R:将 t.test 包装在一个函数中

Question

为什么这个函数执行失败？

my_ttest <- function(df, variable, by){
    variable <- enquo(variable)
    by <- enquo(by)

    t.test(!!variable ~ !!by, df)
}

my_ttest(mtcars, mpg, am) Error in is_quosure(e2) : argument "e2" is missing, with no default

但这个有效

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)

Answer 1

如果我们想在 'my_ttest' 中分别传递参数并在函数内部构造一个公式，请将 quosure ( enquo ) 转换为 'variable' 和 'by' 的符号 ( sym )，然后构造表达式('expr1') 和 eval吃`它

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，可以直接使用 ensym 来完成

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