如何在R中为这个嵌套的for循环编写更有效的代码？
尝试计算p1（实验组中的事件）和p0（对照组中的事件）与or（优势比）的组合1.5。nnt= 需要治疗的人数 (100/( p1- p0))
library(tidyverse)
p1 <- seq(0,1, 0.0001)
p0 <- seq(0,1,0.0001)
or <- 1.5
df <- tibble(p1 = as.numeric(), p0 = as.numeric(), nnt = as.numeric())
for (i in p1) {
for (j in p0) {
or_formula <- round((i/(1-i))/(j/(1-j)),3)
if (or_formula == or & !is.na(or_formula)) {
df <- df %>% add_row(p1 = i, p0 = j, nnt = round(1/(i-j), digits = 0))
}
}
}
回答
我们可以用 outer
or_formula <- function(i, j) round((i/(1-i))/(j/(1-j)), 3)
m1 <- outer(p1, p0, FUN = or_formula)
dim(m1)
#[1] 10001 10001
i1 <- m1 == or & !is.na(m1)
i2 <- which(i1, arr.ind = TRUE)
p1new <- p1[i2[,1]]
p0new <- p0[i2[,2]]
df1 <- tibble(p1 = p1new, p0 = p0new, nnt = round(1/(p1new-p0new), digits = 0))
基准
- 使用外部
system.time({
m1 <- outer(p1, p0, FUN = or_formula)
i1 <- m1 == or & !is.na(m1)
i2 <- which(i1, arr.ind = TRUE)
p1new <- p1[i2[,1]]
p0new <- p0[i2[,2]]
df1 <- tibble(p1 = p1new, p0 = p0new, nnt = round(1/(p1new-p0new), digits = 0))
})
# user system elapsed
# 5.038 1.288 6.319
- 使用 OP 的 for 循环
system.time({
df <- tibble(p1 = as.numeric(), p0 = as.numeric(), nnt = as.numeric())
for (i in p1) {
for (j in p0) {
or_formula <- round((i/(1-i))/(j/(1-j)),3)
if (or_formula == or & !is.na(or_formula)) {
df <- df %>% add_row(p1 = i, p0 = j, nnt = round(1/(i-j), digits = 0))
}
}
}
})
# user system elapsed
#122.391 0.748 123.128
- 测试相等性
identical(df, df1)
#[1] TRUE