将2个连续但不规则的时序数据集与重叠的行合并,从而消除重复的行

时间:2019-02-19 16:22:24

标签: r

使用R,我试图使用相同的字段但具有重叠的行来组合2个连续但不规则的时序数据集。也就是说,一些相同的交易记录同时出现在两个数据集中,我想消除重叠的行。

由于时间间隔是不规则的,因此我在每个数据集中可能有有效的相同行。对于我的示例数据集,我想将数据集1中的第1到12行与数据集2中的第6到11行合并,以获得所需的结果。在此示例中,很明显,数据集2的第1到5行与数据集1的第8到12行相同。我尝试使用unique()函数,但它也消除了相同的有效行。关于如何解决这一难题的任何想法?

数据集1

1  2019-02-19 15:17:14 25886    1                           
2  2019-02-19 15:17:14 25886    1                           
3  2019-02-19 15:17:15 25885    1                           
4  2019-02-19 15:17:16 25886    2                           
5  2019-02-19 15:17:16 25886    1                           
6  2019-02-19 15:17:16 25886    2                           
7  2019-02-19 15:17:16 25886    1                           
8  2019-02-19 15:17:18 25885    4                           
9  2019-02-19 15:17:19 25885    1  
10 2019-02-19 15:17:19 25885    1                            
11 2019-02-19 15:17:20 25885    2                           
12 2019-02-19 15:17:21 25885    1                           

数据集2

1  2019-02-19 15:17:18 25885    4                           
2  2019-02-19 15:17:19 25885    1  
3  2019-02-19 15:17:19 25885    1                          
4  2019-02-19 15:17:20 25885    2                           
5  2019-02-19 15:17:21 25885    1                           
6  2019-02-19 15:17:23 25886    2                           
7  2019-02-19 15:17:23 25886    3                           
8  2019-02-19 15:17:23 25886    3                           
9  2019-02-19 15:17:23 25886    1                           
10 2019-02-19 15:17:23 25886    1                           
11 2019-02-19 15:17:23 25886    2 

我想要的结果是:

1  2019-02-19 15:17:14 25886    1                           
2  2019-02-19 15:17:14 25886    1                           
3  2019-02-19 15:17:15 25885    1                           
4  2019-02-19 15:17:16 25886    2                           
5  2019-02-19 15:17:16 25886    1                           
6  2019-02-19 15:17:16 25886    2                           
7  2019-02-19 15:17:16 25886    1                           
8  2019-02-19 15:17:18 25885    4                           
9  2019-02-19 15:17:19 25885    1   
10 2019-02-19 15:17:19 25885    1                             
11 2019-02-19 15:17:20 25885    2                           
12 2019-02-19 15:17:21 25885    1                      
13 2019-02-19 15:17:23 25886    2                           
14 2019-02-19 15:17:23 25886    3                           
15 2019-02-19 15:17:23 25886    3                           
16 2019-02-19 15:17:23 25886    1                           
17 2019-02-19 15:17:23 25886    1                           
18 2019-02-19 15:17:23 25886    2 

这里的数据集1

structure(list(time = structure(c(1550589434, 1550589434, 1550589435, 
1550589436, 1550589436, 1550589436, 1550589436, 1550589438, 1550589439, 
1550589439, 1550589440, 1550589441), class = c("POSIXct", "POSIXt"
), tzone = "UTC"), price = c(25886, 25886, 25885, 25886, 25886, 
25886, 25886, 25885, 25885, 25885, 25885, 25885), size = c(1, 
1, 1, 2, 1, 2, 1, 4, 1, 1, 2, 1)), row.names = c("1", "2", "3", 
"4", "5", "6", "7", "8", "9", "10", "11", "12"), class = "data.frame")

这里的数据集2

structure(list(time = structure(c(1550589438, 1550589439, 1550589439, 
1550589440, 1550589441, 1550589443, 1550589443, 1550589443, 1550589443, 
1550589443, 1550589443), class = c("POSIXct", "POSIXt"), tzone = "UTC"), 
price = c(25885, 25885, 25885, 25885, 25885, 25886, 25886, 
25886, 25886, 25886, 25886), size = c(4, 1, 1, 2, 1, 2, 3, 
3, 1, 1, 2)), row.names = c("1", "2", "3", "4", "5", "6", 
"7", "8", "9", "10", "11"), class = "data.frame")

1 个答案:

答案 0 :(得分:0)

一个想法是:

library(dplyr)

df2 %>%
  anti_join(df1) %>%
  bind_rows(df1)