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r - 如何解决，问题 w mutate 必须是大小 1，而不是 2

问题描述

我有一个简单的 df，其中包含列（segment、study_name、Effective_Overall_TopboxMean 和其他一些基于数值的字段）。

输入在这里：

structure(list(segment = c("Developers", "lowcode"), study_name = c("DeleteMe", 
    "DeleteMe"), Effective_Overall_TopboxMean = c(0.96875, 0.5), 
        Effective_Effective_TopboxMean = c(1, 0.571428571428571), 
        Effective_Useful_TopboxMean = c(1,0.571428571428571),Effective_Performant_TopboxMean = c(0.916666666666667, 
        0.380952380952381), Efficient_Overall_TopboxMean = c(0.765625, 
        0.357142857142857), Efficient_Efficient_TopboxMean = c(0.78125, 
        0.357142857142857), Efficient_Ease_TopboxMean = c(0.796875, 
        0.321428571428571), Efficient_Learn_TopboxMean = c(0.6875, 
        0.428571428571429), Empowered_Overall_TopboxMean = c(0.8375, 
        0.414285714285714), Empowered_Satisfaction_TopboxMean = c(0.84375, 
        0.321428571428571), Empowered_Empowered_TopboxMean = c(0.75, 
        0.523809523809524), Empowered_Enjoyable_TopboxMean = c(0.916666666666667, 
        0.428571428571429), UXQ_Overall_TopboxMean = c(0.827205882352941, 
        0.411764705882353), USE_Mean = c(0.816825396825397, 0.816825396825397
        ), USE_Usefulness_Mean = c(0.851190476190476, 0.851190476190476
        ), USE_Ease_Mean = c(0.774891774891775, 0.774891774891775
        ), USE_Learning_Mean = c(0.814285714285714, 0.814285714285714
        ), USE_Satisfaction_Mean = c(0.844897959183673, 0.844897959183673
        ), USE_SNOW_Mean = c(0.811904761904762, 0.811904761904762
        ), UXQ_Overall_Mean = c(0.81624649859944, 0.81624649859944
        ), run_date = structure(c(18865, 18865), class = "Date"), 
        TaskCompletionRate = c(0.561904761904762, 0.561904761904762
        ), WeightedScore = c(0, 0), ExperienceGrade = c("F Unacceptable", 
        "F Unacceptable")), row.names = c(NA, -2L), groups = structure(list(
        segment = c("Developers", "lowcode"), .rows = structure(list(
            1L, 2L), ptype = integer(0), class = c("vctrs_list_of", 
        "vctrs_vctr", "list"))), row.names = c(NA, -2L), class = c("tbl_df", 
    "tbl", "data.frame"), .drop = TRUE), class = c("grouped_df", 
    "tbl_df", "tbl", "data.frame"))'

我的 if / else 语句的变异曾经可以工作，但今天它似乎已经坏了，我无法解决错误消息：”

'Error: Problem with `mutate()` column `WeightedEffective`.
ℹ `WeightedEffective = if (...) NULL`.
ℹ `WeightedEffective` must be size 1, not 2.
ℹ The error occurred in group 1: segment = "Developers".
Run `rlang::last_error()` to see where the error occurred.
In addition: Warning messages:
1: Problem with `mutate()` column `WeightedEffective`.
ℹ `WeightedEffective = if (...) NULL`.
ℹ the condition has length > 1 and only the first element will be used
ℹ The warning occurred in group 1: segment = "Developers". 
2: Problem with `mutate()` column `WeightedEffective`.
ℹ `WeightedEffective = if (...) NULL`.
ℹ the condition has length > 1 and only the first element will be used
ℹ The warning occurred in group 1: segment = "Developers". 
3: Problem with `mutate()` column `WeightedEffective`.
ℹ `WeightedEffective = if (...) NULL`.
ℹ the condition has length > 1 and only the first element will be used
ℹ The warning occurred in group 1: segment = "Developers". 
4: Problem with `mutate()` column `WeightedEffective`.
ℹ `WeightedEffective = if (...) NULL`.
ℹ the condition has length > 1 and only the first element will be used
ℹ The warning occurred in group 1: segment = "Developers". '

这是我的脚本：

    UXQ_Boxes <- mutate(UXQ_Boxes,WeightedEffective = 
if (UXQ_Boxes$TaskCompletionRate >.900){
    (UXQ_Boxes$Effective_Overall_TopboxMean)*1
  } else {
    if (UXQ_Boxes$TaskCompletionRate >.790){
      ((UXQ_Boxes$Effective_Overall_TopboxMean)*.9)*100
    } else {
      if (UXQ_Boxes$TaskCompletionRate >.650){
       ((UXQ_Boxes$Effective_Overall_TopboxMean)*.8)*100
      } else {
        if (UXQ_Boxes$TaskCompletionRate >.500){
       ((UXQ_Boxes$Effective_Overall_TopboxMean)*.6)*100
        } else {
          if (UXQ_Boxes$TaskCompletionRate >.250){
       ((UXQ_Boxes$Effective_Overall_TopboxMean)*.4)*100
          } else {
            if (UXQ_Boxes$TaskCompletionRate <.260){
       ((UXQ_Boxes$Effective_Overall_TopboxMean)*.2)*100
            }}}}}}*6)'

任何有关如何解决此问题的建议将不胜感激。

标签： rdplyr

解决方案

使用时if_else代替if使用mutate

UXQ_Boxes2 <- UXQ_Boxes %>% 
  mutate(UWeightedEffective = 6*if_else(TaskCompletionRate > .900, Effective_Overall_TopboxMean*1, 
                                      if_else(TaskCompletionRate > .790, Effective_Overall_TopboxMean*0.9*100,
                                              if_else(TaskCompletionRate > .650, Effective_Overall_TopboxMean*0.8*100,
                                                      if_else(TaskCompletionRate > .500, Effective_Overall_TopboxMean*0.6*100,
                                                              if_else(TaskCompletionRate > .250, Effective_Overall_TopboxMean*0.4*100,
                                                                      Effective_Overall_TopboxMean*0.2*100))))))
#result
UXQ_Boxes2 %>% select(last_col())
Adding missing grouping variables: `segment`
# A tibble: 2 x 2
# Groups:   segment [2]
  segment    UWeightedEffective
  <chr>                   <dbl>
1 Developers               349.
2 lowcode                  180

我注意到在代码的最后乘以 6，我把它移到了代码的开头。我也折叠了最后一个条件UXQ_Boxes$TaskCompletionRate < .260)，因为它没有任何意义，因为之前的条件是UXQ_Boxes$TaskCompletionRate > .250). 我把UXQ_Boxes$Effective_Overall_TopboxMean)*.2)*100_else

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