Question

我有一个数据框，排列如下：

df <- structure(list(NAME1=  c("AAA","CCC","BBB","BBB"), 
                 NAME2    =  c("BBB", "AAA","DDD","AAA"),
                 AMT      = c(40,20,10,50)),.Names=c("NAME1","NAME2","AMT"), 
                 row.names = c("1", "2", "3", "4"), class =("data.frame"))

我想创建一个ID变量作为字符变量NAME1和NAME2的组合，无论顺序如何（即AAA BBB与BBB AAA相同）并总结AMT。

这就是我想要的结果：

df <- structure(list(NAME1 =  c("AAA","CCC", "BBB"), 
                 NAME2     =  c("BBB", "AAA","DDD"),
                 AMT       =  c(90,20,10),
                 ID        =  c(1,2,3)),
                 .Names    =  c("NAME1","NAME2","AMT","ID"), 
                 row.names =  c("1", "2", "3"), class =("data.frame"))

您的意见将非常感谢。

Answer 1

您可以创建两个新的分组变量，这些变量按行对值进行排序，以便AAA, BBB和BBB, AAA处理相同（因为它们按相同的顺序排列）。之后，分组操作非常简单。我选择使用data.table：

library(data.table)

df[,c("NAME1_o","NAME2_o")] <- t(apply(cbind(df$NAME1, df$NAME2), 1, function(x) x[order(x)]))
setDT(df)[, .(AMT = sum(AMT), ID = .GRP), by = .(NAME1_o, NAME2_o)]

#   NAME1_o NAME2_o AMT ID
#1:     AAA     BBB  90  1
#2:     AAA     CCC  20  2
#3:     BBB     DDD  10  3

Answer 2

使用dplyr动词：

df %>% 
   rowwise() %>% 
   mutate(PAIR=paste0(sort(c(NAME1,NAME2)),collapse=" ")) %>% 
   group_by(PAIR) %>% 
   summarise(AMT=sum(AMT)) %>%
   mutate(ID=row_number()) %>%
   separate(PAIR, " ", into=c("NAME1","NAME2"))

  NAME1 NAME2   AMT    ID
1   AAA   BBB    90     1
2   AAA   CCC    20     2
3   BBB   DDD    10     3

基于变量组合的ID变量

2 个答案: