使用基本 R:
df = data.frame(ID = c("a","a","b","b","b","c","d","d"),
day = c("1","2","3","4","5","6","7","8"),
year = c(2016,2017,2017,2016,2017,2016,2017,2016),
stringsAsFactors = F)
> df
ID day year
1 a 1 2016
2 a 2 2017
3 b 3 2017
4 b 4 2016
5 b 5 2017
6 c 6 2016
7 d 7 2017
8 d 8 2016
做:
z = aggregate(df[,2:3],
by = list(id = df$ID),
function(x){ paste0(x, collapse = "/") }
)
结果:
> z
id day year
1 a 1/2 2016/2017
2 b 3/4/5 2017/2016/2017
3 c 6 2016
4 d 7/8 2017/2016
编辑
如果您想避免“崩溃” NA,请执行以下操作:
z = aggregate(df[,2:3],
by = list(id = df$ID),
function(x){ paste0(x[!is.na(x)],collapse = "/") })
对于像这样的数据框:
> df
ID day year
1 a 1 2016
2 a 2 NA
3 b 3 2017
4 b 4 2016
5 b <NA> 2017
6 c 6 2016
7 d 7 2017
8 d 8 2016
结果是:
> z
id day year
1 a 1/2 2016
2 b 3/4 2017/2016/2017
3 c 6 2016
4 d 7/8 2017/2016