重命名重复的列值，按另一列分组。

huangapple
go评论101阅读模式
英文:

Rename duplicated column values group by another column

问题

我有一个数据集，其中相同的id出现在不同的组中：

  1. df <- read.table(text='id   group  
  2.  1         A 
  3.  2         A 
  4.  2         A 
  5.  1         B 
  6.  1         B 
  7.  2         B
  8.  2         C
  9.  2         C 
  10.  1         C 
  11.  2         D         
  12.  1         D
  13.  1         D', header=TRUE)

我想要重命名在另一列group中分组的重复值，位于列id下。预期的输出是：

  1.    id  group
  2.     1     A
  3.     2     A
  4.     2     A
  5.     1_2   B
  6.     1_2   B
  7.     2_2   B
  8.     2_3   C
  9.     2_3   C
  10.     1_3   C
  11.     2_4   D
  12.     1_4   D
  13.     1_4   D

我该如何做到这一点？

英文:

I have a dataset in which the same id appears in different groups:

  1. df &lt;- read.table(text=&#39;id   group  
  2.  1         A 
  3.  2         A 
  4.  2         A 
  5.  1         B 
  6.  1         B 
  7.  2         B
  8.  2         C
  9.  2         C 
  10.  1         C 
  11.  2         D         
  12.  1         D
  13.  1         D&#39;, header=TRUE)

I want to rename the duplicated values under column id that are grouped by another column group. The expected output is:

  1.    id  group
  2.     1     A
  3.     2     A
  4.     2     A
  5.     1_2   B
  6.     1_2   B
  7.     2_2   B
  8.     2_3   C
  9.     2_3   C
  10.     1_3   C
  11.     2_4   D
  12.     1_4   D
  13.     1_4   D

How do I do that?

答案1

得分: 1

这里是使用data.table的方法，利用rleid()为每个唯一的id和group组合生成一个运行长度ID。然后，我们可以将该数字粘贴到现有的id上，当它大于1时。

  1. library(data.table)
  2. setDT(df)
  4. df[, id_num := rleid(group), id][
  5.     ,
  6.     id := fifelse(
  7.         id_num == 1,
  8.         as.character(id),
  9.         paste(id, id_num, sep = "_")
  10.     )
  11. ][, `:=`(id_num = NULL)]
  13. print(df)
  15. #         id  group
  16. #  1:      1      A
  17. #  2:      2      A
  18. #  3:      2      A
  19. #  4:    1_2      B
  20. #  5:    1_2      B
  21. #  6:    2_2      B
  22. #  7:    2_3      C
  23. #  8:    2_3      C
  24. #  9:    1_3      C
  25. # 10:    2_4      D
  26. # 11:    1_4      D
  27. # 12:    1_4      D

请注意，上面的代码段是使用R语言编写的，并且其中的HTML编码（如&lt;和&gt;）已保留，以确保在HTML环境中正确显示。

英文:

Here is a data.table approach using rleid() to generate a run-length id for each unique id and group combination. We can then just paste() that number to the existing id, where it is &gt;1.

  1. library(data.table)
  2. setDT(df)
  4. df[, id_num := rleid(group), id][
  5.     ,
  6.     id := fifelse(
  7.         id_num == 1,
  8.         as.character(id),
  9.         paste(id, id_num, sep = &quot;_&quot;)
  10.     )
  11. ][, `:=`(id_num = NULL)]
  13. print(df)
  15. #         id  group
  16. #     &lt;char&gt; &lt;char&gt;
  17. #  1:      1      A
  18. #  2:      2      A
  19. #  3:      2      A
  20. #  4:    1_2      B
  21. #  5:    1_2      B
  22. #  6:    2_2      B
  23. #  7:    2_3      C
  24. #  8:    2_3      C
  25. #  9:    1_3      C
  26. # 10:    2_4      D
  27. # 11:    1_4      D
  28. # 12:    1_4      D

答案2

得分: 1

感谢@SamR的data.table答案。我能够将他/她的代码转换为使用Chatgpt的tidyverse版本：

  1. df %>%
  2.   mutate(id_num = data.table::rleid(group)) %>%
  3.   mutate(id = ifelse(id_num == 1, as.character(id), paste(id, id_num, sep = "_"))) %>%
  4.   select(-id_num)
英文:

Thanks to @SamR's data.table answer. I was able to convert his/her code to a tidyverse version using Chatgpt:

  1. df %&gt;%
  2.   mutate(id_num = data.table::rleid(group)) %&gt;%
  3.   mutate(id = ifelse(id_num == 1, as.character(id), paste(id, id_num, sep = &quot;_&quot;))) %&gt;%
  4.   select(-id_num)

huangapple
  • 本文由 发表于 2023年7月5日 00:44:24
  • 转载请务必保留本文链接：https://go.coder-hub.com/76614547.html
  • data-manipulation
  • dplyr
  • duplicates
  • r
  • rename
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