Joe*_*Joe 7 r dplyr tidyverse rlang
我正在尝试编写一个自定义函数,它将根据预定义变量向量(例如vector_heavy)的值计算新变量,然后根据提供给函数的参数(例如,custom_name)命名新变量.
这个变量命名是我的技能失败的地方.任何帮助是极大的赞赏.
library(tidyverse)
vector_heavy <- quos(disp, wt, cyl)
cv_compute <- function(data, cv_name, cv_vector){
cv_name <- enquo(cv_name)
data %>%
rowwise() %>%
mutate(!!cv_name = mean(c(!!!cv_vector), na.rm = TRUE)) %>%
ungroup()
}
d <- cv_compute(mtcars, cv_name = custom_name, cv_vector = vector_heavy)
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我的错误消息是:
Error: unexpected '=' in:
" rowwise() %>%
mutate(!!cv_name ="
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删除!!之前cv_name的mutate()将导致一个函数计算一个字面命名的新变量cv_name,并忽略custom_name我作为参数包含的内容.
cv_compute <- function(data, cv_name, cv_vector){
cv_name <- enquo(cv_name)
data %>%
rowwise() %>%
mutate(cv_name = mean(c(!!!cv_vector), na.rm = TRUE)) %>%
ungroup()
}
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我怎样才能使用这个函数来利用custom_name我提供的参数cv_name?
你需要在里面使用:=帮助器mutate.您还需要quo_name将输入转换为字符串.
该mutate那么你的函数的线看起来像
mutate(!!quo_name(cv_name) := mean(c(!!!cv_vector), na.rm = TRUE))
完整的:
cv_compute <- function(data, cv_name, cv_vector){
cv_name <- enquo(cv_name)
data %>%
rowwise() %>%
mutate(!!quo_name(cv_name) := mean(c(!!!cv_vector), na.rm = TRUE)) %>%
ungroup()
}
cv_compute(mtcars, cv_name = custom_name, cv_vector = vector_heavy)
mpg cyl disp hp drat wt qsec vs am gear carb custom_name
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 56.20667
2 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 56.29167
3 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 38.10667
4 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 89.07167
5 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 123.81333
6 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 78.15333
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