如何访问 R 函数外部的变量或函数参数？

Question

我正在创建一个绘图，为了准备数据，我创建了一个需要年份开始值和结束值作为函数参数的函数：


fn_gapminder_benchmark_diff  <- function(year_start = 1952, year_end = 2007){
  
                              year_start = year_start
                              year_end = year_end
                              
                              gapminder_joined %>% 
                              filter(year %in% c(year_start,year_end)) %>% 
                              arrange(country, year) %>%                               
                              group_by(country) %>%                               
                              mutate(benchmark_diff = benchmarked_india[2] - benchmarked_india[1],
                                     max_pop = max(pop)) %>% 
                              
                              ungroup() %>%                               
                              arrange(benchmark_diff) %>%                               
                              filter(max_pop > 30000000) %>%                               
                              mutate(country = droplevels(country)) %>% 
                              select(country, year, continent, benchmarked_india, benchmark_diff)
}

fn_gapminder_benchmark_diff(1987, 2007)

上面的函数在下面的代码中调用

问题是我无法在图表的 title 中使用上述函数 arguments or variable values 来保持图表中的 years 动态标题


# data function
  fn_gapminder_benchmark_diff(1987, 2007) %>% 

# data prep  
    mutate(country = fct_inorder(country)) %>% 
    
    group_by(country) %>% 
      
    mutate(benchmarked_end = benchmarked_india[2],
           benchmarked_start = benchmarked_india[1]  ) %>% 
    
    ungroup() %>% 
  
# plotting
    ggplot() +
      
    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_india, y = country, col = continent), size = 9, alpha = .6) +
    
    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 = "Benchmarked India in blue line \nFor Countries with pop > 30000000 \n(Chart created by ViSa)",
         col = "Continent", x = "GdpPerCap Difference at 1952 & 2007") +
    
# background & theme settings
    theme_classic() +
    
    theme(legend.position = "top", 
          axis.line = element_blank(),
          axis.ticks = element_blank(),
          axis.text = element_blank()
          )

下面是我构建的静态图表的图像，并试图使其标题动态化

Answer 1

很遗憾，您的代码不可重现。也许下次使用 reprex 包来获得更适合您需求的答案。

以 iris 数据为例，我为您提供了一个遵循相同逻辑的快速修复方法，但在我看来，它相当脏：

library(tidyverse)

data_prep <- function(data, species, min_sepal_w) {
  
  species <<- species         # use superassignment operator to assign to global environment
  min_sepal_w <<- min_sepal_w
  
  data %>% 
    filter(species == species,
           Sepal.Width >= min_sepal_w)
}

iris %>% 
  data_prep(species = "setosa", min_sepal_w = 3.0) %>% 
  ggplot(aes(Sepal.Width, Sepal.Length)) +
  geom_point() +
  labs(title = sprintf("Scatterplot of species %s with minimum sepal with of %i)", species, min_sepal_w))

^{由 reprex package (v0.3.0)}

于 2020-09-27 创建

一个更简洁的解决方案是将信息存储在数据中：

data_prep <- function(data, species, min_sepal_w) {
  
  species <<- species
  min_sepal_w <<- min_sepal_w
  
  data %>% 
    filter(species == species,
           Sepal.Width >= min_sepal_w) %>% 
    mutate(species = species,
           min_sepal_w = min_sepal_w)
}

否则上面的代码可以保持不变。

如何访问 R 函数外部的变量或函数参数？

How to access variable or argument of function outside the function in R?

r

user-defined-functions