在R中，通过”S.NO”分组收集数据时，在数据的开头和结尾添加”Age”的NA值。

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英文:

Adding NA for Age at the start and end of the data collected when grouped by S.NO in R

问题

S.NO AGE_1 AGE_2
123 NA 19
123 19 24
123 24 NA
124 NA 18
124 18 21
124 21 28
124 28 35
124 35 NA
125 NA 13
125 13 19
125 19 23
125 23 NA
126 NA 19
126 19 21
126 21 NA
英文:

I have a dataset that looks like this and I am trying to add NA at the beginning and end of the age data collected for each S.NO. as shown in the expected dataset in R.
Would appreciate any help. Thank you.

S.NO AGE_1 AGE_2
123 19 24
124 18 21
124 21 28
124 28 35
125 13 19
125 19 23
126 19 21

Expected Dataset

S.NO AGE_1 AGE_2
123 NA 19
123 19 24
123 24 NA
124 NA 18
124 18 21
124 21 28
124 28 35
124 35 NA
125 NA 13
125 13 19
125 19 23
125 23 NA
126 NA 19
126 19 21
126 21 NA

答案1

得分: 1

使用 tibble::add_row、dplyr::group_split 和 purrr::map_dfr，您可以执行以下操作：

  1. library(dplyr, warn = FALSE)
  2. library(tibble)
  3. library(purrr)
  5. pad_rows <- function(x) {
  6.   x |>
  7.     tibble::add_row(S.NO = first(x$S.NO), AGE_2 = first(x$AGE_1), .before = 1) |>
  8.     tibble::add_row(S.NO = last(x$S.NO), AGE_1 = last(x$AGE_2), .after = Inf)
  9. }
  11. dat |>
  12.   group_split(S.NO) |>
  13.   purrr::map_dfr(pad_rows)
  14. #> # A tibble: 15 × 3
  15. #>     S.NO AGE_1 AGE_2
  16. #>    <int> <int> <int>
  17. #>  1   123    NA    19
  18. #>  2   123    19    24
  19. #>  3   123    24    NA
  20. #>  4   124    NA    18
  21. #>  5   124    18    21
  22. #>  6   124    21    28
  23. #>  7   124    28    35
  24. #>  8   124    35    NA
  25. #>  9   125    NA    13
  26. #> 10   125    13    19
  27. #> 11   125    19    23
  28. #> 12   125    23    NA
  29. #> 13   126    NA    19
  30. #> 14   126    19    21
  31. #> 15   126    21    NA

数据

  1. dat <- data.frame(
  2.         S.NO = c(123L, 124L, 124L, 124L, 125L, 125L, 126L),
  3.        AGE_1 = c(19L, 18L, 21L, 28L, 13L, 19L, 19L),
  4.        AGE_2 = c(24L, 21L, 28L, 35L, 19L, 23L, 21L)
  5. )
英文:

Using tibble::add_row, dplyr::group_split and purrr::map_dfr you could do:

  1. library(dplyr, warn = FALSE)
  2. library(tibble)
  3. library(purrr)
  5. pad_rows &lt;- function(x) {
  6.   x |&gt; 
  7.     tibble::add_row(S.NO = first(x$S.NO), AGE_2 = first(x$AGE_1), .before = 1) |&gt; 
  8.     tibble::add_row(S.NO = last(x$S.NO), AGE_1 = last(x$AGE_2), .after = Inf)
  9. }
  11. dat |&gt; 
  12.   group_split(S.NO) |&gt; 
  13.   purrr::map_dfr(pad_rows)
  14. #&gt; # A tibble: 15 &#215; 3
  15. #&gt;     S.NO AGE_1 AGE_2
  16. #&gt;    &lt;int&gt; &lt;int&gt; &lt;int&gt;
  17. #&gt;  1   123    NA    19
  18. #&gt;  2   123    19    24
  19. #&gt;  3   123    24    NA
  20. #&gt;  4   124    NA    18
  21. #&gt;  5   124    18    21
  22. #&gt;  6   124    21    28
  23. #&gt;  7   124    28    35
  24. #&gt;  8   124    35    NA
  25. #&gt;  9   125    NA    13
  26. #&gt; 10   125    13    19
  27. #&gt; 11   125    19    23
  28. #&gt; 12   125    23    NA
  29. #&gt; 13   126    NA    19
  30. #&gt; 14   126    19    21
  31. #&gt; 15   126    21    NA

DATA

  1. dat &lt;- data.frame(
  2.         S.NO = c(123L, 124L, 124L, 124L, 125L, 125L, 126L),
  3.        AGE_1 = c(19L, 18L, 21L, 28L, 13L, 19L, 19L),
  4.        AGE_2 = c(24L, 21L, 28L, 35L, 19L, 23L, 21L)
  5. )

答案2

得分: 1

使用 group_modify

  1. library(dplyr)
  2. library(tibble)
  3. df %>% 
  4.   group_by(S.NO) %>% 
  5.   group_modify(~ .x %>% 
  6.                 add_row(AGE_2 = first(.x$AGE_1), .before = 1) %>% 
  7.                 add_row(AGE_1 = last(.x$AGE_2))) %>% 
  8.   ungroup

-输出

  1. # A tibble: 15 × 3
  2.     S.NO AGE_1 AGE_2
  3.    <int> <int> <int>
  4.  1   123   NA   19
  5.  2   123   19   24
  6.  3   123   24   NA
  7.  4   124   NA   18
  8.  5   124   18   21
  9.  6   124   21   28
  10.  7   124   28   35
  11.  8   124   35   NA
  12.  9   125   NA   13
  13. 10   125   13   19
  14. 11   125   19   23
  15. 12   125   23   NA
  16. 13   126   NA   19
  17. 14   126   19   21
  18. 15   126   21   NA
英文:

Using group_modify

  1. library(dplyr)
  2. library(tibble)
  3. df %&gt;% 
  4.   group_by(S.NO) %&gt;% 
  5.   group_modify(~ .x %&gt;% 
  6.                 add_row(AGE_2= first(.x$AGE_1), .before = 1) %&gt;% 
  7.                 add_row(AGE_1 = last(.x$AGE_2))) %&gt;% 
  8.   ungroup

-output

  1. # A tibble: 15 &#215; 3
  2.     S.NO AGE_1 AGE_2
  3.    &lt;int&gt; &lt;int&gt; &lt;int&gt;
  4.  1   123    NA    19
  5.  2   123    19    24
  6.  3   123    24    NA
  7.  4   124    NA    18
  8.  5   124    18    21
  9.  6   124    21    28
  10.  7   124    28    35
  11.  8   124    35    NA
  12.  9   125    NA    13
  13. 10   125    13    19
  14. 11   125    19    23
  15. 12   125    23    NA
  16. 13   126    NA    19
  17. 14   126    19    21
  18. 15   126    21    NA

答案3

得分: 0

使用nest_by的方法：

  1. library(dplyr)
  2. library(tidyr) # unnest
  4. df %>%
  5.   nest_by(S.NO) %>%
  6.   mutate(data = list(unique(unlist(data))) %>%
  7.   unnest(data) %>%
  8.   group_by(S.NO) %>%
  9.   reframe(AGE_1 = c(NA, data), AGE_2 = c(data, NA))
  10. # A tibble: 15 × 3
  11.     S.NO AGE_1 AGE_2
  12.    <int> <int> <int>
  13.  1   123    NA    19
  14.  2   123    19    24
  15.  3   123    24    NA
  16.  4   124    NA    18
  17.  5   124    18    21
  18.  6   124    21    28
  19.  7   124    28    35
  20.  8   124    35    NA
  21.  9   125    NA    13
  22. 10   125    13    19
  23. 11   125    19    23
  24. 12   125    23    NA
  25. 13   126    NA    19
  26. 14   126    19    21
  27. 15   126    21    NA

数据

  1. df <- structure(list(S.NO = c(123L, 124L, 124L, 124L, 125L, 125L, 126L
  2. ), AGE_1 = c(19L, 18L, 21L, 28L, 13L, 19L, 19L), AGE_2 = c(24L, 
  3. 21L, 28L, 35L, 19L, 23L, 21L)), class = "data.frame", row.names = c(NA, 
  4. -7L))
英文:

An approach using nest_by

  1. library(dplyr)
  2. library(tidyr) # unnest
  4. df %&gt;% 
  5.   nest_by(S.NO) %&gt;% 
  6.   mutate(data = list(unique(unlist(data)))) %&gt;% 
  7.   unnest(data) %&gt;% 
  8.   group_by(S.NO) %&gt;% 
  9.   reframe(AGE_1 = c(NA, data), AGE_2 = c(data, NA))
  10. # A tibble: 15 &#215; 3
  11.     S.NO AGE_1 AGE_2
  12.    &lt;int&gt; &lt;int&gt; &lt;int&gt;
  13.  1   123    NA    19
  14.  2   123    19    24
  15.  3   123    24    NA
  16.  4   124    NA    18
  17.  5   124    18    21
  18.  6   124    21    28
  19.  7   124    28    35
  20.  8   124    35    NA
  21.  9   125    NA    13
  22. 10   125    13    19
  23. 11   125    19    23
  24. 12   125    23    NA
  25. 13   126    NA    19
  26. 14   126    19    21
  27. 15   126    21    NA

Data

  1. df &lt;- structure(list(S.NO = c(123L, 124L, 124L, 124L, 125L, 125L, 126L
  2. ), AGE_1 = c(19L, 18L, 21L, 28L, 13L, 19L, 19L), AGE_2 = c(24L, 
  3. 21L, 28L, 35L, 19L, 23L, 21L)), class = &quot;data.frame&quot;, row.names = c(NA, 
  4. -7L))

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  • 本文由 发表于 2023年2月7日 02:25:50
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