聚合 R 中每组的唯一字符值

Question

我想按组汇总特定年份的所有关键字。

我有一个如下所示的数据集：

我的主要问题是 Words 列可以在 1 到 52 之间变化！我正在考虑将此列拆分为不同的列，然后使用 group_by。但现在我不确定如何进行。

Answer 1

我们可以把'Words'拆分成vector的list，unnest拆分成'long'格式，去掉重复行，按[分组=22=], 'UID', paste 把 'Words' 变成一个字符串

library(dplyr)
df1 %>% 
    mutate(Words = strsplit(Words, ",")) %>% 
    unnest %>% 
    distinct(Year, UID, Words) %>% 
    group_by(UID, Year) %>% 
    summarise(Words = toString(Words))
# A tibble: 4 x 3
# Groups:   UID [?]
#    UID  Year Words                                    
#  <dbl> <dbl> <chr>                                    
#1    10  2009 ABC, CDEFGH, LMX, ABCD, IJKLM, PQRS, EFGH
#2    11  2010 BDFC, CDE, PQRS, ACCA, IJKLM             
#3    12  2010 ABCD, CADDE                              
#4    12  2011 ABC, CDE, EFGH

数据

df1 <- structure(list(ID = c(1, 2, 3, 4, 5, 6, 5), Year = c(2011, 2011, 
2010, 2010, 2009, 2010, 2009), UID = c(12, 12, 11, 12, 10, 11, 
10), Words = c("ABC,CDE", "EFGH,CDE", "BDFC,CDE,PQRS", "ABCD,CADDE", 
"ABC,CDEFGH,LMX,ABCD,IJKLM,PQRS", "BDFC,ACCA,IJKLM", "EFGH")),
 class = "data.frame", row.names = c(NA, -7L))

Answer 2

基础 R 方法 aggregate:

df <- data.frame(
    id = c(1:5, 6, 5),
    year = c(2011, 2011, 2010, 2010, 2009, 2010, 2009),
    uid = c(12, 12, 11, 12, 10, 11, 10),
    words = c("abc,cde", "efgh,cde", "bdfc,cde,pqrs", "abcd,cadde", "abc,cdefgh,lmx,abcd,ijklm,pqrs","bdfc,acca,ijklm", "efgh"),
    stringsAsFactors = FALSE
)

aggregate(df["words"], df[,c("year", "uid")], function(x) paste0(unique(unlist(strsplit(x, ","))), collapse=","))

聚合 R 中每组的唯一字符值

Aggregating unique character values per group in R

group-by

r

uniqueidentifier

dplyr

数据