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我正在尝试使用标准评估dplyr来计算百分比作为两个分组变量的函数。问题出在我的mutate_ statement.

这是一个数据集：

structure(list(
    var1 = structure(c(2L, 1L, 1L, 2L, 1L, 2L, 1L, 
    2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 
    2L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 2L, 
    2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 
    2L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 
    1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 
    2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 1L
    ), 
    .Label = c("No", "Yes"), class = "factor"), 
    var2 = structure(c(2L, 2L, 1L, 2L, 
    2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 
    1L, 2L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 
    1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 
    2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 
    2L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 
    1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L
    ), 
    .Label = c("Female", "Male"), class = "factor")), 
    .Names = c("var1", "var2"), row.names = c(NA, -100L), class = "data.frame")

这是我正在使用的代码：

for_plots = function(data, var1, var2){
  grouped_data = data %>% group_by_(var1, var2) %>% 
  summarise_(n_in_group = ~n()) %>% 
  mutate_(.dots = setNames(list(
    interp(quote(n_in_group / sum(n_in_group, na.rm = TRUE) * 100),
           n_in_group = as.name(n_in_group)))
    ))
  return(grouped_data)
}

当我运行代码时，我收到一个错误：

setNames(list(interp(quote(n_in_group/sum(n_in_group, na.rm = TRUE) * ) 中的错误：缺少参数“nm”，没有默认值)

有什么想法吗？

4

1 回答 1

2

以下是基于@Frank 回复的一些代码： 

for_plots = function(data, var1, var2) { 
   grouped_data = data %>% group_by_(var1, var2) %>% 
     summarise_(n_in_group = ~n()) %>% 
     mutate(percent = (n_in_group / sum(n_in_group, na.rm = TRUE)) * 100) 
   return(grouped_data) 
}
于 2016-06-15T14:26:46.680 回答