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:"

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

> split(df, df$Condition)
$`-`
  Product Condition Score1 Score2 Score3 Score4
1     AE1         -  0.231  0.968  1e-07  1e-07

$A
  Product Condition Score1 Score2 Score3 Score4
2     AE2         A  0.831  0.168  1e-07  1e-07
3     AE3         A  0.894  0.105  1e-07  1e-07

$B
  Product Condition Score1 Score2 Score3 Score4
4     AE4         B  1e-05  0.239  1e-07  1e-07
5     AE5         B  1e-05  0.149  1e-07  1e-07
6     AE6         B  1e-05  0.125  1e-07  1e-07

$C
  Product Condition Score1 Score2 Score3 Score4
7     AE7         C  0.874  1e-05  1e-05  0.492
8     AE8         C  0.785  1e-05  1e-05  0.500
9     AE9         C  0.879  1e-05  1e-05  0.656

$D
   Product Condition Score1 Score2 Score3 Score4
10    AE10         D  1e-08  0.159  0.855  1e-08
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$C的最大值在Score1,$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:"

Condition    Column
        -    Score1
        A    Score1
        B    Score2
        C    Score1
        D    Score3

"任何建议都有帮助。谢谢。"

英文:

I have a data like this :

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

> split(df, df$Condition)
$`-`
  Product Condition Score1 Score2 Score3 Score4
1     AE1         -  0.231  0.968  1e-07  1e-07

$A
  Product Condition Score1 Score2 Score3 Score4
2     AE2         A  0.831  0.168  1e-07  1e-07
3     AE3         A  0.894  0.105  1e-07  1e-07

$B
  Product Condition Score1 Score2 Score3 Score4
4     AE4         B  1e-05  0.239  1e-07  1e-07
5     AE5         B  1e-05  0.149  1e-07  1e-07
6     AE6         B  1e-05  0.125  1e-07  1e-07

$C
  Product Condition Score1 Score2 Score3 Score4
7     AE7         C  0.874  1e-05  1e-05  0.492
8     AE8         C  0.785  1e-05  1e-05  0.500
9     AE9         C  0.879  1e-05  1e-05  0.656

$D
   Product Condition Score1 Score2 Score3 Score4
10    AE10         D  1e-08  0.159  0.855  1e-08
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

Condition    Column
        -    Score1
        A    Score1
        B    Score2
        C    Score1
        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:

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

请注意,还有更好的替代方法。

英文:

In base R. using your split:

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

library(tidyverse)

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

df |>
  pivot_longer(starts_with("Score"), names_to = "Column") |>
  group_by(Condition) |>
  slice_max(order_by = value, n = 1) |>
  select(Condition, Column)
#> # A tibble: 5 × 2
#> # Groups:   Condition [5]
#>   Condition Column
#>   <chr>     <chr> 
#> 1 -         Score2
#> 2 A         Score1
#> 3 B         Score2
#> 4 C         Score1
#> 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.

library(tidyverse)

df &lt;- data.frame(Product= paste0(&quot;AE&quot;, c(1:11)),
                 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;),
                 Score1=c(0.231,0.831,0.894,1e-05,1e-05,1e-05,0.874,0.785,0.879,1e-08,1e-08),
                 Score2=c(0.968,0.168,0.105,0.239,0.149,0.125,1e-05,1e-05,1e-05,0.159,0.105),
                 Score3=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,1e-05,1e-05,1e-05,0.855,0.827),
                 Score4=c(1e-07,1e-07,1e-07,1e-07,1e-07,1e-07,0.492,0.500,0.656,1e-08,1e-08))


df |&gt; 
  pivot_longer(starts_with(&quot;Score&quot;), names_to = &quot;Column&quot;) |&gt; 
  group_by(Condition) |&gt; 
  slice_max(order_by = value, n = 1) |&gt; 
  select(Condition, Column)
#&gt; # A tibble: 5 &#215; 2
#&gt; # Groups:   Condition [5]
#&gt;   Condition Column
#&gt;   &lt;chr&gt;     &lt;chr&gt; 
#&gt; 1 -         Score2
#&gt; 2 A         Score1
#&gt; 3 B         Score2
#&gt; 4 C         Score1
#&gt; 5 D         Score3

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

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  • 本文由 发表于 2023年7月6日 10:49:14
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