2

考虑dat1在这里创建：

set.seed(123)
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))

dat1有 5 个变量的测量值，观察值 ( IDs) 可以根据 3 个分组变量进行 分组Loc：并保持我的分析整洁。我正在使用该包执行多项操作。以下函数将对我指定的数据进行 Kruskal Wallis 测试，计算效果大小并在单个数据框中返回结果：StateRegionrstatixefszres

library(rstatix)
KruskTest <- function(dat, groupvar, var){
  kt <- dat%>%kruskal_test(get(var) ~ get(groupvar))
  efsz <- dat%>%kruskal_effsize(get(var) ~ get(groupvar))
  res <<- cbind(kt, efsz[,3:5])
  res[1,1] <<- var
  res$groupvar <<- groupvar 
  res <<- res[,c(10,1:9)]
}
KruskTest(dat=dat1, groupvar = "Region", var = "var1")

现在我可以使用该函数循环遍历每个响应变量并Region在单个数据框中获取分组变量的结果（示例显示它），这就是我需要的：

vars <- paste(names(dat1[,5:9]))
a <- data.frame()
for(i in vars){
  KruskTest(dat=dat1, groupvar="Region", var= i)
  a <- rbind(a, res)
}

这对 Kruskal Wallis 测试很有用，现在我想做一个非常相似的函数来做一个 duns 测试，但是看看会发生什么：

dunn <- function(dat, groupvar, var){
  res <<- dat%>%rstatix::dunn_test(get(var) ~ get(groupvar), p.adjust.method = "bonferroni")
}
dunn(dat=dat1, groupvar="Region", var = "var1")

r:Error: Can't extract columns that don't exist. x The column `get(groupvar)` doesn't exist.

dunn_test()在用户编写的函数之外，您可以kruskal_test()以完全相同的方式指定数据。那么在这两个函数中指定变量有什么区别，为什么第一个起作用而第二个不起作用？

4

1 回答 1

1

考虑到@Gregor 关于不写入环境并尝试清理其他一些粗糙边缘的评论，我提出了一个改进建议，尽管 Gregor 是正确的，但你最大的问题只不过是一个错字。

library(rstatix)
library(purrr)

# rewritten to avoid writing to environment

NewKruskTest <- function(dat, groupvar, var) {
  kt <- dat %>% kruskal_test(as.formula(paste(var, "~", groupvar)))
  efsz <- dat %>% kruskal_effsize(as.formula(paste(var, "~", groupvar)))
  results <- cbind(kt, efsz[,3:5])
  results$groupvar <- groupvar 
  results <- results[,c(10,1:9)]
  return(results)
}

# works on a single if you want to test
# NewKruskTest(dat = dat1, groupvar = "Region", var = "var1") 

# No paste needed
vars <- names(dat1[,5:9])

# NewKruskTest will work in your existing for loop but you 
# may find `purrr:map_dfr` cleaner

map_dfr(vars, ~ NewKruskTest(dat = dat1, groupvar = "Region", var = .))
#>   groupvar  .y.   n statistic df      p         method      effsize method.1
#> 1   Region var1 200 3.0520896  1 0.0806 Kruskal-Wallis  0.010364089  eta2[H]
#> 2   Region var2 200 0.5961552  1 0.4400 Kruskal-Wallis -0.002039620  eta2[H]
#> 3   Region var3 200 1.6330090  1 0.2010 Kruskal-Wallis  0.003197015  eta2[H]
#> 4   Region var4 200 3.4031343  1 0.0651 Kruskal-Wallis  0.012137042  eta2[H]
#> 5   Region var5 200 0.7230090  1 0.3950 Kruskal-Wallis -0.001398945  eta2[H]
#>   magnitude
#> 1     small
#> 2     small
#> 3     small
#> 4     small
#> 5     small

# NewDunn rewritten

NewDunn <- function(dat, groupvar, var) {
  results <- dat %>% rstatix::dunn_test(as.formula(paste(var, "~", groupvar)), 
                        p.adjust.method = "bonferroni")
  results$groupvar <- groupvar 
  results <- results[,c(10,1:9)]
  return(results)
}

# works on a single if you want to test
# NewDunn(dat=dat1, groupvar ="Region", var = "var1")

map_dfr(vars, ~ NewDunn(dat = dat1, groupvar = "Region", var = .))
#> # A tibble: 5 x 10
#>   groupvar .y.   group1 group2    n1    n2 statistic      p  p.adj p.adj.signif
#>   <chr>    <chr> <chr>  <chr>  <int> <int>     <dbl>  <dbl>  <dbl> <chr>       
#> 1 Region   var1  r1     r2       100   100    -1.75  0.0806 0.0806 ns          
#> 2 Region   var2  r1     r2       100   100    -0.772 0.440  0.440  ns          
#> 3 Region   var3  r1     r2       100   100    -1.28  0.201  0.201  ns          
#> 4 Region   var4  r1     r2       100   100     1.84  0.0651 0.0651 ns          
#> 5 Region   var5  r1     r2       100   100    -0.850 0.395  0.395  ns

根据您的数据


set.seed(123)
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))
于 2020-06-04T19:54:41.370 回答