Question

library(tidyverse)
library(ggplot2) for diamonds dataset

我无法让我的功能发挥作用。在本例中，我使用钻石数据集格式ggplot2尝试做的是dplyr :: group_by＆＃34; cut＆＃34;和＆＃34; color＆＃34;，然后dplyr :: summarize来获取计数。我使用rlang和purrr将两个计数摘要输出到列表中，然后重命名其中一个列，并用dplyr :: map_df绑定它们。最后，我想重新排序＆＃34; Cut＆＃34;列基于另一个名为＆＃34; Order＆＃34;的向量。该函数有效，直到我尝试合并行重新排序...

这可能对这些数据没有意义，但这只是一个例子，它对我的真实数据有意义。

无论如何，下面的代码有效......

Groups<-list("cut","color")

 Groups<-Groups%>%
 map_df(function(group){

     syms<-syms(group)

     diamonds%>%
         group_by(!!!syms)%>%
         summarise(Count=n())%>%
         set_names(c("Cut","Count"))
 })

接下来，我想根据＆＃34;订单＆＃34;重新排序行。矢量，也有效。

Order<-c("Good","Very Good","Premium","Ideal","Fair","E","F","G","D","H","J","I")

Groups%>%slice(match(Order, Cut))

但是，这就是我被困的地方。我试图在一个功能中完成所有这些，但它似乎不起作用。我觉得我错过了一些小事......

Fun<-function(df){

Order<-c("Good","Very Good","Premium","Ideal","Fair","E","F","G","D","H","J","I")

Groups<-list("cut","color")

 Groups<-Groups%>%
 map_df(function(group){

     syms<-syms(group)

     df%>%
         group_by(!!!syms)%>%
         summarise(Count=n())%>%
         set_names(c("Cut","Count"))%>%
         slice(match(Order,Cut))
return(df)
})
}

这是另一种尝试......

Fun<-function(df){

Order<-c("Good","Very Good","Premium","Ideal","Fair","E","F","G","D","H","J","I")

Groups<-list("cut","color")

 Groups<-Groups%>%
 map_df(function(group){

     syms<-syms(group)

     df%>%
         group_by(!!!syms)%>%
         summarise(Count=n())%>%
         set_names(c("Cut","Count"))

df<-df%>%slice(match(Order,Cut))
return(df)
})
}

我在这里缺少什么？

Answer 1

我们不需要在循环中应用syms。它可以采用长度大于1的向量/列表并将其转换为符号。因此，循环遍历syms，然后使用map在每个符号对象上执行group_by

Fun<-function(df){

Order<-c("Good","Very Good","Premium","Ideal","Fair","E","F","G","D","H","J","I")

Groups<-list("cut","color")

Groups %>%
       syms %>%
       map_df(~ df %>%
               group_by(!!!  .x) %>%
               summarise(Count=n()) %>%
               set_names(c("Cut","Count")) %>%
               slice(match(Order,Cut)) #%>%                    
               #mutate(Cut = as.character(Cut)) 
               #to avoid the warning coercion of factor to character 


      )




}

Fun(diamonds)
# A tibble: 12 x 2
#   Cut       Count
#   <chr>     <int>
# 1 Good       4906
# 2 Very Good 12082
# 3 Premium   13791
# 4 Ideal     21551
# 5 Fair       1610
# 6 E          9797
# 7 F          9542
# 8 G         11292
# 9 D          6775
#10 H          8304
#11 J          2808
#12 I          5422

Answer 2

您在Fun的第一次尝试工作，但结果已分配给Group变量且未返回。请尝试以下

Fun<-function(df){

Order<-c("Good","Very Good","Premium","Ideal","Fair","E","F","G","D","H","J","I")

Groups<-list("cut","color")

 Groups%>%
 map_df(function(group){

     syms<-syms(group)

     df%>%
         group_by(!!!syms)%>%
         summarise(Count=n())%>%
         set_names(c("Cut","Count"))%>%
         slice(match(Order,Cut))
return(df)
})
}

Fun(diamonds)

Answer 3

可能纠正问题。为简单起见，我创建了一个temp_df变量并返回相同的内容。

Fun<-function(df){

  Order<-c("Good","Very Good","Premium","Ideal","Fair","E","F","G","D","H","J","I")

  Groups<-list("cut","color")

  Groups<-Groups%>%
    map_df(function(group){

      syms<-syms(group)

      temp <- df%>%
        group_by(!!!syms)%>%
        summarise(Count=n())%>%
        set_names(c("Cut","Count"))
    })

  temp_df <- Groups%>%slice(match(Order, Cut))
  return(temp_df)
}

> x <- Fun(diamonds)
> x
# A tibble: 12 x 2
   Cut       Count
   <chr>     <int>
 1 Good       4906
 2 Very Good 12082
 3 Premium   13791
 4 Ideal     21551
 5 Fair       1610
 6 E          9797
 7 F          9542
 8 G         11292
 9 D          6775
10 H          8304
11 J          2808
12 I          5422

在如何根据我的函数中的Vector合并数据帧行的重新排序时遇到了困难

3 个答案: