将不同的函数列表传递给 dplyr 总结

Question

是否可以以允许函数列表变化的方式将函数列表传递给 dplyr::summarize？我想创建一个整体函数来创建汇总表，但在输出中允许不同的函数组不同 - [编辑：当函数并非全部应用于同一列时]。

我认为这可以通过创建一个整体函数来完成，其中包含 T/F 参数的一组汇总函数（其中 funA=T/F，funB=T/F 是函数列表，用户可以包括来自 funA、funB 或两者的所有函数），但不是如何编写初始列表函数（funA、funB）- 当函数并非全部应用于同一列时。下面是一个关于它的结构的想法。这可能吗，还是有更好的方法来做到这一点？

#Essentially - how would I write a function to selectively include a group of functions (for example either funA = c(n, min, max) or funB=c(n_na, n_neg), or both).  

extract_all <- function(x){

   x %>% summarize(n=n(), 
                   min = min(disp, na.rm=TRUE), 
                   max = max(disp, na.rm=TRUE),
                   n_na = sum(is.na(wt)),  
                   n_neg = sum(vs < 0, na.rm=TRUE))

}
test <- mtcars %>% group_by(cyl) %>% extract_all()

#Does this structure work?
extract_summaries <- function(x, funA=TRUE, funB=FALSE){
  funAls <- list()  #but how do you write n, min, max in here?
  funBls <- list()  #and n_na, n_neg in here

 funls <- append(funAls[funA], funBls[funB])

 summarize(x, funls)
}

#which could be run with:
test <- mtcars %>% group_by(cyl) %>% extract_summaries(funA=TRUE, funB=TRUE)

}

Answer 1

这是一种选择

extract_summaries <- function(x, colnm, funA=TRUE, funB=FALSE){
  funAls <- list(n = length, min= min, max = max) 
  funBls <- list(n_na = function(y) sum(is.na(y)), 
              n_neg = function(y) sum(y < 0, na.rm=TRUE)) 
 funls <- append(funAls[funA], funBls[funB])

 x %>% 
      summarise_at(vars({{colnm}}), funls)
}


test <- mtcars %>% 
           group_by(cyl) %>%
           extract_summaries(mpg, funA=TRUE, funB=TRUE)



test
# A tibble: 3 x 6
#    cyl     n   min   max  n_na n_neg
#  <dbl> <int> <dbl> <dbl> <int> <int>
#1     4    11  21.4  33.9     0     0
#2     6     7  17.8  21.4     0     0
#3     8    14  10.4  19.2     0     0

---

test <- mtcars %>% 
    group_by(cyl) %>% 
    extract_summaries(mpg, funA = FALSE, funB = TRUE)
test
# A tibble: 3 x 3
#    cyl  n_na n_neg
#  <dbl> <int> <int>
#1     4     0     0
#2     6     0     0
#3     8     0     0