Question

#为什么这段代码在 R 中需要很长时间？

k<-1000
dat <- NULL
for (x in 0:k){
 for (j in 0:k){ 
    Y <- ((x + j)/2)
     dat <- rbind(dat,x,j,Y) 
    }
  }
out <- list(dat=dat)

Answer 1

这是一些方法的基准。我已将问题中的代码修改为我认为您想要的。您可能希望对其他一些 k 值重复此操作。

library(microbenchmark)

k <- 2

microbenchmark(
ques = {
  res_q <- NULL
  for (x in 0:k){
    for (j in 0:k){ 
      Y <- ((x + j)/2)
      res_q <- rbind(res_q, c(x, j, Y)) 
    } 
  }
  res_q
},
expand.grid = {
  g <- expand.grid(x = 0:k, j = 0:k)[2:1]
  res_g <- as.matrix(transform(g, Y = (x + j) / 2))
  dimnames(res_g) <- NULL
  res_g
},
rep = {
  res_r <- cbind(x = rep(x = 0:k, each = k+1), j = 0:k, Y = NA)
  res_r[, "Y"] <- rowSums(res_r[, 1:2]) / 2
  dimnames(res_r) <- NULL
  res_r
})
## Unit: microseconds
##         expr     min       lq      mean   median      uq     max neval cld
##         ques 28829.2 29268.00 30493.273 29712.45 30303.5 56836.1   100   c
##  expand.grid  1719.3  1784.85  1892.908  1882.70  1917.1  3706.7   100  b 
##          rep    70.3    81.55   114.655   118.00   137.9   212.6   100 a

我们验证所有方法都给出了相同的结果：

all.equal(res_q, res_g, res_r)
## [1] TRUE

为什么这段代码在 R 中需要很长时间？

1 个答案: