计算多个配对变量的实际差异和百分比差异同时。

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英文:

Calculating the actual difference and percentage difference for multiple paired variables simultaneously

问题

以下是您要翻译的内容：

  1. 我有以下示例数据框，并希望一次计算多个配对变量（“10”和“20”对应于测试年份）的实际和百分比差异：
  3. 样本数据：
  5. Group| A_10  |    A_20 |  B_10 |  B_20 
  7. 0       20          21        20        23
  8. 1       30          10        19        11
  9. 2       10          53        30        34
  10. 1       22          32        25        20
  11. 2       34          40        32        30
  12. 0       30          50        NA        40
  13. 0       39          40        19        20
  14. 1       40          NA        20        20
  15. 2       50          10        20        10
  16. 0       34          23        30        10
  18. 这是当前的工作代码：
  20. library(dplyr)
  22. # 假设数据框命名为'df'并具有以下结构：
  23. # 'var1_1'，'var1_2'，...代表第一组变量
  24. # 'var2_1'，'var2_2'，...代表第二组变量
  26. # 定义要计算差异的变量对
  27. variable_pairs <- list(
  28.   c("A_10", "A_20"),
  29.   c("B_10", "B_20")) # 我还有另外20个配对变量    
  31. # 计算每个变量对的实际差异和百分比差异
  32. df6 <- df %>%
  33.   mutate(
  34.     across(
  35.       all_of(unlist(variable_pairs)),
  36.       ~ .x - get(variable_pairs[[cur_column()]][2]),
  37.       .names = "{.col}_actual_diff"
  38.     ),
  39.     across(
  40.       all_of(unlist(variable_pairs)),
  41.       ~ (.x - get(variable_pairs[[cur_column()]][2])) / get(variable_pairs[[cur_column()]][2]) * 100,
  42.       .names = "{.col}_percentage_diff"
  43.     )
  44.   )

不幸的是，我在某个地方出错了或者过于复杂。上述代码会出现以下错误：

  1. 错误 in `mutate()`:
  2. ℹ In argument: `across(...)`.
  3. Caused by error in `across()`:
  4. ! Can't compute column `vo2mlkg_12_actual_diff`.
  5. Caused by error in `get()`:
  6. ! invalid first argument
  7. Run `rlang::last_trace()` to see where the error occurred.

有人能提出修复或更简单的解决方案吗？

附加说明：

长数据：

Group| variable | phase | Value |

0 A 10 20
1 B 20 19
2 C 20 30
1 D 10 25
2 E 20 32
0 F 10 NA
0 G 20 19
1 H 10 20
2 I 10 20
0 J 20 30

  2. 感谢@Maël的解决方案：
  4. ```R
  5. library(dplyr)
  6. library(tidyr)
  7. library(magrittr)
  9. df2 <- df[,-2] 
  10. df2 %<>% ...
英文:

I have the following example data frame and would like to calculate the actual and percentage differences across multiple paired variables ("10" and "20" correspond to year tested) at once:

sample data:

  1. Group| A_10  |    A_20 |  B_10 |  B_20 
  3. 0       20          21        20        23
  4. 1       30          10        19        11
  5. 2       10          53        30        34
  6. 1       22          32        25        20
  7. 2       34          40        32        30
  8. 0       30          50        NA        40
  9. 0       39          40        19        20
  10. 1       40          NA        20        20
  11. 2       50          10        20        10
  12. 0       34          23        30        10

This is the current working code:

  1. library(dplyr)
  3. # Assuming data frame is named 'df' and has the following structure:
  4. # 'var1_1', 'var1_2', ... represent the first set of variables
  5. # 'var2_1', 'var2_2', ... represent the second set of variables
  7. # Define the pairs of variables for which you want to calculate the differences
  8. variable_pairs <- list(
  9.   c("A_10", "A_20"),
  10.   c("B_10", "B_20")) # I have another 20 paired variabels    
  12. # Calculate the actual and percentage differences for each variable pair
  13. df6 <- df %>%
  14.   mutate(
  15.     across(
  16.       all_of(unlist(variable_pairs)),
  17.       ~ .x - get(variable_pairs[[cur_column()]][2]),
  18.       .names = "{.col}_actual_diff"
  19.     ),
  20.     across(
  21.       all_of(unlist(variable_pairs)),
  22.       ~ (.x - get(variable_pairs[[cur_column()]][2])) / get(variable_pairs[[cur_column()]][2]) * 100,
  23.       .names = "{.col}_percentage_diff"
  24.     )
  25.   )

Unfortunately I am going wrong somewhere or overcomplicating things. The above code give this error: Error in `mutate()`:
ℹ In argument: `across(...)`.
Caused by error in `across()`:
! Can't compute column `vo2mlkg_12_actual_diff`.
Caused by error in `get()`:
! invalid first argument
Run `rlang::last_trace()` to see where the error occurred.

Can anyone suggest a fix or a simpler solution.

addendum:

long data

  1. Group| variable  | phase |  Value | 
  3. 0       A           10        20        
  4. 1       B           20        19        
  5. 2       C           20        30        
  6. 1       D           10        25        
  7. 2       E           20        32       
  8. 0       F           10        NA        
  9. 0       G           20        19        
  10. 1       H           10        20        
  11. 2       I           10        20        
  12. 0       J           20        30

Solution thanks to @Maël:

  2. library(dplyr)
  3. library(tidyr)
  4. library(magrittr)
  6. df2 <- df[,-2] 
  7. df2 %<>%
  8.   pivot_longer(-group, names_sep = "_", names_to = c("set", ".value")) %>%
  9.   {colnames(.) <- c("group", "set", "pre", "post"); .} %>%
  10.   mutate(
  11.     diff = post - pre,
  12.     diff_perc = ((post - pre) / pre) * 100
  13.   )%>%
  14.   group_by(group, set) %>%
  15.   summarize(
  16.     mean_diff = mean(diff, na.rm = TRUE),
  17.     mean_diff_perc = mean(diff_perc, na.rm = TRUE)
  18.   ) %>%
  19.   pivot_wider(names_from = set, values_from = c(mean_diff, mean_diff_perc))

答案1

得分: 1

你可以使用多个 across 函数来计算差异：

  1. library(dplyr)
  2. df %>%
  3.   mutate(across(matches("_post$"), .names = "{gsub('post','', .col)}diff") - across(matches("_pre$")),
  4.          (across(matches("_post$"), .names = "{gsub('post','', .col)}perc_diff") - across(matches("_pre$"))) / across(matches("_post$")))

或者，可能更简单的方法是，首先对数据进行透视，然后计算差异：

  1. library(tidyr)
  2. df %>%
  3.   pivot_longer(-Group, names_sep = "_", names_to = c("set", ".value")) %>%
  4.   mutate(diff = post - pre,
  5.          diff_perc = (post - pre) / post)
英文:

You can use multiple across:

  1. library(dplyr)
  2. df %>% 
  3.   mutate(across(matches("_post$"), .names = "{gsub('post','', .col)}diff") - across(matches("_pre$")),
  4.          (across(matches("_post$"), .names = "{gsub('post','', .col)}perc_diff") - across(matches("_pre$"))) / across(matches("_post$"))) %>% 
  6. # # A tibble: 10 × 9
  7. #    Group A_pre A_post B_pre B_post A_diff B_diff A_perc_diff B_perc_diff
  8. #    <int> <int>  <int> <int>  <int>  <int>  <int>       <dbl>       <dbl>
  9. #  1     0    20     21    20     23      1      3      0.0476      0.130 
  10. #  2     1    30     10    19     11    -20     -8     -2          -0.727 
  11. #  3     2    10     53    30     34     43      4      0.811       0.118 
  12. #  4     1    22     32    25     20     10     -5      0.312      -0.25  
  13. #  5     2    34     40    32     30      6     -2      0.15       -0.0667
  14. #  6     0    30     50    NA     40     20     NA      0.4        NA     
  15. #  7     0    39     40    19     20      1      1      0.025       0.05  
  16. #  8     1    40     NA    20     20     NA      0     NA           0     
  17. #  9     2    50     10    20     10    -40    -10     -4          -1     
  18. # 10     0    34     23    30     10    -11    -20     -0.478      -2

Or, probably simpler, you can pivot your data first, and then compute the differences:

  1. library(tidyr)
  2. df %>% 
  3.   pivot_longer(-Group, names_sep = "_", names_to = c("set", ".value")) %>% 
  4.   mutate(diff = post - pre,
  5.          diff_perc = (post - pre) / post)
  7. # # A tibble: 20 × 6
  8. #    Group set     pre  post  diff diff_perc
  9. #    <int> <chr> <int> <int> <int>     <dbl>
  10. #  1     0 A        20    21     1    0.0476
  11. #  2     0 B        20    23     3    0.130 
  12. #  3     1 A        30    10   -20   -2     
  13. #  4     1 B        19    11    -8   -0.727 
  14. #  5     2 A        10    53    43    0.811 
  15. #  6     2 B        30    34     4    0.118 
  16. #  7     1 A        22    32    10    0.312 
  17. #  8     1 B        25    20    -5   -0.25  
  18. #  9     2 A        34    40     6    0.15  
  19. # 10     2 B        32    30    -2   -0.0667
  20. # 11     0 A        30    50    20    0.4   
  21. # 12     0 B        NA    40    NA   NA     
  22. # 13     0 A        39    40     1    0.025 
  23. # 14     0 B        19    20     1    0.05  
  24. # 15     1 A        40    NA    NA   NA     
  25. # 16     1 B        20    20     0    0     
  26. # 17     2 A        50    10   -40   -4     
  27. # 18     2 B        20    10   -10   -1     
  28. # 19     0 A        34    23   -11   -0.478 
  29. # 20     0 B        30    10   -20   -2

huangapple
  • 本文由 发表于 2023年5月17日 17:46:09
  • 转载请务必保留本文链接：https://go.coder-hub.com/76270719.html
  • across
  • dplyr
  • mutate
  • percentage
  • r
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