我想将一个函数应用于两个列表元素的每个组合。
library(tidyverse)
map(
.x = c(0, 1)
, .f = function(x) {
qnorm(p = 0.05, mean = x, sd = 1, lower.tail = FALSE)
}
)
[[1]]
[1] 1.644854
[[2]]
[1] 2.644854
map(
.x = c(0, 1)
, .f = function(x) {
qnorm(p = 0.05, mean = x, sd = 2, lower.tail = FALSE)
}
)
[[1]]
[1] 3.289707
[[2]]
[1] 4.289707
现在尝试将两者合并为一(无论如何都不会得到所需的输出(。
map2(
.x = c(0, 1)
, .y = c(1, 2)
, .f = function(x, y) {
qnorm(p = 0.05, mean = x, sd = y, lower.tail = FALSE)
}
)
[[1]]
[1] 1.644854
[[2]]
[1] 4.289707
想知道如何获得所有四种组合的输出?
或具有pmap
和crossing
的另一个选项
library(tidyr)
library(purrr)
library(dplyr)
crossing(v1 = 0:1, v2 = 1:2) %>%
pmap_dbl(~ qnorm(p = 0.05, mean = ..1, sd = ..2, lower.tail = FALSE))
[1] 1.644854 3.289707 2.644854 4.289707
如果我们需要data.frame/tibble,请使用mutate
中的pmap
代码作为新列返回
crossing(v1 = 0:1, v2 = 1:2) %>%
mutate(new = pmap_dbl(., ~ qnorm(p = 0.05,
mean = ..1, sd = ..2, lower.tail = FALSE)))
# A tibble: 4 × 3
v1 v2 new
<int> <int> <dbl>
1 0 1 1.64
2 0 2 3.29
3 1 1 2.64
4 1 2 4.29
注意:如果我们不需要其他列,请使用transmute
而不是mutate
,或者在mutate
中指定.keep = "used"
crossing(v1 = 0:1, v2 = 1:2) %>%
mutate(new = pmap_dbl(., ~ qnorm(p = 0.05,
mean = ..1, sd = ..2, lower.tail = FALSE)), .keep = "used")
# A tibble: 4 × 1
new
<dbl>
1 1.64
2 3.29
3 2.64
4 4.29
您可以使用expand.grid
:
library(purrr)
df1 <- expand.grid(0:1, 1:2)
map2(
.x = df1$Var1,
.y = df1$Var2,
.f = function(x, y) {
qnorm(p = 0.05, mean = x, sd = y, lower.tail = FALSE)
}
)
获取
[[1]]
[1] 1.644854
[[2]]
[1] 2.644854
[[3]]
[1] 3.289707
[[4]]
[1] 4.289707
基于data.table
的解决方案,不使用任何purrr
函数:
library(data.table)
library(magrittr)
setDT(CJ(x = 0:1, y = 1:2))[,
res := qnorm(p = 0.05, mean = x, sd = y, lower.tail = FALSE)] %>% print
#> x y res
#> 1: 0 1 1.644854
#> 2: 0 2 3.289707
#> 3: 1 1 2.644854
#> 4: 1 2 4.289707
另一个基于tidyverse
的解决方案,不使用任何purrr
功能:
library(tidyverse)
data.frame(x = 0:1, y = 1:2) %>%
expand(x,y) %>%
mutate(res = qnorm(p = 0.05, mean = x, sd = y, lower.tail = FALSE))
#> # A tibble: 4 × 3
#> x y res
#> <int> <int> <dbl>
#> 1 0 1 1.64
#> 2 0 2 3.29
#> 3 1 1 2.64
#> 4 1 2 4.29