如何在R中从列表中找到具有最大值的列名

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英文:

How to find the column name with max value from a list in R

问题

以下是要翻译的内容：

"我有这样的数据："

  1. df <- data.frame(Product= paste0("AE", c(1:11)),
  2.                  Condition=c("-","A","A","B","B","B","C","C","C","D","D"),
  3.                  Score1=c(0.231,0.831,0.894,1e-05,1e-05,1e-05,0.874,0.785,0.879,1e-08,1e-08),
  4.                  Score2=c(0.968,0.168,0.105,0.239,0.149,0.125,1e-05,1e-05,1e-05,0.159,0.105),
  5.                  Score3=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,1e-05,1e-05,1e-05,0.855,0.827),
  6.                  Score4=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,0.492,0.500,0.656,1e-08,1e-08))

"使用split(df, df$Condition)后会变成："

  1. > split(df, df$Condition)
  2. $`-`
  3.   Product Condition Score1 Score2 Score3 Score4
  4. 1     AE1         -  0.231  0.968  1e-07  1e-07
  6. $A
  7.   Product Condition Score1 Score2 Score3 Score4
  8. 2     AE2         A  0.831  0.168  1e-07  1e-07
  9. 3     AE3         A  0.894  0.105  1e-07  1e-07
  11. $B
  12.   Product Condition Score1 Score2 Score3 Score4
  13. 4     AE4         B  1e-05  0.239  1e-07  1e-07
  14. 5     AE5         B  1e-05  0.149  1e-07  1e-07
  15. 6     AE6         B  1e-05  0.125  1e-07  1e-07
  17. $C
  18.   Product Condition Score1 Score2 Score3 Score4
  19. 7     AE7         C  0.874  1e-05  1e-05  0.492
  20. 8     AE8         C  0.785  1e-05  1e-05  0.500
  21. 9     AE9         C  0.879  1e-05  1e-05  0.656
  23. $D
  24.    Product Condition Score1 Score2 Score3 Score4
  25. 10    AE10         D  1e-08  0.159  0.855  1e-08
  26. 11    AE11         D  1e-08  0.105  0.827  1e-08

"我想要在每个列表中获取具有最大值的列，如下所示："

"在$`-` 中，$A和$C的最大值在Score1，$B中的最大值在Score2，$D中的最大值在Score3。"

"我想要获得一个表格，如下所示："

  1. Condition    Column
  2.         -    Score1
  3.         A    Score1
  4.         B    Score2
  5.         C    Score1
  6.         D    Score3

"任何建议都有帮助。谢谢。"

英文:

I have a data like this :

  1. df <- data.frame(Product= paste0("AE", c(1:11)),
  2.                  Condition=c("-","A","A","B","B","B","C","C","C","D","D"),
  3.                  Score1=c(0.231,0.831,0.894,1e-05,1e-05,1e-05,0.874,0.785,0.879,1e-08,1e-08),
  4.                  Score2=c(0.968,0.168,0.105,0.239,0.149,0.125,1e-05,1e-05,1e-05,0.159,0.105),
  5.                  Score3=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,1e-05,1e-05,1e-05,0.855,0.827),
  6.                  Score4=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,0.492,0.500,0.656,1e-08,1e-08))

after split(df, df$Condition) will like :

  1. > split(df, df$Condition)
  2. $`-`
  3.   Product Condition Score1 Score2 Score3 Score4
  4. 1     AE1         -  0.231  0.968  1e-07  1e-07
  6. $A
  7.   Product Condition Score1 Score2 Score3 Score4
  8. 2     AE2         A  0.831  0.168  1e-07  1e-07
  9. 3     AE3         A  0.894  0.105  1e-07  1e-07
  11. $B
  12.   Product Condition Score1 Score2 Score3 Score4
  13. 4     AE4         B  1e-05  0.239  1e-07  1e-07
  14. 5     AE5         B  1e-05  0.149  1e-07  1e-07
  15. 6     AE6         B  1e-05  0.125  1e-07  1e-07
  17. $C
  18.   Product Condition Score1 Score2 Score3 Score4
  19. 7     AE7         C  0.874  1e-05  1e-05  0.492
  20. 8     AE8         C  0.785  1e-05  1e-05  0.500
  21. 9     AE9         C  0.879  1e-05  1e-05  0.656
  23. $D
  24.    Product Condition Score1 Score2 Score3 Score4
  25. 10    AE10         D  1e-08  0.159  0.855  1e-08
  26. 11    AE11         D  1e-08  0.105  0.827  1e-08

I want to get column with max value in each list,
like :
In $`-`　, $A and $Cis Score1,in $B is Score2,in $D is Score3.

I want get a table like

  1. Condition    Column
  2.         -    Score1
  3.         A    Score1
  4.         B    Score2
  5.         C    Score1
  6.         D    Score3

Any suggestion is helpful.Thank you.

答案1

得分: 1

在基本的 R 中，使用您的分割方法：

  1. stack(lapply(split(df[-(1:2)], df[2]), \(x)names(x)[col(x)[which.max(unlist(x))]]))
  2.   values ind
  3. 1 Score2   -
  4. 2 Score1   A
  5. 3 Score2   B
  6. 4 Score1   C
  7. 5 Score3   D

请注意，还有更好的替代方法。

英文:

In base R. using your split:

  1. stack(lapply(split(df[-(1:2)], df[2]), \(x)names(x)[col(x)[which.max(unlist(x))]]))
  2.   values ind
  3. 1 Score2   -
  4. 2 Score1   A
  5. 3 Score2   B
  6. 4 Score1   C
  7. 5 Score3   D

Note that there are better alternatives

答案2

得分: 0

一种高效的方法是使用 tidyr::pivot_longer()，以便将所有列数据聚合在一起，然后可以轻松地获取每个 Condition 中的最高值。

  1. library(tidyverse)
  3. df <- data.frame(Product= paste0("AE", c(1:11)),
  4.                  Condition=c("-","A","A","B","B","B","C","C","C","D","D"),
  5.                  Score1=c(0.231,0.831,0.894,1e-05,1e-05,1e-05,0.874,0.785,0.879,1e-08,1e-08),
  6.                  Score2=c(0.968,0.168,0.105,0.239,0.149,0.125,1e-05,1e-05,1e-05,0.159,0.105),
  7.                  Score3=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,1e-05,1e-05,1e-05,0.855,0.827),
  8.                  Score4=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,0.492,0.500,0.656,1e-08,1e-08))
  10. df |>
  11.   pivot_longer(starts_with("Score"), names_to = "Column") |>
  12.   group_by(Condition) |>
  13.   slice_max(order_by = value, n = 1) |>
  14.   select(Condition, Column)
  15. #> # A tibble: 5 × 2
  16. #> # Groups:   Condition [5]
  17. #>   Condition Column
  18. #>   <chr>     <chr> 
  19. #> 1 -         Score2
  20. #> 2 A         Score1
  21. #> 3 B         Score2
  22. #> 4 C         Score1
  23. #> 5 D         Score3

创建于2023年07月05日，使用 reprex v2.0.2。

英文:

An efficient way to get this is to tidyr::pivot_longer() so that you aggregate all the column data together and can easily take the top value in each Condition.

  1. library(tidyverse)
  3. df &lt;- data.frame(Product= paste0(&quot;AE&quot;, c(1:11)),
  4.                  Condition=c(&quot;-&quot;,&quot;A&quot;,&quot;A&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;,&quot;C&quot;,&quot;C&quot;,&quot;C&quot;,&quot;D&quot;,&quot;D&quot;),
  5.                  Score1=c(0.231,0.831,0.894,1e-05,1e-05,1e-05,0.874,0.785,0.879,1e-08,1e-08),
  6.                  Score2=c(0.968,0.168,0.105,0.239,0.149,0.125,1e-05,1e-05,1e-05,0.159,0.105),
  7.                  Score3=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,1e-05,1e-05,1e-05,0.855,0.827),
  8.                  Score4=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,0.492,0.500,0.656,1e-08,1e-08))
  11. df |&gt; 
  12.   pivot_longer(starts_with(&quot;Score&quot;), names_to = &quot;Column&quot;) |&gt; 
  13.   group_by(Condition) |&gt; 
  14.   slice_max(order_by = value, n = 1) |&gt; 
  15.   select(Condition, Column)
  16. #&gt; # A tibble: 5 &#215; 2
  17. #&gt; # Groups:   Condition [5]
  18. #&gt;   Condition Column
  19. #&gt;   &lt;chr&gt;     &lt;chr&gt; 
  20. #&gt; 1 -         Score2
  21. #&gt; 2 A         Score1
  22. #&gt; 3 B         Score2
  23. #&gt; 4 C         Score1
  24. #&gt; 5 D         Score3

<sup>Created on 2023-07-05 with reprex v2.0.2</sup>

huangapple
  • 本文由 发表于 2023年7月6日 10:49:14
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