将变量名称作为参数从外部函数传递到 R 中的内部函数时出现问题？

Question

我正在尝试使流程自动化并创建一个外部函数来运行几个较小的内部函数，但 function 以 variable names 作为参数会导致错误：

当我单独运行以下函数时，它可以正常工作：

gapminder <- read.csv("https://raw.githubusercontent.com/swcarpentry/r-novice-gapminder/gh-pages/_episodes_rmd/data/gapminder-FiveYearData.csv")

################ fn_benchmark_country ################

fn_benchmark_country  <- function(bench_country = India){
  
  bench_country = enquo(bench_country)

  gapminder_benchmarked_wider <- gapminder %>% 
                                  select(country, year, gdpPercap) %>% 
                                  pivot_wider(names_from = country, values_from = gdpPercap) %>% 
                                  arrange(year) %>% 
                                  
                                  mutate(across(-1, ~ . - !!bench_country))
                                  
  # Reshaping back to Longer
  gapminder_benchmarked_longer <- gapminder_benchmarked_wider %>% 
                                  pivot_longer(cols = !year, names_to = "country", values_to = "benchmarked") 
 
  # Joining tables
  gapminder_joined <- left_join(x = gapminder, y = gapminder_benchmarked_longer, by = c("year","country"))

  # converting to factor
  gapminder_joined$country <- as.factor(gapminder_joined$country)
  
  return(gapminder_joined)  
}

################ ----------------------------- ################

# Calling function
fn_benchmark_country(Vietnam) 

country     year     pop  continent  lifeExp    gdpPercap   benchmarked

Afghanistan 1952    8425333 Asia    28.80100    779.4453    232.879565
Afghanistan 1957    9240934 Asia    30.33200    820.8530    230.791034

但是当我在外部函数中运行它时，它会给我错误：

fn_run_all <-function(bench_country = India, year_start = 1952, year_end = 2007){
  
  bench_country = bench_country
  year_start = year_start
  year_end = year_end
  
  fn_benchmark_country(bench_country)
}

fn_run_all()

Error in fn_run_all() : object 'India' not found

如果我将enquo 添加到参数中，我仍然会收到如下所示的错误“

fn_run_all <-function(bench_country = India, year_start = 1952, year_end = 2007){
  
  bench_country = enquo(bench_country)
  year_start = year_start
  year_end = year_end
  
  fn_benchmark_country(bench_country)
}

fn_run_all()

Error: Problem with `mutate()` input `..1`. x Base operators are not defined for quosures. Do you need to unquote the quosure? # Bad: lhs - myquosure # Good: lhs - !!myquosure i Input `..1` is `across(-1, ~. - bench_country)`.

不知道如何解决这个问题，将不胜感激！

从这里添加新的相关问题

由于最后一次内部函数调用fn_create_plot()而出现错误

由于在情节的creating dynamic subtitle 中使用bench_country 而发生相同类型的问题，但这次我使用了{{}} 但仍然遇到问题

fn_run_all <-function(bench_country = India, year_start = 1952, year_end = 2007){
  

  year_start = year_start
  year_end = year_end
  
  fn_benchmark_country({{bench_country}})
  
  fn_year_filter(gapminder_joined, year_start, year_end) %>% 
  
  fn_create_plot(., year_start, year_end, {{bench_country}})
}

fn_run_all(Vietnam, 1967, 2002)

Error in sprintf("Benchmarked %i in blue line \nFor Countries with pop > 30000000 \n(Chart created by ViSa)", : object 'Vietnam' not found

功能代码供参考

################ fn_create_plot ################
  
fn_create_plot <- function(df, year_start, year_end, bench_country ){

                  
                  
                  # plotting
                      ggplot(df) +
                        
                      geom_vline(xintercept = 0, col = "blue", alpha = 0.5) +
    
                      geom_label( label="India - As Benchamrking line", x=0, y="United States",
                        label.padding = unit(0.55, "lines"), # Rectangle size around label
                        label.size = 0.35, color = "black") +
                      
                      geom_segment(aes(x = benchmarked_start, xend = benchmarked_end,
                                       y = country, yend = country,
                                       col = continent), alpha = 0.5, size = 7) +
                      
                      geom_point(aes(x = benchmarked, y = country, col = continent), size = 9, alpha = .6) +
                      
                      geom_text(aes(x = benchmarked_start + 8, y = country,
                                    label = paste(country, round(benchmarked_start))),
                                col = "grey50", hjust = "right") +
                      
                      geom_text(aes(x = benchmarked_end - 4.0, y = country,
                                    label = round(benchmarked_end)),
                                col = "grey50", hjust = "left") +
                    
                      
                      # scale_x_continuous(limits = c(20,85)) +
                      
                      scale_color_brewer(palette = "Pastel2") +
                      
                      labs(title = sprintf("GdpPerCapita Differenece with India (Starting point at %i and Ending at %i)",year_start, year_end),
                           subtitle = sprintf("Benchmarked %i in blue line \nFor Countries with pop > 30000000 \n(Chart created by ViSa)", bench_country),
                           col = "Continent", x = sprintf("GdpPerCap Difference at %i & %i", year_start, year_end) ) +
                      
                  # background & theme settings
                      theme_classic() +
                      
                      theme(legend.position = "top", 
                            axis.line = element_blank(),
                            axis.ticks = element_blank(),
                            axis.text = element_blank()
                            )
}

################ ----------------------------- ################

Answer 1

当您使用enquo() 时，您还需要在函数中调用相关变量时使用!!。这有效：

fn_run_all <-function(bench_country = India, year_start = 1952, year_end = 2007){
  
  bench_country = enquo(bench_country)
  year_start = year_start
  year_end = year_end
  
  fn_benchmark_country(!! bench_country)
}

fn_run_all()

您也可以将enquo() 和!! 替换为隧道{{ }}：

fn_run_all <-function(bench_country = India, year_start = 1952, year_end = 2007){
  
  year_start = year_start
  year_end = year_end
  
  fn_benchmark_country({{ bench_country }})
}

fn_run_all()