r - 具有最大操作的高效双 for 循环

Question

R 中以下双 for 循环的有效实现是什么？

set.seed(1)
u <- rnorm(100, 1)
v <- rnorm(100, 2)
x <- rnorm(100, 3)
y <- rnorm(100, 4)
sum = 0
for (i in 1:100){
  for (j in 1:100) {
    sum = sum + (1 - max(u[i], v[j])) * (1 - max(x[i], y[j]))
  }
}

特别是对于非常长的向量，评估需要相当长的时间，但我想知道是否有办法向量化这个双 for 循环？非常感谢。

Answer 1

我的速度更快。它使用 outer 而不是循环，这就是它的意思。

首先是不需要外部包的函数，OP，user20650's comment 中的函数和我的。

original <- function(u, v, x, y){
  sum1 = 0
  for (i in seq_along(u)){
    for (j in seq_along(v)) {
      sum1 = sum1 + (1 - max(u[i], v[j])) * (1 - max(x[i], y[j]))
    }
  }
  sum1
}

comment <- function(u, v, x, y){
  sum1 = 0
  for (i in seq_along(u)){
    sum1 = sum1 + (1 - pmax(u[i], v)) * (1 - pmax(x[i], y))
  }
  sum(sum1)
}

rui <- function(u, v, x, y){
  tmp1 <- outer(u, v, pmax)
  tmp2 <- outer(x, y, pmax)
  sum((1 - tmp1) * (1 - tmp2))
}

现在 www's answer 中的函数在IceCreamToucan's answer .

library(tidyverse)

www <- function(u, v, x, y){
  dat <- data_frame(u, v, x, y)
  dat2 <- dat %>% complete(nesting(u, x), nesting(v, y))
  SUM2 <- sum(with(dat2, (1 - pmax(u, v)) * (1 - pmax(x, y))))
  SUM2
}

IceCream <- function(u, v, x, y){
  uv <- expand.grid(u, v)
  xy <- expand.grid(x, y)
  sum((1 - do.call(pmax, uv))*(1 - do.call(pmax, xy)))
}

全部测试，看结果是否相同。请注意，存在浮点问题。

set.seed(1234)

u <- rnorm(1e2, 1)
v <- rnorm(1e2, 2)
x <- rnorm(1e2, 3)
y <- rnorm(1e2, 4)

o <- original(u, v, x, y)
c <- comment(u, v, x, y)
w <- www(u, v, x, y)
i <- IceCream(u, v, x, y)
r <- rui(u, v, x, y)

all.equal(o, c)
all.equal(o, w)
all.equal(o, i)
all.equal(o, r)

o - c
o - w
o - r
w - r
i - r
c - r

现在是速度测试。

library(microbenchmark)
library(ggplot2)

mb <- microbenchmark(
  loop = original(u, v, x, y),
  pmax = comment(u, v, x, y),
  tidy = www(u, v, x, y),
  ice = IceCream(u, v, x, y),
  outer = rui(u, v, x, y)
)

autoplot(mb)