如何将列名称传递给函数dplyr

Question

我正在尝试创建一个简单的汇总函数来加速报告多列数据以用于R Markdown文件.

var1是数据的分类列,t_var是表示数据四分之一的整数,dt是完整数据.

summarise_data_categorical <- function(var1, t_var, dt){

  print(var1)
  print(t_var)

  #Select the columns to aggregate
  group_func <- dt %>% 
    select(one_of(t_var, var1)) %>%
    group_by(t_var,var1)

  #create simple count summary
  count_table <- group_func %>%
    summarise(count = n()) %>%
    spread(t_var, count)

  #create a frequency version of the same table...
  freq <- dt %>%
    select(t_var, var1) %>%
    group_by(t_var,var1) %>%
    summarise(count = n()) %>%
    mutate(freq = round(count / sum(count),3)*100) %>%
    select(-count)

  #Present that table
  freq_table <- freq %>%
    spread(t_var, freq)

  #Create the chart to do the same thing..
  freq_chart <- freq %>%
    ggplot()+
    geom_line(mapping=aes(x=t_var, y = freq, colour=var1))

  #Compile outputs as a list
  results <- list(count_table, freq_table, freq_chart)

  #Return list
  results

}

说我有一个框架:

fr <- data.frame(lets = sample(LETTERS, 100, replace=TRUE),
           `quarter type` = sample(1:4, 100, replace=TRUE))

如果我运行该函数,那么:

summarise_data_categorical("lets", "quarter type", fr)

初始产出很有希望:

[1] "lets"
[1] "quarter type"

(注意:在尝试重新创建数据时,出于某种原因我也会收到警告:

未知变量:quarter type,虽然这不会出现在我的原始数据中)

主要是我得到一个错误:

Error in resolve_vars(new_groups, tbl_vars(.data)) : unknown variable to group by : t_var

来自Python,我仍然对如何引用列感到困惑.有人可以解释我如何解决我的错误吗？

Answer 1

我们可以使用devel版本的新版本dplyr(很快将在0.6.0发布)

summarise_data_categorical <- function(var1, t_var, dt){

  var1 <- enquo(var1)
  t_var <- enquo(t_var)
  v1 <- quo_name(var1)
  v2 <- quo_name(t_var) 

  dt %>%
    select(one_of(v1, v2)) %>%
    group_by(!!t_var, !!var1) %>%
    summarise(count = n()) 

}
summarise_data_categorical(lets, quartertype, fr)
#Source: local data frame [65 x 3]
#Groups: quartertype [?]

#   quartertype   lets count
#         <int> <fctr> <int>
#1            1      A     1
#2            1      F     2
#3            1      G     2
#4            1      H     1
#5            1      I     1
#6            1      J     4
#7            1      M     3
#8            1      N     1
#9            1      P     1
#10           1      S     5
# ... with 55 more rows

在enquo做了类似的功能substitute,从base R采取的输入参数,并将其转换为quosures.在one_of需要一个字符串参数,所以quosures可以转换为字符串quo_name.在group_by/summarise/mutate等内部,我们可以通过unquote(UQ或!!)评估quosure

在quosures似乎是工作的罚款dplyr,虽然我们在实现相同的有一定的难度tidyr的功能.以下代码应适用于完整代码

 summarise_data_categorical <- function(var1, t_var, dt){

  var1 <- enquo(var1)
  t_var <- enquo(t_var)

  v1 <- quo_name(var1)
  v2 <- quo_name(t_var) 

  Summ_func <- dt %>%
                    select(one_of(v1, v2)) %>%
                  group_by(!!t_var, !!var1) %>%
                    summarise(count = n())

   count_table <- Summ_func %>%
                  spread_(v2, "count") 

   freq <-  Summ_func %>%
                  mutate(freq = round(count / sum(count),3)*100) %>%
              select(-count)

   freq_table <- freq %>%
                    spread_(v2, "freq")

   freq_chart <- freq %>%
             ggplot()+
               geom_line(mapping=aes_string(x= v2 , y = "freq", colour= v1)) 

   results <- list(count_table, freq_table, freq_chart)
   results

    }
summarise_data_categorical(lets, quartertype, fr)
#[[1]]
# A tibble: 24 × 5
#     lets   `1`   `2`   `3`   `4`
#*  <fctr> <int> <int> <int> <int>
#1       A    NA    NA     1     2
#2       B     2    NA    NA     1
#3       C     1     5     1     2
#4       E     1     1    NA    NA
#5       G    NA     1     2     2
#6       H     1    NA     1     1
#7       I    NA     1     1     2
#8       J     2     1     1     1
#9       K     1     1     2     1
#10      L    NA     2    NA    NA
# ... with 14 more rows

#[[2]]
# A tibble: 24 × 5
#     lets   `1`   `2`   `3`   `4`
#*  <fctr> <dbl> <dbl> <dbl> <dbl>
#1       A    NA    NA   3.1   9.5
#2       B   8.7    NA    NA   4.8
#3       C   4.3  20.8   3.1   9.5
#4       E   4.3   4.2    NA    NA
#5       G    NA   4.2   6.2   9.5
#6       H   4.3    NA   3.1   4.8
#7       I    NA   4.2   3.1   9.5
#8       J   8.7   4.2   3.1   4.8
#9       K   4.3   4.2   6.2   4.8
#10      L    NA   8.3    NA    NA
## ... with 14 more rows

#[[3]]