在将 df 和列作为输入的函数上使用 lapply - 按百分比重新排序

Question

我有一个教育数据框，其中包含大约 40 个特征，我想将这些特征与名为 'Tutor' 的单个列进行交叉制表并生成百分比。然后我希望该行从每一列的 "Yes" 百分比属性

降序排列

library(purrr)
library(tidyverse)
library(janitor)
toydat <- data.frame(ID = c(1:12), Learning_mode = rep(c("External","Internal"), times =6), 
                     Subject = rep(c("Maths","English","Chemistry", "Physics"), each = 3 ),
                      Grade = runif(12, 0,100), Tutor = rep(c("Yes","No"), times = 6))


toydat %>%
  select_if(~is.factor(.) |is.character(.)) %>%
  select (-Tutor) %>%
  imap(.f = ~janitor::tabyl(toydat, !!sym(.y), Tutor, sort = TRUE)) %>%
  adorn_totals(c("row", "col")) %>%
  adorn_percentages("row") %>% 
  adorn_pct_formatting(rounding = "half up", digits = 0) %>%
  adorn_ns() %>%
  adorn_title("combined")

我最终会将其放入 Rmarkdown 报告中，该报告将表明功能如何影响参与辅导。输出是这样的：

$Learning_mode
 Learning_mode/Tutor       No      Yes     Total
            External   0% (0) 100% (6) 100%  (6)
            Internal 100% (6)   0% (0) 100%  (6)
               Total  50% (6)  50% (6) 100% (12)

$Subject
 Subject/Tutor      No     Yes     Total
     Chemistry 33% (1) 67% (2) 100%  (3)
       English 67% (2) 33% (1) 100%  (3)
         Maths 33% (1) 67% (2) 100%  (3)
       Physics 67% (2) 33% (1) 100%  (3)
         Total 50% (6) 50% (6) 100% (12)

但我希望它像这样 - 按每个属性的 "Yes" 百分比列下降：

$Learning_mode
 Learning_mode/Tutor       No      Yes     Total
            External   0% (0) 100% (6) 100%  (6)
            Internal 100% (6)   0% (0) 100%  (6)
               Total  50% (6)  50% (6) 100% (12)

$Subject
 Subject/Tutor      No     Yes     Total
     Chemistry 33% (1) 67% (2) 100%  (3)
         Maths 33% (1) 67% (2) 100%  (3)
       English 67% (2) 33% (1) 100%  (3)
       Physics 67% (2) 33% (1) 100%  (3)  
         Total 50% (6) 50% (6) 100% (12)

我尝试制作一个函数，根据针对 Tutor 的百分比重新调整每个字符属性，并且它有效，但我无法使用 lapply 将其应用于我的数据框。要重新调整的函数：

newlevels <-function(x){
  tab <- table(toydat[[x]], toydat$Tutor)
  tab <-as.data.frame(prop.table(tab, 1))
  tab <-as.data.frame(tab)
  tab <- tab %>% filter(Var2 =="Yes")
  tab <-  tab[order(-tab$Freq),]
  ordered <- as.character(tab$Var1)
}

toydat$Subject <- factor(toydat$Subject, level = newlevels("Subject"))

如何应用 newlevels 函数对数据框中的所有非数字列重新排序。目前它不工作：

cols <- c("Subject","Learning_mode")
toydat[cols] <-lapply(toydat[cols], function(x) {factor(x), levels = newlevels(x)})

我收到以下错误：

Error: unexpected ',' in "toydat[cols] <-lapply(toydat[cols], function(x) {factor(x),"

Answer 1

可能最容易在管道中使用 arrange()：

toydat %>%
  select_if(~is.factor(.) |is.character(.)) %>%
  select (-Tutor) %>%
  imap(.f = ~janitor::tabyl(toydat, !!sym(.y), Tutor, sort = TRUE)) %>%
  map(~arrange(.x, desc(Yes))) %>%
  adorn_totals(c("row", "col")) %>%
  adorn_percentages("row") %>% 
  adorn_pct_formatting(rounding = "half up", digits = 0) %>%
  adorn_ns() %>%
  adorn_title("combined") 

$Learning_mode
 Learning_mode/Tutor       No      Yes     Total
            External   0% (0) 100% (6) 100%  (6)
            Internal 100% (6)   0% (0) 100%  (6)
               Total  50% (6)  50% (6) 100% (12)

$Subject
 Subject/Tutor      No     Yes     Total
     Chemistry 33% (1) 67% (2) 100%  (3)
         Maths 33% (1) 67% (2) 100%  (3)
       English 67% (2) 33% (1) 100%  (3)
       Physics 67% (2) 33% (1) 100%  (3)
         Total 50% (6) 50% (6) 100% (12)

在将 df 和列作为输入的函数上使用 lapply - 按百分比重新排序

Use lapply on a function that takes df and column as inputs - reorder by percentage

r

dplyr

janitor