我有一个像这样的R数据框：
blogger; date; word
joe; 2016-11-19; ears
paul; 2017-04-02; dilapidated
joe; 2016-11-21; night
joe; 2016-11-20; girl
paul; 2017-04-01; crumpled
paul; 2017-04-03; bellow
我想先按blogger然后按date对其进行排序。然后，我想为每个博客添加一个包含行号的列。
我有这个：
# make linenumbers
df <- df %>%
arrange(blogger,date) %>%
mutate(linenumber = row_number())
但显然，对于每个博主来说，这并不适用于亚麻1号。我得到的输出是：
blogger; date; word; linenumber
joe; 2016-11-19; ears; 1
joe; 2016-11-20; girl; 2
joe; 2016-11-21; night; 3
paul; 2017-04-01; crumpled; 4
paul; 2017-04-02; dilapidated; 5
paul; 2017-04-03; bellow; 6
但我想要的输出是：
blogger; date; word; linenumber
joe; 2016-11-19; ears; 1
joe; 2016-11-20; girl; 2
joe; 2016-11-21; night; 3
paul; 2017-04-01; crumpled; 1
paul; 2017-04-02; dilapidated; 2
paul; 2017-04-03; bellow; 3
答案 0 :(得分：2)
如果不需要对行进行排序，您可以执行以下操作：
d$linenumber <- ave(as.integer(d$blogger), d$blogger, FUN=seq_along)
# > d
# blogger date word linenumber
# 1 joe 2016-11-19 ears 1
# 2 paul 2017-04-02 dilapidated 1
# 3 joe 2016-11-21 night 2
# 4 joe 2016-11-20 girl 3
# 5 paul 2017-04-01 crumpled 2
# 6 paul 2017-04-03 bellow 3
d <- read.table(header=TRUE, sep=';', text=
"blogger; date; word
joe; 2016-11-19; ears
paul; 2017-04-02; dilapidated
joe; 2016-11-21; night
joe; 2016-11-20; girl
paul; 2017-04-01; crumpled
paul; 2017-04-03; bellow")
答案 1 :(得分：1)
在博客上试用group_by
df %>% arrange(blogger, date) %>%
group_by(blogger) %>%
mutate(linenumber = row_number()) %>%
ungroup()
答案 2 :(得分：1)
我们可以使用rowid
data.table）来完成此操作
library(data.table)
setDT(d)[, linenumber := rowid(blogger)]
d
# blogger date word linenumber
#1: joe 2016-11-19 ears 1
#2: paul 2017-04-02 dilapidated 1
#3: joe 2016-11-21 night 2
#4: joe 2016-11-20 girl 3
#5: paul 2017-04-01 crumpled 2
#6: paul 2017-04-03 bellow 3