快速提问-
我有一个包含一些重复项的数据框，我想仅在type == 'c1'时将其删除。因此，举例来说，我只想在每个type == 'c1'的{{1}}中保持 1 行，有没有办法用dplyr做到这一点？我本打算使用id，但是却转了一圈。
case_when所需的输出-
sample_df <- data.frame(id = c(14129, 14129, 14129, 29102, 29102, 2191, 2191, 2191, 2191, 2192, 2192, 1912, 1912, 1912)
, date = c("2018-06-15 00:15:42","2018-10-08 12:44:44",
"2018-07-09 18:14:58", "2018-06-15 00:15:40",
"2018-06-15 00:19:42", "2018-10-15 08:17:47",
"2018-09-29 10:16:34", "2018-07-09 18:28:25",
"2018-07-09 18:28:25", "2018-07-09 18:20:32",
"2018-08-30 13:06:45", "2018-10-08 11:32:55",
"2018-10-05 11:32:55", "2018-10-08 09:09:56")
, color = c("blue", "blue", "green", "red", "red", "red", "green", "blue", "green", "purple", "blue", "blue", "red", "red")
, day = rep("c1", times = 14)
, happy = c(1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1))
sample_df$date <- as.POSIXct(sample_df$date)
sample_df_2 <- sample_df %>%
gather(key, type, color:day) %>%
mutate(happy = case_when(key == "color" ~ 0, TRUE ~ as.numeric(happy))) %>%
select(-key) %>%
arrange(id)
> sample_df_2
id date happy type
1 1912 2018-10-08 11:32:55 0 blue
2 1912 2018-10-05 11:32:55 0 red
3 1912 2018-10-08 09:09:56 0 red
4 1912 2018-10-08 11:32:55 0 c1
5 1912 2018-10-05 11:32:55 0 c1
6 1912 2018-10-08 09:09:56 1 c1
7 2191 2018-10-15 08:17:47 0 red
8 2191 2018-09-29 10:16:34 0 green
9 2191 2018-07-09 18:28:25 0 blue
10 2191 2018-07-09 18:28:25 0 green
11 2191 2018-10-15 08:17:47 1 c1
12 2191 2018-09-29 10:16:34 0 c1
13 2191 2018-07-09 18:28:25 1 c1
14 2191 2018-07-09 18:28:25 0 c1
15 2192 2018-07-09 18:20:32 0 purple
16 2192 2018-08-30 13:06:45 0 blue
17 2192 2018-07-09 18:20:32 0 c1
18 2192 2018-08-30 13:06:45 1 c1
19 14129 2018-06-15 00:15:42 0 blue
20 14129 2018-10-08 12:44:44 0 blue
21 14129 2018-07-09 18:14:58 0 green
22 14129 2018-06-15 00:15:42 1 c1
23 14129 2018-10-08 12:44:44 0 c1
24 14129 2018-07-09 18:14:58 0 c1
25 29102 2018-06-15 00:15:40 0 red
26 29102 2018-06-15 00:19:42 0 red
27 29102 2018-06-15 00:15:40 0 c1
28 29102 2018-06-15 00:19:42 1 c1
答案 0 :(得分：4)
基本R
sample_df_2[ !duplicated(sample_df_2[c("id","type")]) | sample_df_2$type != "c1", ]
# id date happy type
# 1 1912 2018-10-08 11:32:55 0 blue
# 2 1912 2018-10-05 11:32:55 0 red
# 3 1912 2018-10-08 09:09:56 0 red
# 4 1912 2018-10-08 11:32:55 0 c1
# 7 2191 2018-10-15 08:17:47 0 red
# 8 2191 2018-09-29 10:16:34 0 green
# 9 2191 2018-07-09 18:28:25 0 blue
# 10 2191 2018-07-09 18:28:25 0 green
# 11 2191 2018-10-15 08:17:47 1 c1
# 15 2192 2018-07-09 18:20:32 0 purple
# 16 2192 2018-08-30 13:06:45 0 blue
# 17 2192 2018-07-09 18:20:32 0 c1
# 19 14129 2018-06-15 00:15:42 0 blue
# 20 14129 2018-10-08 12:44:44 0 blue
# 21 14129 2018-07-09 18:14:58 0 green
# 22 14129 2018-06-15 00:15:42 1 c1
# 25 29102 2018-06-15 00:15:40 0 red
# 26 29102 2018-06-15 00:19:42 0 red
# 27 29102 2018-06-15 00:15:40 0 c1
Tidyverse：
library(dplyr)
sample_df_2 %>%
filter(!duplicated(cbind(id,type)) | type != "c1")
# id date happy type
# 1 1912 2018-10-08 11:32:55 0 blue
# 2 1912 2018-10-05 11:32:55 0 red
# 3 1912 2018-10-08 09:09:56 0 red
# 4 1912 2018-10-08 11:32:55 0 c1
# 5 2191 2018-10-15 08:17:47 0 red
# 6 2191 2018-09-29 10:16:34 0 green
# 7 2191 2018-07-09 18:28:25 0 blue
# 8 2191 2018-07-09 18:28:25 0 green
# 9 2191 2018-10-15 08:17:47 1 c1
# 10 2192 2018-07-09 18:20:32 0 purple
# 11 2192 2018-08-30 13:06:45 0 blue
# 12 2192 2018-07-09 18:20:32 0 c1
# 13 14129 2018-06-15 00:15:42 0 blue
# 14 14129 2018-10-08 12:44:44 0 blue
# 15 14129 2018-07-09 18:14:58 0 green
# 16 14129 2018-06-15 00:15:42 1 c1
# 17 29102 2018-06-15 00:15:40 0 red
# 18 29102 2018-06-15 00:19:42 0 red
# 19 29102 2018-06-15 00:15:40 0 c1
答案 1 :(得分：0)
使用dplyr：
sample_df_2 %>%
group_by(id) %>%
filter(!duplicated(type) | type!="c1")