R:如何在没有for循环的情况下计算回归

Question

set.seed(1)
n <- 100
ret <- rnorm(n, 0, 0.02)
ret[1] <- 0
price <- cumprod(1+ret)*100
maxi <- 0
drawdown <- rep(0, n)

for (i in 1 : n){
    maxi <- max(price[1 : i])
    drawdown[i] <- price[i] / maxi - 1
}

你好，

是否可以加快计算速度？也许删除 for 循环？

问候

Answer 1

R 有一个向量化的 cummax 函数，除法和加法运算也是向量化的，所以你可以这样做:

price/cummax(price) - 1

n <- 10000 时的效率比较:

library(microbenchmark)
microbenchmark(
  OP= {
    drawdown <- rep(0, n)
    for (i in 1 : n){
      maxi <- max(price[1 : i])
      drawdown[i] <- price[i] / maxi - 1
    }
  },
  me={
    drawdown2 <- price/cummax(price) - 1
  }, times=10)

# Unit: microseconds
#  expr        min         lq        mean     median         uq        max neval
#    OP 456216.519 483387.361 536067.7521 550912.471 565453.555 663352.635    10
#    me     98.075    102.067    107.5978    105.203    112.331    127.726    10

identical(drawdown, drawdown2)
# [1] TRUE