使用 dplyr::filter 创建 R 函数问题

我查看了其他答案，但找不到使下面的代码起作用的解决方案。基本上，我正在创建一个函数inner_join两个数据框和filter基于函数中输入的列。

问题是filter部分功能不起作用。但是，如果我将过滤器从函数中取出并附加它，它就会起作用mydiff("a") %>% filter(a.x != a.y)

任何建议都有帮助。

请注意，我是函数输入用引号引起来

library(dplyr)

# fake data
df1<- tibble(id = seq(4,19,2), 
             a = c("a","b","c","d","e","f","g","h"), 
             b = c(rep("foo",3), rep("bar",5)))
df2<- tibble(id = seq(10, 20, 1), 
             a = c("d","a", "e","f","k","m","g","i","h", "a", "b"),
             b = c(rep("bar", 7), rep("foo",4)))

# What I am trying to do
dplyr::inner_join(df1, df2, by = "id") %>% select(id, b.x, b.y) %>% filter(b.x!=b.y)

#> # A tibble: 1 x 3
#>      id b.x   b.y  
#>   <dbl> <chr> <chr>
#> 1    18 bar   foo

# creating a function so that I can filter by difference in column if I have more columns
mydiff <- function(filteron, df_1 = df1, df_2 = df2){
  require(dplyr, warn.conflicts = F)
  col_1 = paste0(quo_name(filteron), "x")
  col_2 = paste0(quo_name(filteron), "y")
  my_df<- inner_join(df_1, df_2, by = "id", suffix = c("x", "y"))
  my_df %>% select(id, col_1, col_2) %>% filter(col_1 != col_2)
}

# the filter part is not working as expected. 
# There is no difference whether i pipe filter or leave it out
mydiff("a")

#> # A tibble: 5 x 3
#>      id ax    ay   
#>   <dbl> <chr> <chr>
#> 1    10 d     d    
#> 2    12 e     e    
#> 3    14 f     k    
#> 4    16 g     g    
#> 5    18 h     h

它在你原来的功能中不起作用的原因是col_1 was string but dplyr::filter()LHS 的预期“未加引号”输入变量。因此，您需要首先转换col_1变量使用sym()然后取消里面的引号filter using !!（嘭嘭）。

rlang有非常好的功能qq_show显示引用/取消引用实际发生的情况（参见下面的输出）

另请参阅类似的question https://stackoverflow.com/q/51738267/

library(rlang)
library(dplyr)

# creating a function that can take either string or symbol as input
mydiff <- function(filteron, df_1 = df1, df_2 = df2) {

  col_1 <- paste0(quo_name(enquo(filteron)), "x")
  col_2 <- paste0(quo_name(enquo(filteron)), "y")

  my_df <- inner_join(df_1, df_2, by = "id", suffix = c("x", "y"))

  cat('\nwithout sym and unquote\n')
  qq_show(col_1 != col_2)

  cat('\nwith sym and unquote\n')
  qq_show(!!sym(col_1) != !!sym(col_2))
  cat('\n')

  my_df %>% 
    select(id, col_1, col_2) %>% 
    filter(!!sym(col_1) != !!sym(col_2))
}

### testing: filteron as a string
mydiff("a")
#> 
#> without sym and unquote
#> col_1 != col_2
#> 
#> with sym and unquote
#> ax != ay
#> 
#> # A tibble: 1 x 3
#>      id ax    ay   
#>   <dbl> <chr> <chr>
#> 1    14 f     k

### testing: filteron as a symbol
mydiff(a)
#> 
#> without sym and unquote
#> col_1 != col_2
#> 
#> with sym and unquote
#> ax != ay
#>  
#> # A tibble: 1 x 3
#>      id ax    ay   
#>   <dbl> <chr> <chr>
#> 1    14 f     k

^{Created on 2018-09-28 by the reprex package https://reprex.tidyverse.org (v0.2.1.9000)}