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_rowdplyr::group_splitpurrr::map_dfr,您可以执行以下操作:

library(dplyr, warn = FALSE)
library(tibble)
library(purrr)

pad_rows <- function(x) {
  x |>
    tibble::add_row(S.NO = first(x$S.NO), AGE_2 = first(x$AGE_1), .before = 1) |>
    tibble::add_row(S.NO = last(x$S.NO), AGE_1 = last(x$AGE_2), .after = Inf)
}

dat |>
  group_split(S.NO) |>
  purrr::map_dfr(pad_rows)
#> # A tibble: 15 × 3
#>     S.NO AGE_1 AGE_2
#>    <int> <int> <int>
#>  1   123    NA    19
#>  2   123    19    24
#>  3   123    24    NA
#>  4   124    NA    18
#>  5   124    18    21
#>  6   124    21    28
#>  7   124    28    35
#>  8   124    35    NA
#>  9   125    NA    13
#> 10   125    13    19
#> 11   125    19    23
#> 12   125    23    NA
#> 13   126    NA    19
#> 14   126    19    21
#> 15   126    21    NA

数据

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

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

library(dplyr, warn = FALSE)
library(tibble)
library(purrr)

pad_rows &lt;- function(x) {
  x |&gt; 
    tibble::add_row(S.NO = first(x$S.NO), AGE_2 = first(x$AGE_1), .before = 1) |&gt; 
    tibble::add_row(S.NO = last(x$S.NO), AGE_1 = last(x$AGE_2), .after = Inf)
}

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

DATA

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

答案2

得分: 1

使用 group_modify

library(dplyr)
library(tibble)
df %>% 
  group_by(S.NO) %>% 
  group_modify(~ .x %>% 
                add_row(AGE_2 = first(.x$AGE_1), .before = 1) %>% 
                add_row(AGE_1 = last(.x$AGE_2))) %>% 
  ungroup

-输出

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

Using group_modify

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

-output

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

答案3

得分: 0

使用nest_by的方法:

library(dplyr)
library(tidyr) # unnest

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

数据

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

An approach using nest_by

library(dplyr)
library(tidyr) # unnest

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

Data

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

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  • 本文由 发表于 2023年2月7日 02:25:50
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