我有数据，如下所示，每个let favoriteSong: String?
if let favoriteSong = favoriteSong {
print("My favorite song is \(favoriteSong)")
} else {
print("I don't have a favorite song")
}
每个id都有一个status。我想添加一个列time，在前一个prev_status中显示status的值。
time我可以为每个人set.seed(10); library(dplyr); library(data.table)
df <- data.table(time = sample(1:3, 20, T),
status = sample(letters[1:15], 20, T)
)[order(time)
][, id := 1:.N, by = time]
time status id
1: 1 j 1
2: 1 g 2
3: 1 k 3
4: 1 m 4
5: 1 d 5
6: 1 c 6
7: 1 m 7
8: 1 o 8
9: 2 m 1
10: 2 l 2
11: 2 l 3
12: 2 f 4
13: 2 i 5
14: 2 b 6
15: 2 n 7
16: 2 g 8
17: 2 l 9
18: 2 k 10
19: 3 f 1
20: 3 h 2
执行此操作，并使用下面的联接（time = 2）。
time有更好的方法可以为整个data.frame创建df1 <- df %>% filter(time == 1)
df2 <- df %>% filter(time == 2)
df2 %>%
left_join(df1, by = 'id') %>%
select(-time.y) %>%
rename(status = status.x,
prev_status = status.y,
time = time.x)
time status id prev_status
1 2 m 1 j
2 2 l 2 g
3 2 l 3 k
4 2 f 4 m
5 2 i 5 d
6 2 b 6 c
7 2 n 7 m
8 2 g 8 o
9 2 l 9 <NA>
10 2 k 10 <NA>
吗？我可以使用prev_status和dplyr解决方案（以及基础R）。
答案 0 :(得分：1)
lag()具有df %>%
arrange(time) %>%
group_by(id) %>%
mutate(prev_status=lag(status))
功能，可以轻松实现
thread::create