Question

我正在使用以下示例数据开发一个闪亮的应用程序：

library(tidyr)
library(dplyr)

df <- data.frame(Year = rep(2014:2017, each = 10),
                 ID = rep(1:10, times = 4),
                 Score1 = runif(40),
                 Score2 = runif(40),
                 Score3 = runif(40)) %>% 
  gather(Score1, Score2, Score3, key = "Measure", value = "Value") %>% 
  unite(Measure, Year, col = "Measure", sep = "_") %>% 
  spread(Measure, Value)

给出了：

> glimpse(df)
Observations: 10
Variables: 13
$ ID          <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
$ Score1_2014 <dbl> 0.03936843, 0.62027828, 0.56994489, 0.94410280, 0.98747476, 0.78021699, 0.5...
$ Score1_2015 <dbl> 0.456492381, 0.881373411, 0.601315132, 0.003073382, 0.436619197, 0.49193024...
$ Score1_2016 <dbl> 0.4937857, 0.4414206, 0.6716621, 0.2483740, 0.2376593, 0.4231311, 0.5250772...
$ Score1_2017 <dbl> 0.6824536, 0.1020127, 0.9973474, 0.4304465, 0.9194684, 0.8938086, 0.9133654...
$ Score2_2014 <dbl> 0.01550399, 0.03318784, 0.31463461, 0.99324685, 0.19417234, 0.10408623, 0.9...
$ Score2_2015 <dbl> 0.7631779, 0.4471922, 0.9119910, 0.5792838, 0.8458717, 0.9716529, 0.9580503...
$ Score2_2016 <dbl> 0.78565372, 0.20382477, 0.04103231, 0.33246223, 0.65301709, 0.03227641, 0.3...
$ Score2_2017 <dbl> 0.320235691, 0.211477745, 0.575208127, 0.290498894, 0.696220903, 0.94622610...
$ Score3_2014 <dbl> 0.93234031, 0.40570043, 0.07134056, 0.83916278, 0.57897129, 0.59457072, 0.3...
...

我想创建一个用户定义的函数，允许选择分数类型（例如Score1，Score2或Score3），即开始年份year_from ）和结束年份（year_to），并计算年份之间的差异。例如，选择Score1，2015和2016会给出：

   ID Score1_2016 Score1_2015        Diff
1   1   0.4937857 0.456492381  0.03729332
2   2   0.4414206 0.881373411 -0.43995279
3   3   0.6716621 0.601315132  0.07034700
4   4   0.2483740 0.003073382  0.24530064
5   5   0.2376593 0.436619197 -0.19895987
6   6   0.4231311 0.491930246 -0.06879918
7   7   0.5250772 0.596241541 -0.07116431
8   8   0.1416265 0.019224651  0.12240182
9   9   0.7573208 0.073456457  0.68386434
10 10   0.3575724 0.566328136 -0.20875574

我已阅读programming with dplyr上的文档，但对quosures的使用不太自信。尝试以下公式失败：

selectr <- function(data, value, year_from, year_to){
  recent <- max(year_from, year_to) # determine earlier year
  older <- min(year_from, year_to)
  recent.name <- paste(value, recent, sep = "_") # Create column names from original df
  older.name <- paste(value, older, sep = "_")
  recent.name <- enquo(recent.name) 
  older.name <- enquo(older.name)

  data %>% 
    select(ID, !!recent.name, !!older.name) %>% 
    mutate(Diff = !!recent.name - !!older.name)
}


selectr(data = df, 
        value = "Score1", 
        year_from = 2015, 
        year_to = 2016)

产生错误：Error in !older.name : invalid argument type

如果省略mutate(Diff = !!recent.name - !!older.name)，则函数的其余部分可以正常工作，但我确实需要公式中的差值计算。

Answer 1

我认为你需要改变两件事来使你的功能发挥作用：

您想将recent.name和older.name从字符串转换为符号。这可以通过as.name()函数实现。函数enquo()将“promise”对象转换为quosure（symbol）。
mutate步骤中的运算符优先级似乎存在问题。如果您将!!更改为UQ()（相当于），问题就解决了。

这是更正版本：

selectr <- function(data, value, year_from, year_to) {

  recent <- max(year_from, year_to) 
  older <- min(year_from, year_to)
  recent.name <- paste(value, recent, sep = "_") 
  older.name <- paste(value, older, sep = "_")
  recent.name <- as.name(recent.name) 
  older.name <- as.name(older.name)

  data %>% 
    select(ID, UQ(recent.name), UQ(older.name)) %>% 
    mutate(Diff = UQ(recent.name) - UQ(older.name))

}

selectr(data = df, 
        value = "Score1", 
        year_from = 2015, 
        year_to = 2016)

使用dplyr的用户定义函数 - 基于组合参数的mutate列

1 个答案: