r - 使用 dplyr 中的 mutate 对 R 中的自定义函数中的分组数据使用数据框和列作为参数

Question

我在 R 中创建了一个自定义函数来准备绘图数据。我将一个数据帧和两列(来自该数据帧)传递给我的函数，然后使用 dplyr。该函数需要按分类变量(在本例中为age.group)进行分组，并且在数据仍然分组时创建连续变量的分箱版本(to.be.binned)并获取该组的计数。我尝试使用 mutate 来完成这两个任务。

此函数中的代码在函数外部工作，但我将数据帧和变量传递给函数(使用大括号，因为它是 dplyr)。

我收到以下错误:

错误:无法修改列“age.group”，因为它是分组变量

我认为我的代码没有做任何修改这个变量的事情。我需要按组进行计数，以便获得每个组的百分比，因此我无法先取消分组(这是对其他遇到相同错误的人的建议)。

如有任何建议，我们将不胜感激!

代表:

library(tidyverse)

simple.df <- data.frame(
  age.group = c("18-30","Under 18","Over 30",
                "Over 30","Over 30","Under 18","18-30","18-30",
                "Over 30","Under 18","18-30","18-30","18-30","18-30",
                "Under 18","18-30","Under 18","18-30","Under 18",
                "Under 18","Under 18","Over 30","Over 30","Over 30",
                "Over 30","Over 30","18-30","Under 18","Over 30",
                "Under 18"),
  to.be.binned = c(98.415794,32.35116,73.29943,
                   81.92012,99.61144,29.665798,97.652885,94.94358,
                   77.798035,24.110243,99.110245,98.415794,99.80469,94.24913,
                   79.665794,98.415794,72.02691,96.332466,94.94358,
                   97.02691,97.02691,92.860245,98.415794,97.02691,
                   90.082466,99.110245,99.80469,98.415794,99.55236,99.110245)
)



bin_by_group <- function(df, my.grouping, bin.this) {
  
  bw = 25
  
  new.df <- df %>%
    group_by({{my.grouping}}) %>%
    mutate(this.binned = cut(as.numeric({{bin.this}}),
                             breaks = seq(0, 100, bw),
                             labels = seq(0 + bw, 100, bw)-(bw/2)),
           n = n()) %>%
    group_by({{my.grouping}}, this.binned) %>%
    summarise(p = n()/n[1]) %>%
    ungroup() %>%
    mutate(this.binned = as.numeric(as.character(this.binned)))
  
  return(new.df)
  
}


test.df <- bin_by_group(simple.df, "age.group", "to.be.binned")
#> Warning in cut(as.numeric(~"to.be.binned"), breaks = seq(0, 100, bw), labels =
#> seq(0 + : NAs introduced by coercion
#> Error: Column `"age.group"` can't be modified because it's a grouping variable

Answer 1

只是我们需要传递不带引号的参数，因为 {{}} 希望它不带引号，因为 {{}} 相当于 enquo + !!.

bin_by_group(simple.df, age.group, to.be.binned)

-输出

# A tibble: 7 x 3
#  age.group this.binned     p
#  <chr>           <dbl> <dbl>
#1 18-30            87.5   1  
#2 Over 30          62.5   0.1
#3 Over 30          87.5   0.9
#4 Under 18         12.5   0.1
#5 Under 18         37.5   0.2
#6 Under 18         62.5   0.1
#7 Under 18         87.5   0.6

---

如果我们想要传递带引号或不带引号的值，请使用 ensym 进行转换，然后评估 (!!)

bin_by_group <- function(df, my.grouping, bin.this) {
  
  bw = 25
  my.grouping <- ensym(my.grouping)
  bin.this <- ensym(bin.this)
  new.df <- df %>%
    group_by(!! my.grouping) %>%
    mutate(this.binned = cut(as.numeric(!!bin.this),
                             breaks = seq(0, 100, bw),
                             labels = seq(0 + bw, 100, bw)-(bw/2)),
           n = n()) %>%
    group_by(!! my.grouping, this.binned) %>%
    summarise(p = n()/n[1], .groups = 'drop') %>%
    ungroup() %>%
    mutate(this.binned = as.numeric(as.character(this.binned)))
  
  return(new.df)
  
}

-测试

 bin_by_group(simple.df, "age.group", "to.be.binned")
# A tibble: 7 x 3
  age.group this.binned     p
  <chr>           <dbl> <dbl>
1 18-30            87.5   1  
2 Over 30          62.5   0.1
3 Over 30          87.5   0.9
4 Under 18         12.5   0.1
5 Under 18         37.5   0.2
6 Under 18         62.5   0.1
7 Under 18         87.5   0.6

bin_by_group(simple.df, age.group, to.be.binned)
# A tibble: 7 x 3
  age.group this.binned     p
  <chr>           <dbl> <dbl>
1 18-30            87.5   1  
2 Over 30          62.5   0.1
3 Over 30          87.5   0.9
4 Under 18         12.5   0.1
5 Under 18         37.5   0.2
6 Under 18         62.5   0.1
7 Under 18         87.5   0.6