获取每个样本的平均值，按另一列的ID进行排列

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英文:

Get mean values per sample, arranged by another column of ID's

问题

这是一个我想使用tidyverse解决的问题。很难解释，但示例将其清晰展示出来。

我有一个包含多个名称的数据集。每个名称由1个或多个ID表示（在实际数据中范围为1-5）。每个ID都有5个值，按抽取编号排序。

我想要获取每个名称nms的每个抽取的平均值。这意味着要获取每个个体id的每个抽取的平均val。一些名称（如A）只有一个ID，因此平均val或mean_val将相同，但由于B和C有多个ID，mean_val将对每个抽取进行平均。我希望结果看起来像这样：

  1. # A tibble: 15 × 3
  2. #   nms    draw   mean_val
  3. #   <chr>  <int>     <dbl>
  4. # 1 A         1        3  
  5. # 2 A         2        4  
  6. # 3 A         3        2  
  7. # 4 A         4        4  
  8. # 5 A         5        1  
  9. # 6 B         1        1.5
  10. # 7 B         2        1.5
  11. # 8 B         3        2.5
  12. # 9 B         4        2  
  13. #10 B         5        2  
  14. #11 C         1        1.5
  15. #12 C         2        2.5
  16. #13 C         3        2.5
  17. #14 C         4        2.5
  18. #15 C         5        4

希望这可以帮助你解决问题。

英文:

Here is a problem that I would like to solve using the tidyverse. It's difficult to explain but the example lays it out.

I have a dataset with several names. Each name is represented by 1 or more ID's (ranges from 1-5 in the real data). Each ID has 5 values, ordered by the draw number.

  1. library(tidyverse)
  3. set.seed(190)
  5. dplyr::tibble(nms = c(rep(&quot;A&quot;, 5), rep(&quot;B&quot;,10), rep(&quot;C&quot;, 10)),
  6.               draw = c(rep(1:5, times = 5)),
  7.               id = c(rep(c(&quot;A1&quot;, &quot;B1&quot;, &quot;B2&quot;, &quot;C1&quot;, &quot;C2&quot;), each = 5)),
  8.               val = c(sample(1:4, replace = T, size = 25))) %&gt;%  print(n=25)
  9. #&gt; # A tibble: 25 &#215; 4
  10. #&gt;    nms    draw id      val
  11. #&gt;    &lt;chr&gt; &lt;int&gt; &lt;chr&gt; &lt;int&gt;
  12. #&gt;  1 A         1 A1        3
  13. #&gt;  2 A         2 A1        4
  14. #&gt;  3 A         3 A1        2
  15. #&gt;  4 A         4 A1        4
  16. #&gt;  5 A         5 A1        1
  17. #&gt;  6 B         1 B1        2
  18. #&gt;  7 B         2 B1        1
  19. #&gt;  8 B         3 B1        3
  20. #&gt;  9 B         4 B1        3
  21. #&gt; 10 B         5 B1        2
  22. #&gt; 11 B         1 B2        1
  23. #&gt; 12 B         2 B2        2
  24. #&gt; 13 B         3 B2        2
  25. #&gt; 14 B         4 B2        1
  26. #&gt; 15 B         5 B2        2
  27. #&gt; 16 C         1 C1        2
  28. #&gt; 17 C         2 C1        1
  29. #&gt; 18 C         3 C1        1
  30. #&gt; 19 C         4 C1        3
  31. #&gt; 20 C         5 C1        4
  32. #&gt; 21 C         1 C2        1
  33. #&gt; 22 C         2 C2        4
  34. #&gt; 23 C         3 C2        4
  35. #&gt; 24 C         4 C2        2
  36. #&gt; 25 C         5 C2        4

<sup>Created on 2023-05-31 with reprex v2.0.2</sup>

I would like to get the average of each draw for each name nms. This means getting the average val of each draw for every individual id. Some names like A only have one id, so the average val, or mean_val, will be the same, but since B and C have multiple id's, the mean_val will be averaged for each draw. I would like the results to look like this

  1. # A tibble: 15 &#215; 3
  2. #   nms    draw   mean_val
  3. #   &lt;chr&gt; &lt;int&gt; &lt;dbl&gt;
  4. # 1 A         1   3  
  5. # 2 A         2   4  
  6. # 3 A         3   2  
  7. # 4 A         4   4  
  8. # 5 A         5   1  
  9. # 6 B         1   1.5
  10. # 7 B         2   1.5
  11. # 8 B         3   2.5
  12. # 9 B         4   2  
  13. #10 B         5   2  
  14. #11 C         1   1.5
  15. #12 C         2   2.5
  16. #13 C         3   2.5
  17. #14 C         4   2.5
  18. #15 C         5   4

答案1

得分: 0

我认为这将提供您想要的结果，或至少让您接近目标。

  1. output <- df %>%
  2.   group_by(nms, draw) %>%
  3.   summarize(mean_val = mean(val))
  1. # 一个 tibble: 15 × 3
  2. # 组:   nms [3]
  3.    nms    draw mean_val
  4.    <chr> <int>    <dbl>
  5.  1 A         1      3  
  6.  2 A         2      4  
  7.  3 A         3      2  
  8.  4 A         4      4  
  9.  5 A         5      1  
  10.  6 B         1      1.5
  11.  7 B         2      1.5
  12.  8 B         3      2.5
  13.  9 B         4      2  
  14. 10 B         5      2  
  15. 11 C         1      1.5
  16. 12 C         2      2.5
  17. 13 C         3      2.5
  18. 14 C         4      2.5
  19. 15 C         5      4
英文:

I think this will give you what you want or at least get you close.

  1. output &lt;- df %&gt;% 
  2.   group_by(nms, draw) %&gt;% 
  3.   summarize(mean_val = mean(val))
  1. # A tibble: 15 &#215; 3
  2. # Groups:   nms [3]
  3.    nms    draw mean_val
  4.    &lt;chr&gt; &lt;int&gt;    &lt;dbl&gt;
  5.  1 A         1      3  
  6.  2 A         2      4  
  7.  3 A         3      2  
  8.  4 A         4      4  
  9.  5 A         5      1  
  10.  6 B         1      1.5
  11.  7 B         2      1.5
  12.  8 B         3      2.5
  13.  9 B         4      2  
  14. 10 B         5      2  
  15. 11 C         1      1.5
  16. 12 C         2      2.5
  17. 13 C         3      2.5
  18. 14 C         4      2.5
  19. 15 C         5      4

huangapple
  • 本文由 发表于 2023年6月1日 06:13:02
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  • r
  • tidyverse
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