基于另一列重新排序的 Forcats 解决方案

Question

好的，我知道 fct_reorder() 允许您根据另一列对因子重新排序，但据我所知，您必须提供一些对第二列进行运算的函数（均值、中位数等）列以了解如何对列进行排序。但是，如果您有另一列按照您希望的方式对列进行排序，那会怎样 leveled/ordered 原样 ?

比如我有一个列ACADEMIC_PERIOD_DESC，用英文给出学年："Fall 2019," "Spring 2020,"等，我有一个对应的列，ACADEMIC_PERIOD ，那是对应学年的数字代码：“201940”，“202020”等。这是我要ACADEMIC_PERIOD_DESC被拉平的栏目。

数据

df <- structure(list(ACADEMIC_PERIOD = c("200810", "200820", "200830", 
"200840", "200910", "200920", "200930", "200940", "201010", "201020"
), ACADEMIC_PERIOD_DESC = structure(1:10, .Label = c("J-Term 2008", 
"Spring 2008", "Summer 2008", "Fall 2008", "J-Term 2009", "Spring 2009", 
"Summer 2009", "Fall 2009", "J-Term 2010", "Spring 2010", "Summer 2010", 
"Fall 2010", "J-Term 2011", "Spring 2011", "Summer 2011", "Fall 2011", 
"J-Term 2012", "Spring 2012", "Summer 2012", "Fall 2012", "J-Term 2013", 
"Spring 2013", "Summer 2013", "Fall 2013", "Spring 2014", "Summer 2014", 
"Fall 2014", "J-Term 2015", "Spring 2015", "Summer 2015", "Fall 2015", 
"J-Term 2016", "Spring 2016", "Summer 2016", "Fall 2016", "J-Term 2017", 
"Spring 2017", "Summer 2017", "Fall 2017", "J-Term 2018", "Spring 2018", 
"Summer 2018", "Fall 2018", "J-Term 2019", "Spring 2019", "Summer 2019", 
"Fall 2019", "J-Term 2020", "Spring 2020"), class = "factor")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L))

即使不必要地应用中位数，我是否应该只执行以下操作？

df %>% 
    mutate(ACADEMIC_PERIOD_DESC = fct_reorder(ACADEMIC_PERIOD_DESC, as.integer(ACADEMIC_PERIOD)))

我也知道我可以像这样使用 base R:

df$ACADEMIC_PERIOD_DESC <- reorder(df$ACADEMIC_PERIOD_DESC, df$ACADEMIC_PERIOD)

有没有更优雅的forcats/tidyverse解决方案？我是不是漏掉了什么？

谢谢！

Answer 1

我们可以将 .fun 从默认值 median 更改为 I，即按原样获取值

library(dplyr)
library(forcats)
df %>% 
   mutate(ACADEMIC_PERIOD_DESC = fct_reorder(ACADEMIC_PERIOD_DESC, 
           as.integer(ACADEMIC_PERIOD), .fun = I))

基于另一列重新排序的 Forcats 解决方案

Forcats solution for reordering based on another column

r

tidyverse

forcats

数据