R如何使用带有filter或filter_的curl curl？

Question

我在评论者建议!! ensym的地方回答这个question，我认为这可能是使用 curl {{但我无法使其正常工作(也许不适用？)的好地方。

在不使用filter_，eval/parse或quote-unquote的情况下，如何执行此过滤器操作？会帮忙吗？

我的解决方案(1g)使用filter_和使用粘贴建立的条件。
1a有效(但是可以使用{{}}吗？)

如果我们想按多个变量过滤该怎么办？这是您看到2g在下面工作的地方(而2a不再工作了)。

library(tidyverse)
set.seed(1234)
A <- matrix(rnorm(30),nrow = 10, ncol = 3) %>% as_tibble() %>% set_names(paste("var", seq(1:3), sep = ""))
varnames_1 <- c("var2")

(expected_result_1 <- filter(A, var2 > 0))
#> # A tibble: 3 x 3
#>     var1   var2   var3
#>    <dbl>  <dbl>  <dbl>
#> 1 -2.35  0.0645  0.460
#> 2  0.429 0.959  -0.694
#> 3 -0.890 2.42   -0.936

(answer_1a <- filter(A,!!ensym(varnames_1) > 0)) # works (thanks joran and aosmith)
#> # A tibble: 3 x 3
#>     var1   var2   var3
#>    <dbl>  <dbl>  <dbl>
#> 1 -2.35  0.0645  0.460
#> 2  0.429 0.959  -0.694
#> 3 -0.890 2.42   -0.936

(answer_1b <- filter_(A, varnames_1 > 0)) # filter_ not doing what I thought it might
#> Warning: filter_() is deprecated. 
#> Please use filter() instead
#> 
#> The 'programming' vignette or the tidyeval book can help you
#> to program with filter() : https://tidyeval.tidyverse.org
#> This warning is displayed once per session.
#> # A tibble: 10 x 3
#>      var1    var2    var3
#>     <dbl>   <dbl>   <dbl>
#>  1 -1.21  -0.477   0.134 
#>  2  0.277 -0.998  -0.491 
#>  3  1.08  -0.776  -0.441 
#>  4 -2.35   0.0645  0.460 
#>  5  0.429  0.959  -0.694 
#>  6  0.506 -0.110  -1.45  
#>  7 -0.575 -0.511   0.575 
#>  8 -0.547 -0.911  -1.02  
#>  9 -0.564 -0.837  -0.0151
#> 10 -0.890  2.42   -0.936

(answer_1c <- filter(A, {{varnames_1}} > 0)) # curly curly not doing what I thought it might
#> # A tibble: 10 x 3
#>      var1    var2    var3
#>     <dbl>   <dbl>   <dbl>
#>  1 -1.21  -0.477   0.134 
#>  2  0.277 -0.998  -0.491 
#>  3  1.08  -0.776  -0.441 
#>  4 -2.35   0.0645  0.460 
#>  5  0.429  0.959  -0.694 
#>  6  0.506 -0.110  -1.45  
#>  7 -0.575 -0.511   0.575 
#>  8 -0.547 -0.911  -1.02  
#>  9 -0.564 -0.837  -0.0151
#> 10 -0.890  2.42   -0.936
(answer_1d <- filter(A, {{varnames_1 > 0}})) # curly curly not doing what I thought it might
#> `arg` must be a symbol

conditions_1 <- paste(varnames_1, "> 0")
(answer_1e <- filter(A, conditions_1)) # does not work
#> Error: Argument 2 filter condition does not evaluate to a logical vector
(answer_1f <- filter(A, {{conditions_1}})) # curly curly not doing what I thought it might
#> Error: Argument 2 filter condition does not evaluate to a logical vector
(answer_1g <- filter_(A, conditions_1)) # works
#> # A tibble: 3 x 3
#>     var1   var2   var3
#>    <dbl>  <dbl>  <dbl>
#> 1 -2.35  0.0645  0.460
#> 2  0.429 0.959  -0.694
#> 3 -0.890 2.42   -0.936

# what if we wanted to filter multiple variables?

varnames_2 <- c("var2", "var3")
(expected_result_2 <- filter(A, var2 > 0 & var3 > 0))
#> # A tibble: 1 x 3
#>    var1   var2  var3
#>   <dbl>  <dbl> <dbl>
#> 1 -2.35 0.0645 0.460

(answer_2a <- filter(A,!!ensym(varnames_2) > 0)) # does not work
#> Only strings can be converted to symbols

conditions_2 <- paste(paste(varnames_2, "> 0"), collapse = " & ")
(answer_2f <- filter(A, {{conditions_2}})) # curly curly not doing what I thought it might
#> Error: Argument 2 filter condition does not evaluate to a logical vector
(answer_2g <- filter_(A, conditions_2)) # works
#> # A tibble: 1 x 3
#>    var1   var2  var3
#>   <dbl>  <dbl> <dbl>
#> 1 -2.35 0.0645 0.460

由 reprex package(v0.3.0)创建于2019-08-28

Answer 1

正如莱昂内尔(Lionel)指出的那样，curl-curly在函数内部起作用。要将它与filter一起使用，您必须将调用包装在一个函数中。

f <- function(.df, v) { 
  filter(.df, {{ v }} > 0) 
}

# Curly-curly provides automatic NSE support
f( A, var2 )
# # A tibble: 3 x 3
#     var1   var2   var3
#    <dbl>  <dbl>  <dbl>
# 1 -2.35  0.0645  0.460
# 2  0.429 0.959  -0.694
# 3 -0.890 2.42   -0.936

# Strings have to be first converted to symbols
f( A, !!sym("var3") )
# # A tibble: 3 x 3
#     var1    var2  var3
#    <dbl>   <dbl> <dbl>
# 1 -1.21  -0.477  0.134
# 2 -2.35   0.0645 0.460
# 3 -0.575 -0.511  0.575

Curly-curly旨在引用单个参数。您可以在 purrr::reduce的帮助下通过顺序应用将其扩展为使用多个变量。 (不要忘记首先将您的字符串转换为实际的变量名!):

syms(varnames_2) %>% reduce(f, .init=A)
# # A tibble: 1 x 3
#    var1   var2  var3
#   <dbl>  <dbl> <dbl>
# 1 -2.35 0.0645 0.460