如何解决在R中将变量名作为参数从外部函数传递到内部函数时出现问题吗？
我试图使一个过程自动化，并创建一个外部函数来运行几个较小的内部函数，但是以function作为参数的variable names会导致错误：
当我自己运行以下功能时，它将正常工作：
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，它给了我错误：
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添加到自变量中，则仍然会出现错误，如下所示”
fn_run_all <-function(bench_country = India,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)`.
不确定如何解决此问题，将不胜感激！
从此处添加新的相关问题
现在由于上一次内部函数调用fn_create_plot()
由于在情节的bench_country中使用了creating dynamic subtitle，因此发生了相同类型的问题，但是这次我使用了{{}}，但仍然出现问题
fn_run_all <-function(bench_country = India,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_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
功能代码参考
################ fn_create_plot ################
fn_create_plot <- function(df,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,size = 9,alpha = .6) +
geom_text(aes(x = benchmarked_start + 8,label = paste(country,round(benchmarked_start))),col = "grey50",hjust = "right") +
geom_text(aes(x = benchmarked_end - 4.0,label = round(benchmarked_end)),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_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_end) ) +
# background & theme settings
theme_classic() +
theme(legend.position = "top",axis.line = element_blank(),axis.ticks = element_blank(),axis.text = element_blank()
)
}
################ ----------------------------- ################
解决方法
使用enquo()时，在函数中调用有问题的变量时，也需要使用!!。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_end = 2007){
year_start = year_start
year_end = year_end
fn_benchmark_country({{ bench_country }})
}
fn_run_all()
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