一次聚合数字和字符答案

【问题标题】：Aggregate numerical and character at once一次聚合数字和字符
【发布时间】：2020-10-15 15:57:05
【问题描述】：

我需要一次聚合不同类的多个变量。

test<- data.frame (name = c("anna", "joe", "anna"), 
                   party = c("red", "blue", "red"),
                   text = c("hey there", "we ate an apple", "i took a walk"), 
                   numberofwords = c(2, 4, 4), 
                   score1 = 1:3, 
                   score2 = 4:6)

现在是这个样子

#   name    party      text           numberofwords score1 score2
#1  anna    red       hey there             2         1      4
#2  joe     blue    we ate an apple         4         2      5
#3  anna    red      i took a walk          4         3      6

我想根据姓名和派对聚合 score1、score2、numberofwords、文本变量。

想要的结果是：

#   name  party            text                  numberofwords score1 score2
#1  anna  red           hey there i took a walk       6           4      10
#2   joe  blue           we ate an apple              4           2      5

【问题讨论】：

标签： r dataframe dplyr tidyverse

【解决方案1】：

使用最新版本的dplyr 和across：

test %>%
  group_by(name, party) %>%
  summarize(
    across(text, paste, collapse = " "),
    across(where(is.numeric), sum)
  )
# # A tibble: 2 x 6
#   name  party text                    numberofwords score1 score2
#   <chr> <chr> <chr>                           <dbl>  <int>  <int>
# 1 anna  red   hey there i took a walk             6      4     10
# 2 joe   blue  we ate an apple                     4      2      5

旧版本，保持第一party值：

test %>%
  group_by(name) %>%
  summarize(
    across(party, first),
    across(text, paste, collapse = " "),
    across(where(is.numeric), sum)
  )
# # A tibble: 2 x 6
#   name  party text                    numberofwords score1 score2
#   <chr> <chr> <chr>                           <dbl>  <int>  <int>
# 1 anna  red   hey there i took a walk             6      4     10
# 2 joe   blue  we ate an apple                     4      2      5

【讨论】：

好，我还没习惯across。这比使用if else 更好。
一开始我总是被绊倒，因为我想使用summarize(across(..cols..), fun)，不明白为什么这个函数是inside。然后我意识到你可以在同一个summarize 中做多个acrosses，一切都点击了，我喜欢它。

【解决方案2】：

我们可以根据dplyr中每一列的类做一个条件summarise。

library(dplyr)

test %>% 
  mutate_at("text", as.character) %>% 
  group_by(name) %>% 
  summarise_all(list(~if(is.numeric(.)) sum(., na.rm = TRUE)  
                      else if(is.factor(.)) first(.) 
                      else paste(., collapse = " ")))

#> # A tibble: 2 x 6
#>   name  party text                    numberofwords score1 score2
#>   <fct> <fct> <chr>                           <dbl>  <int>  <int>
#> 1 anna  red   hey there i took a walk             6      4     10
#> 2 joe   blue  we ate an apple                     4      2      5

【讨论】：

【解决方案3】：

在base R 中，我们可以使用aggregate 和merge 来做到这一点

out1 <- aggregate(cbind(numberofwords, score1, score2) ~ name + party, test, sum)
out2 <- aggregate(text ~ name + party, test, paste, collapse=' ')
merge(out1, out2)

-输出

# name party numberofwords score1 score2                    text
#1 anna   red             6      4     10 hey there i took a walk
#2  joe  blue             4      2      5         we ate an apple

【讨论】：

【解决方案4】：

尝试这种方法，首先聚合文本变量，然后聚合连续变量。之后合并所有。这里使用dplyr的代码：

library(dplyr)
#Data
test<- data.frame (name = c("anna", "joe", "anna"), 
                   party =c("red", "blue", "red"),
                   text = c("hey there", "we ate an apple", "i took a walk"),
                   numberofwords = c(2,4,4),
                   score1 = 1:3, score2= 4:6,stringsAsFactors = F)
#First aggregate text after that aggregate continuous variables and merge
new <- test %>% 
  group_by(name,party) %>% summarise(text=paste0(text,collapse = ' ')) %>%
  left_join(
    test %>% select(-text) %>%
      group_by(name,party) %>%
      summarise_all(sum,na.rm=T)
  )

输出：

# A tibble: 2 x 6
# Groups:   name [2]
  name  party text                    numberofwords score1 score2
  <chr> <chr> <chr>                           <dbl>  <int>  <int>
1 anna  red   hey there i took a walk             6      4     10
2 joe   blue  we ate an apple                     4      2      5

【讨论】：