使用dplyr在R中如何创建一个新列，该列的名称是包含最大值的列的名称？

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英文:

How to create new column with name of column that contains maximum value using dplyr in R?

问题

以下是代码部分的翻译：

  1. 我有这样一个数据框：
  3.     dat <- data.frame(var1 = rnorm(10), var2 = rnorm(10), var3 = rnorm(10), var4 = rnorm(10))
  4.     > dat
  5.              var1        var2        var3       var4
  6.     1  -1.3784414  1.06816022  1.46578217 -0.4141153
  7.     2  -0.3272332 -0.69470574  0.02220395 -0.5502878
  8.     3   0.2559891 -0.06964848 -0.34745180  0.6399705
  9.     4   0.6029044  1.23680560 -0.72392358 -0.1990832
  10.     5   1.3097174 -0.58028595 -0.01487186 -0.8765290
  11.     6  -1.2356668  0.41330063 -1.00375989 -1.1974204
  12.     7  -0.4126320  3.83320678 -1.42059022 -0.6747575
  13.     8   1.7339653  0.58610348  0.40200428  1.4582103
  14.     9   1.2994859  1.65355306  0.75985071  0.6455882
  15.     10 -0.2353356  2.04468739 -0.11521602  0.3251901
  17. 目标是创建一个新的列，该列包含每行中在var2、var3和var4列中包含最大值的列的名称。
  19. 使用以下命令不会得到正确的输出：
  21.     library(dplyr)
  22.     dat %>%
  23.       rowwise() %>%
  24.       mutate(var.max = colnames(.)[which.max(c_across(var2:var4))])
  26.     # A tibble: 10 x 5
  27.     # Rowwise: 
  28.          var1    var2    var3   var4 var.max
  29.         <dbl>   <dbl>   <dbl>  <dbl> <chr>  
  30.      1 -1.38   1.07    1.47   -0.414 var2   
  31.      2 -0.327 -0.695   0.0222 -0.550 var2   
  32.      3  0.256 -0.0696 -0.347   0.640 var3   
  33.      4  0.603  1.24   -0.724  -0.199 var1   
  34.      5  1.31  -0.580  -0.0149 -0.877 var2   
  35.      6 -1.24   0.413  -1.00   -1.20  var1   
  36.      7 -0.413  3.83   -1.42   -0.675 var1   
  37.      8  1.73   0.586   0.402   1.46  var3   
  38.      9  1.30   1.65    0.760   0.646 var1   
  39.     10 -0.235  2.04   -0.115   0.325 var1  
  41. 但是，如果从数据中排除列var1，它可以正常工作：
  43.     dat %>%
  44.       select(-var1) %>%
  45.       rowwise() %>%
  46.       mutate(var.max = colnames(.)[which.max(c_across(var2:var4))])
  48.     # A tibble: 10 x 4
  49.     # Rowwise: 
  50.           var2    var3   var4 var.max
  51.          <dbl>   <dbl>  <dbl> <chr>  
  52.      1  1.07    1.47   -0.414 var3   
  53.      2 -0.695   0.0222 -0.550 var3   
  54.      3 -0.0696 -0.347   0.640 var4   
  55.      4  1.24   -0.724  -0.199 var2   
  56.      5 -0.580  -0.0149 -0.877 var3   
  57.      6  0.413  -1.00   -1.20  var2   
  58.      7  3.83   -1.42   -0.675 var2   
  59.      8  0.586   0.402   1.46  var4   
  60.      9  1.65    0.760   0.646 var2   
  61.     10 2.04   -0.115   0.325 var2  
  63. 就像当var1在最后位置时一样：
  65.     dat %>%
  66.       select(var2, var3, var4, var1) %>%
  67.       rowwise() %>%
  68.       mutate(var.max = colnames(.)[which.max(c_across(var2:var4))])
  70.     # A tibble: 10 x 5
  71.     # Rowwise: 
  72.           var2    var3   var4   var1 var.max
  73.          <dbl>   <dbl>  <dbl>  <dbl> <chr>  
  74.      1  1.07    1.47   -0.414 -1.38  var3   
  75.      2 -0.695   0.0222 -0.550 -0.327 var3   
  76.      3 -0.0696 -0.347   0.640  0.256 var4   
  77.      4  1.24   -0.724  -0.199  0.603 var2   
  78.      5 -0.580  -0.0149 -0.877  1.31  var3   
  79.      6  0.413  -1.00   -1.20  -1.24  var2   
  80.      7 3.83   -1.42   -0.675 -0.413 var2   
  81.      8 0.586   0.402   1.46   1.73  var4   
  82.      9 1.65    0.760   0.646  1.30  var2   
  83.     10 2.04   -0.115   0.325 -0.235 var2 
  85. 我在这里漏掉了什么？
英文:

I have such a data frame:

  1. dat &lt;- data.frame(var1 = rnorm(10), var2 = rnorm(10), var3 = rnorm(10), var4 = rnorm(10))
  2. &gt; dat
  3. var1        var2        var3       var4
  4. 1  -1.3784414  1.06816022  1.46578217 -0.4141153
  5. 2  -0.3272332 -0.69470574  0.02220395 -0.5502878
  6. 3   0.2559891 -0.06964848 -0.34745180  0.6399705
  7. 4   0.6029044  1.23680560 -0.72392358 -0.1990832
  8. 5   1.3097174 -0.58028595 -0.01487186 -0.8765290
  9. 6  -1.2356668  0.41330063 -1.00375989 -1.1974204
  10. 7  -0.4126320  3.83320678 -1.42059022 -0.6747575
  11. 8   1.7339653  0.58610348  0.40200428  1.4582103
  12. 9   1.2994859  1.65355306  0.75985071  0.6455882
  13. 10 -0.2353356  2.04468739 -0.11521602  0.3251901

The aim is to create a new column with the name of the column that contains the maximum value in each row within columns var2, var3 and var4.

Using the following command does not result in the correct output:

  1. library(dplyr)
  2. dat %&gt;%
  3. rowwise() %&gt;%
  4. mutate(var.max = colnames(.)[which.max(c_across(var2:var4))])
  5. # A tibble: 10 x 5
  6. # Rowwise: 
  7. var1    var2    var3   var4 var.max
  8. &lt;dbl&gt;   &lt;dbl&gt;   &lt;dbl&gt;  &lt;dbl&gt; &lt;chr&gt;  
  9. 1 -1.38   1.07    1.47   -0.414 var2   
  10. 2 -0.327 -0.695   0.0222 -0.550 var2   
  11. 3  0.256 -0.0696 -0.347   0.640 var3   
  12. 4  0.603  1.24   -0.724  -0.199 var1   
  13. 5  1.31  -0.580  -0.0149 -0.877 var2   
  14. 6 -1.24   0.413  -1.00   -1.20  var1   
  15. 7 -0.413  3.83   -1.42   -0.675 var1   
  16. 8  1.73   0.586   0.402   1.46  var3   
  17. 9  1.30   1.65    0.760   0.646 var1   
  18. 10 -0.235  2.04   -0.115   0.325 var1

But if the column var1 is excluded from the data it works:

  1. dat %&gt;%
  2. select(-var1) %&gt;%
  3. rowwise() %&gt;%
  4. mutate(var.max = colnames(.)[which.max(c_across(var2:var4))])
  5. # A tibble: 10 x 4
  6. # Rowwise: 
  7. var2    var3   var4 var.max
  8. &lt;dbl&gt;   &lt;dbl&gt;  &lt;dbl&gt; &lt;chr&gt;  
  9. 1  1.07    1.47   -0.414 var3   
  10. 2 -0.695   0.0222 -0.550 var3   
  11. 3 -0.0696 -0.347   0.640 var4   
  12. 4  1.24   -0.724  -0.199 var2   
  13. 5 -0.580  -0.0149 -0.877 var3   
  14. 6  0.413  -1.00   -1.20  var2   
  15. 7  3.83   -1.42   -0.675 var2   
  16. 8  0.586   0.402   1.46  var4   
  17. 9  1.65    0.760   0.646 var2   
  18. 10  2.04   -0.115   0.325 var2

.. just like when var1 is at the last position:

  1. dat %&gt;%
  2. select(var2, var3, var4, var1) %&gt;%
  3. rowwise() %&gt;%
  4. mutate(var.max = colnames(.)[which.max(c_across(var2:var4))])
  5. # A tibble: 10 x 5
  6. # Rowwise: 
  7. var2    var3   var4   var1 var.max
  8. &lt;dbl&gt;   &lt;dbl&gt;  &lt;dbl&gt;  &lt;dbl&gt; &lt;chr&gt;  
  9. 1  1.07    1.47   -0.414 -1.38  var3   
  10. 2 -0.695   0.0222 -0.550 -0.327 var3   
  11. 3 -0.0696 -0.347   0.640  0.256 var4   
  12. 4  1.24   -0.724  -0.199  0.603 var2   
  13. 5 -0.580  -0.0149 -0.877  1.31  var3   
  14. 6  0.413  -1.00   -1.20  -1.24  var2   
  15. 7  3.83   -1.42   -0.675 -0.413 var2   
  16. 8  0.586   0.402   1.46   1.73  var4   
  17. 9  1.65    0.760   0.646  1.30  var2   
  18. 10  2.04   -0.115   0.325 -0.235 var2

What am I missing here?

答案1

得分: 3

为了继续你的逻辑，并且你只是移除了第一列，只需将 1 添加到 which.max()，即：

  1. library(dplyr)
  3. dat %>%
  4.  rowwise() %>%
  5.  mutate(max_col = names(dat)[which.max(c_across(var2:var4)) + 1])

如果你想指定要考虑的列，可以这样做：

  1. my_cols <- c('var2', 'var3', 'var4')
  3. dat %>%
  4.  rowwise() %>%
  5.  mutate(max_col = names(dat)[which.max(c_across(names(dat)[names(dat) %in% my_cols])) + (ncol(dat) - length(my_cols))])
英文:

To continue your logic and since you are only removing the firsrt column, just add 1 to which.max(), i.e.

  1. library(dplyr)
  2. dat %&gt;% 
  3. rowwise() %&gt;% 
  4. mutate(max_col = names(dat)[which.max(c_across(var2:var4)) + 1])
  5. # A tibble: 10 &#215; 5
  6. # Rowwise: 
  7. var1     var2     var3     var4 max_col
  8. &lt;dbl&gt;    &lt;dbl&gt;    &lt;dbl&gt;    &lt;dbl&gt; &lt;chr&gt;  
  9. 1 -1.09     0.768    0.251   -2.67    var2   
  10. 2 -0.822   -1.37     0.901    1.83    var4   
  11. 3  0.0280  -0.00555 -0.0709   0.729   var4   
  12. 4  1.45    -0.132   -2.47     1.45    var4   
  13. 5  0.506   -1.31    -2.75    -0.264   var4   
  14. 6 -0.00538  1.31    -0.368    0.00679 var2   
  15. 7 -0.166   -0.976   -1.42     1.50    var4   
  16. 8 -0.377   -0.101    0.135    0.784   var4   
  17. 9  0.535    0.438    0.0597   0.924   var4   
  18. 10  0.281   -0.481   -0.00177 -0.601   var3

If you want to do it by specifying which columns to consider then,

  1. my_cols &lt;- c(&#39;var2&#39;, &#39;var3&#39;, &#39;var4&#39;)
  2. dat %&gt;%
  3. rowwise() %&gt;%
  4. mutate(max_col = names(dat)[which.max(c_across(names(dat)[names(dat) %in% my_cols])) + (ncol(dat) - length(my_cols))])

答案2

得分: 1

以下是代码部分的翻译：

  1. library(dplyr)
  3. max_col_name <- function(...) {
  4.   row_dat <- across(c(...)) # 如果 dplyr 版本 >= 1.1，请使用 `pick()` 代替 `across()`
  5.   names(row_dat)[which.max(row_dat)]
  6. }
  8. dat %>%
  9.   rowwise() %>%
  10.   mutate(max_col = max_col_name(var2:var4))

数据来源于 OP：

  1. set.seed(123)
  2. dat <- data.frame(var1 = rnorm(10), var2 = rnorm(10), var3 = rnorm(10), var4 = rnorm(10))

创建于 2023-02-23，使用 reprex 包 (v2.0.1)

英文:

If you want to avoid adding the number of columns which are left out (in the above case +1) then we can write a custom function max_col_name() using across() or pick():

  1. library(dplyr)
  3. max_col_name &lt;- function(...) {
  4.   row_dat &lt;- across(c(...)) # if dplyr v &gt;= v 1.1. use `pick()` instead of `across()`
  5.   names(row_dat)[which.max(row_dat)]
  6. }
  8. dat %&gt;% 
  9.   rowwise() %&gt;% 
  10.   mutate(max_col = max_col_name(var2:var4))
  12. #&gt; # A tibble: 10 x 5
  13. #&gt; # Rowwise: 
  14. #&gt;       var1   var2   var3    var4 max_col
  15. #&gt;      &lt;dbl&gt;  &lt;dbl&gt;  &lt;dbl&gt;   &lt;dbl&gt; &lt;chr&gt;  
  16. #&gt;  1 -0.560   1.22  -1.07   0.426  var2   
  17. #&gt;  2 -0.230   0.360 -0.218 -0.295  var2   
  18. #&gt;  3  1.56    0.401 -1.03   0.895  var4   
  19. #&gt;  4  0.0705  0.111 -0.729  0.878  var4   
  20. #&gt;  5  0.129  -0.556 -0.625  0.822  var4   
  21. #&gt;  6  1.72    1.79  -1.69   0.689  var2   
  22. #&gt;  7  0.461   0.498  0.838  0.554  var3   
  23. #&gt;  8 -1.27   -1.97   0.153 -0.0619 var3   
  24. #&gt;  9 -0.687   0.701 -1.14  -0.306  var2   
  25. #&gt; 10 -0.446  -0.473  1.25  -0.380  var3

Data from the OP

  1. set.seed(123)
  2. dat &lt;- data.frame(var1 = rnorm(10), var2 = rnorm(10), var3 = rnorm(10), var4 = rnorm(10))

<sup>Created on 2023-02-23 by the reprex package (v2.0.1)</sup>

huangapple
  • 本文由 发表于 2023年2月23日 20:13:16
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