r - 无法在 dplyr 中使用多字变量，还是我遗漏了什么？

Question

与 beta.linalool 相比，为什么 dplyr 不喜欢我的函数中的“beta linalool”格式？

我花了几个小时进行故障排除才找出问题所在。有没有一种方法可以使用变量被标记为多个单词的数据，或者我应该将所有内容都移至 beta.linalool 类型格式？

我学到的一切都来自Programming with dplyr 。

library(ggplot2)
library(readxl)
library(dplyr)
library(magrittr)

Data3<- read_excel("Desktop/Data3.xlsx")

Data3 %>% filter(Variety=="CS 420A"&`Red Blotch`=="-")%>% group_by(`Time Point`)%>%
  summarise(m=mean(`beta linalool`),SD=sd(`beta linalool`))
# A tibble: 4 x 3
  `Time Point`       m         SD
  <chr>           <dbl>      <dbl>
1 End          0.00300  0.000117  
2 Mid          0.00385  0.000353  
3 Must         0.000254 0.00000633
4 Start        0.000785 0.000283  

现在当我把它变成一个函数时:

cwine<-function(df,v,rb,c){
  c<-enquo(c)
  df %>% filter(Variety==v&`Red Blotch`==rb)%>% 
    group_by(`Time Point`) %>%
    summarise_(m=mean(!!c),SD=sd(!!c)) %>% 
}
cwine(Data3,"CS 420A","-",'beta linalool')
# A tibble: 4 x 3
  `Time Point`     m    SD
  <chr>        <dbl> <dbl>
1 End             NA    NA
2 Mid             NA    NA
3 Must            NA    NA
4 Start           NA    NA
Warning messages:
1: In mean.default(~"beta linalool") :
  argument is not numeric or logical: returning NA #this statement is repeated 4 more times
5: In var(if (is.vector(x) || is.factor(x)) x else as.double(x), na.rm = na.rm) :
  NAs introduced by coercion #this statement is repeated 4 more times

问题在于 beta 芳樟醇被输入为“beta linalool”。我通过在虹膜数据集上尝试这种方法并发现 Petal.Length 不是“花瓣宽度”来解决这个问题:

my_function<-function(ds,x,y,c){
  c<-enquo(c)
  ds %>%filter(Sepal.Length>x&Sepal.Width<y) %>% 
    group_by(Species) %>% 
    summarise(m=mean(!!c),SD=sd(!!c))
}
my_function2(iris,5,4,Petal.Length)
# A tibble: 3 x 3
  Species        m    SD
  <fct>      <dbl> <dbl>
1 setosa      1.53 0.157
2 versicolor  4.32 0.423
3 virginica   5.57 0.536

事实上，我的函数在不同的变量上运行良好:

> cwine(Data2,"CS 420A","-",nerol)
# A tibble: 4 x 3
  `Time Point`        m        SD
  <chr>           <dbl>     <dbl>
1 End          0.000453 0.0000338
2 Mid          0.000659 0.0000660
3 Must         0.000560 0.0000234
4 Start        0.000927 0.0000224

dplyr 真的那么敏感吗？还是我遗漏了什么？

Answer 1

一个选项是将其转换为 symbol 并对其进行评估

library(tidyverse)
cwine <- function(df,v,rb,c){
  
  df %>% 
      filter(Variety==v & `Red Blotch` == rb)%>% 
      group_by(`Time Point`) %>%
       summarise(m = mean(!!rlang::sym(c)),
                 SD = sd(!! rlang::sym(c))) 
}

cwine(Data3,"CS 420A","-",'beta linalool')
# A tibble: 2 x 3
#  `Time Point`       m    SD
#         <int>   <dbl> <dbl>
#1            2 -2.11    2.23
#2            4  0.0171 NA  

<小时/>

此外，如果我们想通过转换为 quosure (enquo) 来传递它，当我们传递带有反引号的变量名称时，它就可以工作(通常，不带引号的版本可以工作，但这里有一个空格)在单词之间并按原样评估它，需要反引号)

cwine <- function(df,v,rb,c){
  c1 <- enquo(c)
  df %>% 
      filter(Variety==v & `Red Blotch` == rb)%>% 
      group_by(`Time Point`) %>%
       summarise(m = mean(!! c1 ),
                 SD = sd(!! c1)) 
}

cwine(Data3,"CS 420A","-",`beta linalool`)
# A tibble: 2 x 3
#   `Time Point`       m    SD
#         <int>   <dbl> <dbl>
#1            2 -2.11    2.23
#2            4  0.0171 NA   

数据

set.seed(24)
Data3 <- tibble(Variety = sample(c("CS 420A", "CS 410A"), 20, replace = TRUE),
`Red Blotch` = sample(c("-", "+"), 20, replace = TRUE), 
`Time Point` = sample(1:4, 20, replace = TRUE),
`beta linalool` = rnorm(20))