根据ID反转列

huangapple
go评论93阅读模式
英文:

Reverse Column based on ID

问题

我有一个包含一些重复ID的数据集（ID的频率不同）。还有一列显示了每个ID的第n次出现。我试图找到一种快速的方法来为每个ID创建一个新列，该列将反转该时间列。例如，对于以下向量，我希望它们看起来像这样：

  1. (1, 1, 1, 2, 3, 3, 3, 3, 3) = id
  2. (1, 2, 3, 1, 1, 2, 3, 4, 5) = time
  3. (3, 2, 1, 1, 5, 4, 3, 2, 1) = time_reverse

我在下面提供了一个可重现的示例：

  1. df <- data.frame(
  2.   id = rep(c(1, 2, 3, 4, 5), times = c(3, 1, 5, 2, 7)),
  3.   time = c(1, 2, 3, 1, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 4, 5, 6, 7), 
  4.   time_reverse = sample(NA, size = 18, replace = TRUE)
  5. )

谢谢！

英文:

I have a dataset with some recurring IDs (there are different frequencies of IDs). There also is a column which shows the nth occurrence of each ID.
I am trying to find a quick way to create a new column which will reverse that time column for each ID.
For example for the following vectors i would expect them to look like this:

  1. (1, 1, 1, 2, 3, 3, 3, 3, 3) = id  
  2. (1, 2, 3, 1, 1, 2, 3, 4, 5) = time  
  3. (3, 2, 1, 1, 5, 4, 3, 2, 1) = time_reverse

I posted a reproducible example below.

  1. df &lt;- data.frame(
  2.   id = rep(c(1, 2, 3, 4, 5), times = c(3, 1, 5, 2, 7)),
  3.   time = c(1, 2, 3, 1, 1, 2,  3, 4, 5, 1, 2, 1, 2, 3, 4, 5, 6, 7), 
  4.   time_reverse = sample(NA, size = 18, replace = TRUE)
  5. )

Thanks in advance 根据ID反转列

答案1

得分: 2

dplyr

你可以使用 row_number 和 rev 来按组进行操作：

  1. library(dplyr)
  2. df %>%
  3.   group_by(id) %>%
  4.   mutate(time = row_number(), 
  5.          time_reverse = rev(time))

base R

  1. transform(df,
  2.           time = ave(id, id, FUN = seq_along),
  3.           time_reverse = ave(time, id, FUN = rev))
  1.       id  time time_reverse
  2.  1     1     1            3
  3.  2     1     2            2
  4.  3     1     3            1
  5.  4     2     1            1
  6.  5     3     1            5
  7.  6     3     2            4
  8.  7     3     3            3
  9.  8     3     4            2
  10.  9     3     5            1
  11. 10     4     1            2
  12. 11     4     2            1
  13. 12     5     1            7
  14. 13     5     2            6
  15. 14     5     3            5
  16. 15     5     4            4
  17. 16     5     5            3
  18. 17     5     6            2
  19. 18     5     7            1
英文:

dplyr

You can use row_number and rev by group:

  1. library(dplyr)
  2. df %&gt;% 
  3.   group_by(id) %&gt;% 
  4.   mutate(time = row_number(), 
  5.          time_reverse = rev(time))

base R

  1. transform(df,
  2.           time = ave(id, id, FUN = seq_along),
  3.           time_reverse = ave(time, id, FUN = rev))
  1.       id  time time_reverse
  2.  1     1     1            3
  3.  2     1     2            2
  4.  3     1     3            1
  5.  4     2     1            1
  6.  5     3     1            5
  7.  6     3     2            4
  8.  7     3     3            3
  9.  8     3     4            2
  10.  9     3     5            1
  11. 10     4     1            2
  12. 11     4     2            1
  13. 12     5     1            7
  14. 13     5     2            6
  15. 14     5     3            5
  16. 15     5     4            4
  17. 16     5     5            3
  18. 17     5     6            2
  19. 18     5     7            1

huangapple
  • 本文由 发表于 2023年3月7日 17:41:19
  • 转载请务必保留本文链接：https://go.coder-hub.com/75660232.html
  • r
  • tidyverse
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定