为什么此功能无法执行?
my_ttest <- function(df, variable, by){
variable <- enquo(variable)
by <- enquo(by)
t.test(!!variable ~ !!by, df)
}
my_ttest(mtcars,mpg,上午) is_quosure(e2)中的错误:参数“ e2”丢失,没有默认值
但是这个有效
my_mean <- function(df, variable, by){
variable <- enquo(variable)
by <- enquo(by)
df %>% group_by(!!by) %>% summarize(mean(!!variable))
}
my_mean(mtcars, mpg, am)
# A tibble: 2 x 2
am `mean(mpg)`
<dbl> <dbl>
1 0 17.1
2 1 24.4
(dplyr_0.8.0.1)
答案 0 :(得分:2)
并非每个函数(和程序包)都可以进行整洁的评估。 t.test
将数字向量x,y
作为参数或公式。在您的示例中,您可以提供公式和数据框,尽管实际上看起来并不比直接调用t.test
更有效。
my_ttest <- function(df, frma) {
t.test(frma, df)
}
my_ttest(mtcars, mpg ~ am)
#>
#> Welch Two Sample t-test
#>
#> data: mpg by am
#> t = -3.7671, df = 18.332, p-value = 0.001374
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -11.280194 -3.209684
#> sample estimates:
#> mean in group 0 mean in group 1
#> 17.14737 24.39231
由reprex package(v0.2.1)于2019-03-23创建
答案 1 :(得分:2)
如果我们想分别在'my_ttest'中传递参数并在函数内部构造一个公式,则将'variable',''的quosure(enquo
)转换为符号(sym
)。 by”,然后构造表达式('expr1')并对其进行eval
uate
my_ttest <- function(df, variable, by, env = parent.frame()){
variable <- rlang::sym(rlang::as_label(rlang::enquo(variable)))
by <- rlang::sym(rlang::as_label(rlang::enquo(by)))
exp1 <- rlang::expr(!! variable ~ !! by)
t.test(formula = eval(exp1), data = df)
}
my_ttest(mtcars, mpg, am)
#Welch Two Sample t-test
#data: mpg by am
#t = -3.7671, df = 18.332, p-value = 0.001374
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -11.280194 -3.209684
#sample estimates:
#mean in group 0 mean in group 1
# 17.14737 24.39231
或者如评论中提到的@lionel一样,可以直接通过ensym
my_ttest <- function(df, variable, by, env = parent.frame()){
exp1 <- expr(!!ensym(variable) ~ !!ensym(by))
t.test(formula = eval(exp1), data = df)
}
my_ttest(mtcars, mpg, am)
编辑:基于@lionel的评论