * t * DEL
fac2 <- s * sqrt(DEL)
for (j in 1:k) {
  S_t[, j] <- S_0[j] * exp(fac1[j] + fac2[j] * sumzv[, j])
}

The fully vectorized version of the loop on j above is the one-liner below. The transposes are needed because R is column major and we are multiplying by row vectors indexed on j = 1:k .

S_t2 <- t(S_0 * exp(fac1 + fac2 * t(sumzv)))