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我正在尝试编写一个函数来为我提供两个变量的数据透视表。在这里扩展我的问题，我想包括预测变量和目标之间关系的卡方检验的 p 值。我应该如何更改功能？

library(dplyr)
mean_mpg <- mean(mtcars$mpg)

# creating a new variable that shows that Miles/(US) gallon is greater than the mean or not

mtcars <-
mtcars %>%
  mutate(mpg_cat = ifelse(mpg > mean_mpg, 1,0))

mtcars %>%
  group_by(as.factor(cyl)) %>%
  summarise(sum=sum(mpg_cat),total=n()) %>%
  mutate(percentage=sum*100/total)

# Note: needs installation of rlang 0.4.0 or later
get_pivot <- function(data, predictor,target) {
  result <-
    data %>%
    group_by(as.factor( {{ predictor }} )) %>%
    summarise(sum=sum( {{ target }} ),total=n()) %>%
    mutate(percentage=sum*100/total);

  print(result)
}

这是我的工作示例：

mtcars %>%
  group_by(as.factor(cyl)) %>%
  summarise(sum=sum(mpg_cat),total=n(),
            pvalue= chisq.test(as.factor(.$mpg_cat), as.factor(.$cyl))$p.value) %>% 
  mutate(percentage=sum*100/total)

我尝试了以下功能，但没有奏效。

get_pivot <- function(data, predictor,target) {
  result <-
    data %>%
    group_by( {{ predictor }} ) %>%
    summarise(clicks=sum( {{ target }} ),total=n(),
              pvalue= chisq.test(.$target, .$predictor)$p.value) %>%
    mutate(percentage=clicks*100/total);

  print(result)
}
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1 回答 1

2

{{...}}curly-curly 插值运算符是引用-取消引用的便捷方式。但是，它不会在所有情况下都有效。$在 OP 的函数中，用ie提取一列。该部分.$target或.$predictor将不起作用。相反，我们可以将其转换为character ( as_name)，然后使用[[

library(rlang)
library(dplyr)

get_pivot <- function(data, predictor,target) {

     data %>%
     group_by( {{ predictor }} ) %>%
     summarise(clicks=sum( {{ target }} ),total=n(),
               pvalue= chisq.test(.[[as_name(enquo(target))]], 
                       .[[as_name(enquo(predictor))]])$p.value) %>%
     mutate(percentage=clicks*100/total);


 }

get_pivot(mtcars, cyl, mpg_cat)
# A tibble: 3 x 5
#    cyl clicks total     pvalue percentage
#  <dbl>  <dbl> <int>      <dbl>      <dbl>
#1     4     11    11 0.00000366      100  
#2     6      3     7 0.00000366       42.9
#3     8      0    14 0.00000366        0
于 2019-07-15T05:01:12.137 回答