如何在数据框架中迭代因子水平并添加来自列表的值的列？

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英文:

How can I add a column with values from a list, iterating in a data frame over factor levels?

问题

I can help you translate the code-related parts into Chinese. Here it is:

给定数据框架 df：

  1. df <- data.frame(1,
  2.                  1:12,
  3.                  groups = factor(rep(LETTERS[1:4], each = 3)))
  4. df

以及列表 list_group：

  1. list_group <- list(
  2.   A = c("Jack","John","Joe"),
  3.   B = c("Ed","Edd","Eddy"),
  4.   C = c("Gianni","Franco","Ugo"),
  5.   D = c("Bob","Rob","Frank"))
  6. list_group

我想要得到以下结果：

如何在数据框架中迭代因子水平并添加来自列表的值的列？

因此，添加列 person，其中包含与 df$groups 水平对应的名称。这需要将行多次复制到因子水平，但要与 list_group 中相应的名称相乘（即 df$groups 水平 "A" = "Jack","John","Joe"; df$groups 水平 "B" = "Ed","Edd","Eddy"; 等等）。

请注意，我只提供了代码的翻译部分，没有包含其他内容。

英文:

I am struggling in performing and iterating the following operation in the R environment.

Given a data frame df:

  1. df &lt;- data.frame(1,
  2.                  1:12,
  3.                  groups = factor(rep(LETTERS[1:4], each = 3)))
  4. df

and the list list_group:

  1. list_group &lt;- list(
  2.   A = c(&quot;Jack&quot;,&quot;John&quot;,&quot;Joe&quot;),
  3.   B = c(&quot;Ed&quot;,&quot;Edd&quot;,&quot;Eddy&quot;),
  4.   C = c(&quot;Gianni&quot;,&quot;Franco&quot;,&quot;Ugo&quot;),
  5.   D = c(&quot;Bob&quot;,&quot;Rob&quot;,&quot;Frank&quot;))
  6. list_group

I would like to get the following result:

如何在数据框架中迭代因子水平并添加来自列表的值的列？

Thus, add the column person with the names in correspondence of the df$groups level. This requires to multiple the rows over the factor levels, but specifically with the corresponding names in the list_group (i.e., df$groups level "A" = "Jack","John","Joe"; df$groups level "B" = "Ed","Edd","Eddy"; etc.)

I went through the function of the Tidyverse, in particular dplyr,tidyr and purrr, but I did not find specific functions for my problem.

Also, I tried to dig into the theory behind Iteration in R, mainly following the guide 21 Iteration, but I did not succeed in that.

答案1

得分: 2

你可以使用 stack + merge：

  1. merge(df, stack(list_group), by.x = "groups", by.y = "ind")

或者使用 dplyr 的等效方法：

  1. library(dplyr)
  2. full_join(df, stack(list_group), by = c("groups" = "ind"), relationship = "many-to-many")

输出：

  1. #    groups X1 X1.12 values
  2. # 1       A  1     1   Jack
  3. # 2       A  1     1   John
  4. # 3       A  1     1    Joe
  5. # 4       A  1     2   Jack
  6. # 5       A  1     2   John
  7. # 6       A  1     2    Joe
  8. # 7       A  1     3   Jack
  9. # 8       A  1     3   John
  10. # 9       A  1     3    Joe
  11. # 10      B  1     4     Ed
  12. # 11      B  1     4    Edd
  13. # 12      B  1     4   Eddy
  14. # 13      B  1     5     Ed
  15. # 14      B  1     5    Edd
  16. # 15      B  1     5   Eddy
  17. # 16      B  1     6     Ed
  18. # 17      B  1     6    Edd
  19. # 18      B  1     6   Eddy
  20. # 19      C  1     7 Gianni
  21. # 20      C  1     7 Franco
  22. # 21      C  1     7    Ugo
  23. # 22      C  1     8 Gianni
  24. # 23      C  1     8 Franco
  25. # 24      C  1     8    Ugo
  26. # 25      C  1     9 Gianni
  27. # 26      C  1     9 Franco
  28. # 27      C  1     9    Ugo
  29. # 28      D  1    10    Bob
  30. # 29      D  1    10    Rob
  31. # 30      D  1    10  Frank
  32. # 31      D  1    11    Bob
  33. # 32      D  1    11    Rob
  34. # 33      D  1    11  Frank
  35. # 34      D  1    12    Bob
  36. # 35      D  1    12    Rob
  37. # 36      D  1    12  Frank
英文:

You can use stack + merge:

  1. merge(df, stack(list_group), by.x = &quot;groups&quot;, by.y = &quot;ind&quot;)

Or the dplyr equivalent:

  1. library(dplyr)
  2. full_join(df, stack(list_group), by = c(&quot;groups&quot; = &quot;ind&quot;), relationship = &quot;many-to-many&quot;)

output

  1. #    groups X1 X1.12 values
  2. # 1       A  1     1   Jack
  3. # 2       A  1     1   John
  4. # 3       A  1     1    Joe
  5. # 4       A  1     2   Jack
  6. # 5       A  1     2   John
  7. # 6       A  1     2    Joe
  8. # 7       A  1     3   Jack
  9. # 8       A  1     3   John
  10. # 9       A  1     3    Joe
  11. # 10      B  1     4     Ed
  12. # 11      B  1     4    Edd
  13. # 12      B  1     4   Eddy
  14. # 13      B  1     5     Ed
  15. # 14      B  1     5    Edd
  16. # 15      B  1     5   Eddy
  17. # 16      B  1     6     Ed
  18. # 17      B  1     6    Edd
  19. # 18      B  1     6   Eddy
  20. # 19      C  1     7 Gianni
  21. # 20      C  1     7 Franco
  22. # 21      C  1     7    Ugo
  23. # 22      C  1     8 Gianni
  24. # 23      C  1     8 Franco
  25. # 24      C  1     8    Ugo
  26. # 25      C  1     9 Gianni
  27. # 26      C  1     9 Franco
  28. # 27      C  1     9    Ugo
  29. # 28      D  1    10    Bob
  30. # 29      D  1    10    Rob
  31. # 30      D  1    10  Frank
  32. # 31      D  1    11    Bob
  33. # 32      D  1    11    Rob
  34. # 33      D  1    11  Frank
  35. # 34      D  1    12    Bob
  36. # 35      D  1    12    Rob
  37. # 36      D  1    12  Frank

答案2

得分: 0

你可以使用基础的stack()函数，然后使用整洁的full_join(multiple = "all")函数：

  1. df %>% full_join(
  2.   y = list_group %>% stack(),
  3.   by = join_by(groups == ind),
  4.   multiple = "all"
  5. )

结果：

  1.    X1 X1.12 groups values
  2. 1   1     1      A   Jack
  3. 2   1     1      A   John
  4. 3   1     1      A    Joe
  5. 4   1     2      A   Jack
  6. 5   1     2      A   John
  7. 6   1     2      A    Joe
  8. 7   1     3      A   Jack
  9. 8   1     3      A   John
  10. 9   1     3      A    Joe
  11. 10  1     4      B     Ed
  12. 11  1     4      B    Edd
  13. 12  1     4      B   Eddy
  14. 13  1     5      B     Ed
  15. 14  1     5      B    Edd
  16. 15  1     5      B   Eddy
  17. 16  1     6      B     Ed
  18. 17  1     6      B    Edd
  19. 18  1     6      B   Eddy
  20. 19  1     7      C Gianni
  21. 20  1     7      C Franco
  22. 21  1     7      C    Ugo
  23. 22  1     8      C Gianni
  24. 23  1     8      C Franco
  25. 24  1     8      C    Ugo
  26. 25  1     9      C Gianni
  27. 26  1     9      C Franco
  28. 27  1     9      C    Ugo
  29. 28  1    10      D    Bob
  30. 29  1    10      D    Rob
  31. 30  1    10      D  Frank
  32. 31  1    11      D    Bob
  33. 32  1    11      D    Rob
  34. 33  1    11      D  Frank
  35. 34  1    12      D    Bob
  36. 35  1    12      D    Rob
  37. 36  1    12      D  Frank
英文:

You can use base stack() and then a tidy full_join(multiple = &quot;all&quot;):

  1. df %&gt;% full_join(
  2.   y = list_group %&gt;% stack(),
  3.   by = join_by(groups == ind),
  4.   multiple = &quot;all&quot;
  5. )

Result:

  1.    X1 X1.12 groups values
  2. 1   1     1      A   Jack
  3. 2   1     1      A   John
  4. 3   1     1      A    Joe
  5. 4   1     2      A   Jack
  6. 5   1     2      A   John
  7. 6   1     2      A    Joe
  8. 7   1     3      A   Jack
  9. 8   1     3      A   John
  10. 9   1     3      A    Joe
  11. 10  1     4      B     Ed
  12. 11  1     4      B    Edd
  13. 12  1     4      B   Eddy
  14. 13  1     5      B     Ed
  15. 14  1     5      B    Edd
  16. 15  1     5      B   Eddy
  17. 16  1     6      B     Ed
  18. 17  1     6      B    Edd
  19. 18  1     6      B   Eddy
  20. 19  1     7      C Gianni
  21. 20  1     7      C Franco
  22. 21  1     7      C    Ugo
  23. 22  1     8      C Gianni
  24. 23  1     8      C Franco
  25. 24  1     8      C    Ugo
  26. 25  1     9      C Gianni
  27. 26  1     9      C Franco
  28. 27  1     9      C    Ugo
  29. 28  1    10      D    Bob
  30. 29  1    10      D    Rob
  31. 30  1    10      D  Frank
  32. 31  1    11      D    Bob
  33. 32  1    11      D    Rob
  34. 33  1    11      D  Frank
  35. 34  1    12      D    Bob
  36. 35  1    12      D    Rob
  37. 36  1    12      D  Frank

huangapple
  • 本文由 发表于 2023年6月8日 15:51:07
  • 转载请务必保留本文链接：https://go.coder-hub.com/76429705.html
  • dataframe
  • iteration
  • list
  • r
  • tidyverse
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