r dplyr非标准评估 - 在函数

时间:2017-08-03 07:25:35

标签: r function ggplot2 dplyr non-standard-evaluation

我已阅读http://dplyr.tidyverse.org/articles/programming.html关于dplyr中的非标准评估,但仍然无法使用。

plot_column< - " columnA"

raw_data %>%
    group_by(.dots = plot_column) %>%
    summarise (percentage = mean(columnB)) %>%
    filter(percentage > 0) %>%
    arrange(percentage) %>%
    # mutate(!!plot_column := factor(!!plot_column, !!plot_column))%>%
    ggplot() + aes_string(x=plot_column, y="percentage")  +
  geom_bar(stat="identity", width = 0.5) +
  coord_flip()
禁用mutate语句时,

正常工作。但是,当启用它以按高度排序时,只返回一个条。

如何将上述语句转换为函数/使用变量,但仍然按照大小排列多个条形图。

数据集的示例可能是:

columnA,columnB
a, 1
a, 0.4
a, 0.3
b, 0.5

修改

样本:

mtcars %>%
  group_by(mpg) %>%
  summarise (mean_col = mean(cyl)) %>%
  filter(mean_col > 0) %>%
  arrange(mean_col) %>%
  mutate(mpg := factor(mpg, mpg))%>%
    ggplot() + aes(x=mpg, y=mean_col)  +
  geom_bar(stat="identity")
  coord_flip()

将输出有序的条形图。 如何将其包装到可以替换列的函数中,并且我得到多个条形图?

1 个答案:

答案 0 :(得分:2)

This works with dplyr 0.7.0 and ggplot 2.2.1:

rm(list = ls())
library(ggplot2)
library(dplyr)
raw_data <- tibble(columnA = c("a", "a", "b", "b"), columnB = c(1, 0.4, 0.3, 0.5))

plot_col <- function(df, plot_column, val_column){

  pc <- enquo(plot_column)
  vc <- enquo(val_column)
  pc_name <- quo_name(pc) # generate a name from the enquoted statement!

  df <- df %>%
   group_by(!!pc) %>%
   summarise (percentage = mean(!!vc)) %>%
   filter(percentage > 0) %>%
   arrange(percentage) %>%
   mutate(!!pc_name := factor(!!pc, !!pc)) # insert pc_name here!

  ggplot(df) + aes_(y = ~percentage, x = substitute(plot_column)) +
    geom_bar(stat="identity", width = 0.5) +
    coord_flip()
}
plot_col(raw_data, columnA, columnB)
plot_col(mtcars, mpg, cyl)

Problem I ran into was kind of that ggplot and dplyr use different kinds of non-standard evaluation. I got the answer at this question: Creating a function using ggplot2

编辑:参数化了值列(例如columnB / cyl)并添加了mtcars示例。