按重叠类别进行分组（一个类别应该分别归入另外两个不同的类别）。

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英文:

Group by an overlapping category (a category should be grouped in 2 different other categories)

问题

我有一个数据集，其中有'gr1'、'gr2'和'both'这些组。基本上，我想通过c("gr1", "both")和c("gr2", "both")来分组'gr'列。

这里我提出了一个使用简单数据框的解决方案，但我想知道是否有一种方法可以进行'复杂'的分组，例如group_by(gr ''using c("gr1", "both") and c("gr2", "both") as groups'')。是否有一种方法可以在dplyr中指定要分组在一起的内容，而不是像下面所示那样手动操作？

  1. library(tidyverse)
  2. set.seed(1234)
  4. df = data.frame(x = 1:10, id = sample(LETTERS[1:3], size = 10, replace = TRUE),
  5.            gr = c(rep("gr1",3), rep("gr2",4),rep("both",3)))
  7. sum.gr1 = df %>%
  8.   filter(gr %in% c("gr1", "both")) %>%
  9.   group_by(id) %>%
  10.   summarize(x.sum = sum(x)) %>%
  11.   mutate(gr.filt = "gr1.both")
  13. sum.gr2 = df %>%
  14.   filter(gr %in% c("gr2", "both")) %>%
  15.   group_by(id) %>%
  16.   summarize(x.sum = sum(x))%>%
  17.   mutate(gr.filt = "gr2.both")
  19. df.gr = rbind(sum.gr1, sum.gr2)
  20. df.gr

希望这能满足你的需求。

英文:

I have a dataset where there are groups 'gr1', 'gr2', and 'both'. Basically, I'd like to group the 'gr' column by c("gr1", "both") and c("gr2", "both").

Here I'm proposing a solution with a simple data frame, but I'd like to know if there is a way to make 'complex' grouping such as group_by(gr ''using c("gr1", "both") and c("gr2", "both") as groups'' ). Is there a way to specify what to group together in dplyr instead of doing the rind like shown below?

  1. library(tidyverse)
  2. set.seed(1234)
  4. df = data.frame(x = 1:10, id = sample(LETTERS[1:3], size = 10, replace = TRUE),
  5.            gr = c(rep("gr1",3), rep("gr2",4),rep("both",3)))
  7. df
  8.     x id   gr
  9. 1   1  B  gr1
  10. 2   2  B  gr1
  11. 3   3  A  gr1
  12. 4   4  C  gr2
  13. 5   5  A  gr2
  14. 6   6  A  gr2
  15. 7   7  B  gr2
  16. 8   8  B both
  17. 9   9  C both
  18. 10 10  B both
  20. sum.gr1 = df %>% 
  21.   filter(gr %in% c("gr1", "both")) %>% 
  22.   group_by(id) %>% 
  23.   summarize(x.sum = sum(x)) %>% 
  24.   mutate(gr.filt = "gr1.both")
  26. sum.gr2 = df %>% 
  27.   filter(gr %in% c("gr2", "both")) %>% 
  28.   group_by(id) %>% 
  29.   summarize(x.sum = sum(x))%>% 
  30.   mutate(gr.filt = "gr2.both")
  32. df.gr = rbind(sum.gr1, sum.gr2)
  33. df.gr
  35. # A tibble: 6 × 3
  36. id    x.sum gr.filt 
  37. <chr> <int> <chr>   
  38. 1 A         3 gr1.both
  39. 2 B        21 gr1.both
  40. 3 C         9 gr1.both
  41. 4 A        11 gr2.both
  42. 5 B        25 gr2.both
  43. 6 C        13 gr2.both

答案1

得分: 3

以下是使用 map_df 的版本：

  1. library(dplyr)
  2. library(purrr)
  4. map_df(list(c("gr1", "both"), c("gr2", "both")), ~df %>%
  5.                         filter(gr %in% .x) %>%
  6.                         group_by(id) %>%
  7.                         summarize(x.sum = sum(x)) %>%
  8.                         mutate(gr.filt = paste(.x, collapse = ".")))
  1.  id    x.sum gr.filt 
  2.   <chr> <int> <chr>   
  3. 1 A         3 gr1.both
  4. 2 B        21 gr1.both
  5. 3 C         9 gr1.both
  6. 4 A        11 gr2.both
  7. 5 B        25 gr2.both
  8. 6 C        13 gr2.both
英文:

Update after clarification:

Here is a version using map_df:

  1. library(dplyr)
  2. library(purrr)
  4. map_df(list(c(&quot;gr1&quot;, &quot;both&quot;), c(&quot;gr2&quot;, &quot;both&quot;)), ~df %&gt;% 
  5.                         filter(gr %in% .x) %&gt;% 
  6.                         group_by(id) %&gt;% 
  7.                         summarize(x.sum = sum(x)) %&gt;% 
  8.                         mutate(gr.filt = paste(.x, collapse = &quot;.&quot;)))
  1.  id    x.sum gr.filt 
  2.   &lt;chr&gt; &lt;int&gt; &lt;chr&gt;   
  3. 1 A         3 gr1.both
  4. 2 B        21 gr1.both
  5. 3 C         9 gr1.both
  6. 4 A        11 gr2.both
  7. 5 B        25 gr2.both
  8. 6 C        13 gr2.both

答案2

得分: 3

这里我将“non-both”数据与“both”数据的一个版本组合，其中每一行都复制到“non-both”组中。

  1. library(dplyr)
  2. bind_rows(
  3.   df %>% filter(gr != "both"),
  4.   df %>% filter(gr == "both") %>% select(-gr) %>%
  5.     tidyr::crossing(gr = unique(df$gr[df$gr != "both"]))
  6. ) %>%
  7.   count(gr = paste0(gr, ".both"), id, wt = x, name = "x.sum")

结果

  1.          gr id x.sum
  2. 1  gr1.both  A     3
  3. 2  gr1.both  B    21
  4. 3  gr1.both  C     9
  5. 4  gr2.both  A    11
  6. 5  gr2.both  B    25
  7. 6  gr2.both  C    13
英文:

Here I combine the "non-both" data with a version of the "both" data where each row has been copied to each of the "non-both" groups.

  1. library(dplyr)
  2. bind_rows(
  3.   df |&gt; filter(gr != &quot;both&quot;),
  4.   df |&gt; filter(gr == &quot;both&quot;) |&gt; select(-gr) |&gt;
  5.     tidyr::crossing(gr = unique(df$gr[df$gr != &quot;both&quot;]))
  6. ) |&gt;
  7.   count(gr = paste0(gr, &quot;.both&quot;), id, wt = x, name = &quot;x.sum&quot;)

Result

  1.         gr id x.sum
  2. 1 gr1.both  A     3
  3. 2 gr1.both  B    21
  4. 3 gr1.both  C     9
  5. 4 gr2.both  A    11
  6. 5 gr2.both  B    25
  7. 6 gr2.both  C    13

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  • 本文由 发表于 2023年5月25日 02:37:07
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  • dplyr
  • group-by
  • r
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