使用dplyr找到数据框中所有右侧列都为零的最左列的方法。

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英文:

How to find the leftmost column of a data frame with all zeros in columns to the right with dplyr

问题

以下是您要翻译的内容：

  1. I have a data frame like this:
  3. df <- data.frame(v1=c(0,1), v2=c(0,1),v3=c(0,1), v4=c(0,1) ) %>% 
  4.     expand(v1,v2,v3,v4)
  5. df
  7. # A tibble: 16 x 4
  8.     var1  var2  var3  var4
  9.    <dbl> <dbl> <dbl> <dbl>
  10.  1     0     0     0     0
  11.  2     0     0     0     1
  12.  3     0     0     1     0
  13.  4     0     0     1     1
  14.  5     0     1     0     0
  15.  6     0     1     0     1
  16.  7     0     1     1     0
  17.  8     0     1     1     1
  18.  9     1     0     0     0
  19. 10     1     0     0     1
  20. 11     1     0     1     0
  21. 12     1     0     1     1
  22. 13     1     1     0     0
  23. 14     1     1     0     1
  24. 15     1     1     1     0
  25. 16     1     1     1     1

For each row, I would like to identify the leftmost column that contains a zero, such that all columns to the right also contain 0s. The expected output is as follows:

  1. # A tibble: 16 x 5
  2.     var1  var2  var3  var4 result
  3.    <dbl> <dbl> <dbl> <dbl> <chr> 
  4.  1     0     0     0     0 var1  
  5.  2     0     0     0     1 NA    
  6.  3     0     0     1     0 var4  
  7.  4     0     0     1     1 NA    
  8.  5     0     1     0     0 var3  
  9.  6     0     1     0     1 NA    
  10.  7     0     1     1     0 var4  
  11.  8     0     1     1     1 NA    
  12.  9     1     0     0     0 var2  
  13. 10     1     0     0     1 NA    
  14. 11     1     0     1     0 var4  
  15. 12     1     0     1     1 NA    
  16. 13     1     1     0     0 var3  
  17. 14     1     1     0     1 NA    
  18. 15     1     1     1     0 var4  
  19. 16     1     1     1     1 NA

If possible, I would prefer a tidyverse solution.

  2. <details>
  3. <summary>英文:</summary>
  5. I have a data frame like this:

df <- data.frame(v1=c(0,1), v2=c(0,1),v3=c(0,1), v4=c(0,1) ) %>%
expand(v1,v2,v3,v4)
df

A tibble: 16 x 4

  1. var1  var2  var3  var4

<dbl> <dbl> <dbl> <dbl>
1 0 0 0 0
2 0 0 0 1
3 0 0 1 0
4 0 0 1 1
5 0 1 0 0
6 0 1 0 1
7 0 1 1 0
8 0 1 1 1
9 1 0 0 0
10 1 0 0 1
11 1 0 1 0
12 1 0 1 1
13 1 1 0 0
14 1 1 0 1
15 1 1 1 0
16 1 1 1 1

  4. For each row, I would like to identify the leftmost column that contains a zero, such that all columns to the right also contain 0s. The expected output is as follows:

A tibble: 16 x 5

  1. var1  var2  var3  var4 result

<dbl> <dbl> <dbl> <dbl> <chr>
1 0 0 0 0 var1
2 0 0 0 1 NA
3 0 0 1 0 var4
4 0 0 1 1 NA
5 0 1 0 0 var3
6 0 1 0 1 NA
7 0 1 1 0 var4
8 0 1 1 1 NA
9 1 0 0 0 var2
10 1 0 0 1 NA
11 1 0 1 0 var4
12 1 0 1 1 NA
13 1 1 0 0 var3
14 1 1 0 1 NA
15 1 1 1 0 var4
16 1 1 1 1 NA

  2. If possible, I would prefer a tidyverse solution.
  4. </details>
  7. # 答案1
  8. **得分**: 3
  10. 这是您提供的代码的中文翻译部分：
  12. ```R
  13. 我对tidyverse解决方案一无所知，但这里有一个基本的R答案：
  15. chk <- simplify2array(rev(Reduce(`+`, rev(df), accumulate=TRUE))) == 0
  16. df$name <- names(df)[max.col(chk, "first")]
  17. df$name[rowSums(chk) == 0] <- NA
  18. df

这段代码的作用是将数据框df中的列进行一些操作，然后将结果存储在df的"name"列中。

英文:

I've no idea about a tidyverse solution, but here's a base R answer:

  1. chk &lt;- simplify2array(rev(Reduce(`+`, rev(df), accumulate=TRUE))) == 0
  2. df$name &lt;- names(df)[max.col(chk, &quot;first&quot;)]
  3. df$name[rowSums(chk) == 0] &lt;- NA
  4. df
  6. ### A tibble: 16 &#215; 5
  7. ##      v1    v2    v3    v4 name 
  8. ##   &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt;
  9. ## 1     0     0     0     0 v1   
  10. ## 2     0     0     0     1 NA   
  11. ## 3     0     0     1     0 v4   
  12. ## 4     0     0     1     1 NA   
  13. ## 5     0     1     0     0 v3   
  14. ## 6     0     1     0     1 NA   
  15. ## 7     0     1     1     0 v4   
  16. ## 8     0     1     1     1 NA   
  17. ## 9     1     0     0     0 v2   
  18. ##10     1     0     0     1 NA   
  19. ##11     1     0     1     0 v4   
  20. ##12     1     0     1     1 NA   
  21. ##13     1     1     0     0 v3   
  22. ##14     1     1     0     1 NA   
  23. ##15     1     1     1     0 v4   
  24. ##16     1     1     1     1 NA

答案2

得分: 3

这个解决方案将每一行的数值传递给一个辅助函数，使用 dplyr::c_across()；然后辅助函数使用 rle() 来返回最后一组零值的起始位置，如果最后一个值是零，否则返回 NA。

  1. library(dplyr)
  3. find_zero <- function(x, cols) {
  4.   x_rle <- rle(x)
  5.   if (tail(x_rle$values, 1) != 0) NA
  6.   else rev(cols)[[tail(x_rle$lengths, 1)]]
  7. }
  9. df %>%
  10.   rowwise() %>%
  11.   mutate(result = find_zero(c_across(v1:v4), names(.))) %>%
  12.   ungroup()
  1. # A tibble: 16 × 5
  2.       v1    v2    v3    v4 result
  3.    <dbl> <dbl> <dbl> <dbl> <chr> 
  4.  1     0     0     0     0 v1    
  5.  2     0     0     0     1 <NA>  
  6.  3     0     0     1     0 v4    
  7.  4     0     0     1     1 <NA>  
  8.  5     0     1     0     0 v3    
  9.  6     0     1     0     1 <NA>  
  10.  7     0     1     1     0 v4    
  11.  8     0     1     1     1 <NA>  
  12.  9     1     0     0     0 v2    
  13. 10     1     0     0     1 <NA>  
  14. 11     1     0     1     0 v4    
  15. 12     1     0     1     1 <NA>  
  16. 13     1     1     0     0 v3    
  17. 14     1     1     0     1 <NA>  
  18. 15     1     1     1     0 v4    
  19. 16     1     1     1     1 <NA>
英文:

This solution passes each row of values to a helper function using dplyr::c_across(); the helper function then uses rle() to return the start of the last run of zeros, if the last value is zero, and NA otherwise.

  1. library(dplyr)
  3. find_zero &lt;- function(x, cols) {
  4.   x_rle &lt;- rle(x)
  5.   if (tail(x_rle$values, 1) != 0) NA
  6.   else rev(cols)[[tail(x_rle$lengths, 1)]]
  7. }
  9. df %&gt;%
  10.   rowwise() %&gt;%
  11.   mutate(result = find_zero(c_across(v1:v4), names(.))) %&gt;%
  12.   ungroup()
  1. # A tibble: 16 &#215; 5
  2.       v1    v2    v3    v4 result
  3.    &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt; 
  4.  1     0     0     0     0 v1    
  5.  2     0     0     0     1 &lt;NA&gt;  
  6.  3     0     0     1     0 v4    
  7.  4     0     0     1     1 &lt;NA&gt;  
  8.  5     0     1     0     0 v3    
  9.  6     0     1     0     1 &lt;NA&gt;  
  10.  7     0     1     1     0 v4    
  11.  8     0     1     1     1 &lt;NA&gt;  
  12.  9     1     0     0     0 v2    
  13. 10     1     0     0     1 &lt;NA&gt;  
  14. 11     1     0     1     0 v4    
  15. 12     1     0     1     1 &lt;NA&gt;  
  16. 13     1     1     0     0 v3    
  17. 14     1     1     0     1 &lt;NA&gt;  
  18. 15     1     1     1     0 v4    
  19. 16     1     1     1     1 &lt;NA&gt;

huangapple
  • 本文由 发表于 2023年2月24日 10:27:07
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