如何按月重新排序数据的行？

Question

我的目标是每年按照月份的顺序(从12月、11月、10月等到1月、2月、3月等)重新排序数据的行。

目前，我有以下数据

            Date            Title  Article
1   1 December 2000 PRESS CONFERENCE Article1
2   1 November 2000 PRESS CONFERENCE Article2
3    1 October 2000 PRESS CONFERENCE Article3
4  1 September 2000 PRESS CONFERENCE Article1
5     1 August 2000 PRESS CONFERENCE Article2
6       1 July 2000 PRESS CONFERENCE Article3
7       1 June 2000 PRESS CONFERENCE Article1
8             1 May PRESS CONFERENCE Article2
9      1 April 2000 PRESS CONFERENCE Article3
10     1 March 2000 PRESS CONFERENCE Article1
11  1 February 2000 PRESS CONFERENCE Article2
12   1 January 2000 PRESS CONFERENCE Article3
13   1 December 2001 PRESS CONFERENCE Article1
14   1 November 2001 PRESS CONFERENCE Article2
15   1 October 2001 PRESS CONFERENCE Article3
.
.
.
n    1 January 2001 PRESS CONFERENCE Article

这种结构从2000年一直延续到2019年。现在，我想扭转这种局面，从2000年到2019年，每年都有这样的情况：

Date            Title  Article
1   1 January 2000 PRESS CONFERENCE Article1
2   1 February 2000 PRESS CONFERENCE Article2
3    1 March 2000 PRESS CONFERENCE Article3
4  1 April 2000 PRESS CONFERENCE Article1
5     1 May 2000 PRESS CONFERENCE Article2
6       1 June 2000 PRESS CONFERENCE Article3
7       1 July 2000 PRESS CONFERENCE Article1
8       1 August 2000 PRESS CONFERENCE Article2
9      1 September 2000 PRESS CONFERENCE Article3
10     1 October 2000 PRESS CONFERENCE Article1
11  1 November 2000 PRESS CONFERENCE Article2
12   1 December 2000 PRESS CONFERENCE Article3
13   1 January 2001 PRESS CONFERENCE Article1
14   1 February 2001 PRESS CONFERENCE Article2
15   1 March 2001 PRESS CONFERENCE Article3
.
.
.
n    1 December 2001 PRESS CONFERENCE Article

有人能帮我吗？我该怎么做呢？

非常感谢!

Answer 1

将Date转换为正确的格式，然后从dplyr中使用arrange

library(dplyr)
df %>% arrange(lubridate::dmy(Date))

             Date Title
1  1 October 2000 PRESS
2 1 November 2000 PRESS
3 1 December 2000 PRESS
4  1 October 2001 PRESS
5 1 November 2001 PRESS
6 1 December 2001 PRESS

使用基R

df[order(as.Date(df$Date, "%d %B %Y")),]

数据

#Providing small dataset using `dput` will make it easier for other to help
df <- structure(list(Date = c("1 December 2000", "1 November 2000", 
"1 October 2000", "1 December 2001", "1 November 2001", "1 October 2001"
), Title = c("PRESS", "PRESS", "PRESS", "PRESS", "PRESS", "PRESS"
)), class = "data.frame", row.names = c("1", "2", "3", "13", 
"14", "15"))

Answer 2

我们可以使用来自anydate的anytime

library(dplyr)
library(anytime)
df %>%
    arrange(anydate(Date))