一次对多个变量进行 shapiro 测试的函数

Question

我在下面创建了类似于dat1 的数据框：

dat1 <- data.frame(Region = rep(c("r1","r2"), each = 100),
                   State = rep(c("NY","MA","FL","GA"), each = 10),
                   Loc = rep(c("a","b","c","d","e","f","g","h"),each = 5),
                   ID = rep(c(1:10), each=2),
                   var1 = rnorm(200),
                   var2 = rnorm(200),
                   var3 = rnorm(200),
                   var4 = rnorm(200),
                   var5 = rnorm(200))

我正在尝试创建一个函数，允许我指定数据、分组变量（Region、State 或 Loc）和变量列（例如，var1:var5）。我想指定 var 列作为与我指定的任何数据框相关联的列号（因此，如果我想在此处执行所有 5 个 vars，则将给出参数 5:9）。我希望该函数从 rstatix 包中进行单变量 shapiro 测试 shapiro_test()，并在 . 到目前为止，我的函数看起来像这样，主要反映了我用来在函数之外完成这项工作的工作流程：

library(rstatix)
UnivShapiro <- function(dat, groupvar, columns){
  var.list <- paste(names(dat[columns]),collapse=",")
  result<-
    dat%>%
    group_by(paste(groupvar))%>%
    shapiro_test(vars=paste(var.list))%>%
    arrange(variable)%>%
    dplyr::filter(p<0.05)
  result
}

我已经尝试了几种不同的变体，但是当我尝试运行该函数时：UnivShapiro(dat=dat1, groupvar="Region", columns = 5:9)，我收到一条错误消息：Error: Can't subset columns that don't exist. x The column var1,var2,var3,var4,var5 doesn't exist. 我无法弄清楚发生了什么。

Answer 1

变量在函数中的评估方式不同。这将是获得所需结果的一种方法（出于说明目的使用 p

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
set.seed(1)
dat1 <- data.frame(Region = rep(c("r1","r2"), each = 100),
                   State = rep(c("NY","MA","FL","GA"), each = 10),
                   Loc = rep(c("a","b","c","d","e","f","g","h"),each = 5),
                   ID = rep(c(1:10), each=2),
                   var1 = rnorm(200),
                   var2 = rnorm(200),
                   var3 = rnorm(200),
                   var4 = rnorm(200),
                   var5 = rnorm(200))
library(rstatix)
#> 
#> Attaching package: 'rstatix'
#> The following object is masked from 'package:stats':
#> 
#>     filter
UnivShapiro <- function(dat, groupvar, columns){
  var.list <- rlang::syms(names(dat[columns]))
  result<-
    dat%>%
    group_by(!!sym(groupvar))%>%
    shapiro_test(vars=var.list)%>%
    arrange(variable)%>%
    dplyr::filter(p < 0.5)
  result
}
UnivShapiro(dat1, groupvar="Region", columns=5:9)
#> # A tibble: 4 x 4
#>   Region variable statistic      p
#>   <chr>  <chr>        <dbl>  <dbl>
#> 1 r2     var1         0.974 0.0476
#> 2 r2     var2         0.985 0.339 
#> 3 r2     var4         0.977 0.0782
#> 4 r2     var5         0.981 0.162

^{由reprex package (v0.3.0) 于 2020 年 5 月 26 日创建}