我想重新排列我的数据顺序.我的数据集如下所示:
Animal<-c("bird","Bird ","Dog","Cat F","Lion","Lion","Lion","dog","Horse","cat", "Lion")
A_date<-c("02-08-2020","20-06-2018","01-01-2015","10-07-2021","20-06-2018","15-08-2019","05-08-2013","20-06-2010","15-11-2016","22-03-2022","15-05-2019")
ID<-c("T1", "T1","T1","T2","T2","T3","T3","T4","T4","T15","T15")
Mydata<-data.frame(Animal, A_date,ID)
Animal A_date ID
bird 02-08-2020 T1
Bird 20-06-2018 T1
Dog 01-01-2015 T1
Cat F 10-07-2021 T2
Lion 20-06-2018 T2
Lion 15-08-2019 T3
lion 05-08-2013 T3
dog 20-06-2010 T4
Horse 15-11-2016 T4
cat 22-03-2022 T15
Lion 15-05-2019 T15
我使用下面的代码删除了重复的行,这样就只剩下带有最新日期pr.ID的行.
library(lubridate)
library(dplyr)
Mydata %>%
mutate(Animal = trimws(toupper(Animal)), A_date = lubridate::dmy(A_date)) %>%
group_by(ID, Animal) %>%
arrange(ID, Animal, -desc(A_date)) %>%
slice(1)
输出如下所示:
A tibble: 9 x 3
# Groups: ID, Animal [9]
Animal A_date ID
<chr> <date> <chr>
1 BIRD 2018-06-20 T1
2 DOG 2015-01-01 T1
3 CAT 2022-03-22 T15
4 LION 2019-05-15 T15
5 CAT F 2021-07-10 T2
6 LION 2018-06-20 T2
7 LION 2013-08-05 T3
8 DOG 2010-06-20 T4
9 HORSE 2016-11-15 T4
我想更改订单,以便订单应该是最高ID、最新日期的pr.ID.
Animal A_date ID
<chr> <date> <chr>
CAT 2022-03-22 T15
LION 2019-05-15 T15
HORSE 2016-11-15 T4
DOG 2010-06-20 T4
LION 2019-08-15 T3
CAT F 2021-07-10 T2
LION 2018-06-20 T2
BIRD 2020-08-02 T1
DOG 2015-01-01 T1
关于如何得到这个结果有什么建议吗?