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如果听起来没有很多类似的问题，很难定义这个问题！

我有一个函数，我希望其中一个参数是函数名称，它将传递给 dplyr::summarise，例如“mean”或“sum”：

data(mtcars)
  f <- function(x = mtcars,
                groupcol = "cyl",
                zCol = "disp",
                zFun = "mean") {
    
    zColquo = quo_name(zCol)
    
    cellSummaries <- x %>%
      group_by(gear, !!sym(groupcol)) %>% # 1 preset grouper, 1 user-defined
      summarise(Count = n(), # 1 preset summary, 1 user defined
                !!zColquo := mean(!!sym(zColquo))) # mean should be zFun, user-defined
    ungroup
  }

（这按 gear 和 cyl 分组，然后返回，每组，count 和 mean(disp)）

根据我的笔记，我希望“意思”是动态的，执行 zFun 定义的功能，但我一生都无法弄清楚如何去做！提前感谢您的任何建议。

4

2 回答 2

1

您可以使用match.fun使函数动态化。我也删除zColquo了，因为它不需要。

library(dplyr)
library(rlang)

f <- function(x = mtcars,
              groupcol = "cyl",
              zCol = "disp",
              zFun = "mean") {

  cellSummaries <- x %>%
                   group_by(gear, !!sym(groupcol)) %>% 
                   summarise(Count = n(), 
                             !!zCol := match.fun(zFun)(!!sym(zCol))) %>%
                   ungroup

  return(cellSummaries)
}

然后您可以检查输出

f()

# A tibble: 8 x 4
#   gear   cyl Count  disp
#  <dbl> <dbl> <int> <dbl>
#1     3     4     1  120.
#2     3     6     2  242.
#3     3     8    12  358.
#4     4     4     8  103.
#5     4     6     4  164.
#6     5     4     2  108.
#7     5     6     1  145 
#8     5     8     2  326 

f(zFun = "sum")

# A tibble: 8 x 4
#   gear   cyl Count  disp
#  <dbl> <dbl> <int> <dbl>
#1     3     4     1  120.
#2     3     6     2  483 
#3     3     8    12 4291.
#4     4     4     8  821 
#5     4     6     4  655.
#6     5     4     2  215.
#7     5     6     1  145 
#8     5     8     2  652
于 2020-07-15T01:33:51.417 回答
1

我们可以用get

library(dplyr)    
f <- function(x = mtcars,
            groupcol = "cyl",
            zCol = "disp",
            zFun = "mean") {


  zColquo = quo_name(zCol)
  x %>%
  group_by(gear, !!sym(groupcol)) %>% # 1 preset grouper, 1 user-defined
  summarise(Count = n(), # 1 preset summary, 1 user defined
            !!zColquo := get(zFun)(!!sym(zCol))) %>% 
ungroup
 }

f()
# A tibble: 8 x 4
#   gear   cyl Count  disp
#  <dbl> <dbl> <int> <dbl>
#1     3     4     1  120.
#2     3     6     2  242.
#3     3     8    12  358.
#4     4     4     8  103.
#5     4     6     4  164.
#6     5     4     2  108.
#7     5     6     1  145 
#8     5     8     2  326 


f(zFun = "sum")
# A tibble: 8 x 4
#   gear   cyl Count  disp
#  <dbl> <dbl> <int> <dbl>
#1     3     4     1  120.
#2     3     6     2  483 
#3     3     8    12 4291.
#4     4     4     8  821 
#5     4     6     4  655.
#6     5     4     2  215.
#7     5     6     1  145 
#8     5     8     2  652

此外，sym如果我们group_by用summariseacross

f <- function(x = mtcars,
            groupcol = "cyl",
            zCol = "disp",
            zFun = "mean") {



 x %>%
    group_by(across(c(gear, groupcol))) %>% # 1 preset grouper, 1 user-defined
    summarise(Count = n(), # 1 preset summary, 1 user defined
            across(zCol, ~ get(zFun)(.))) %>% 
    ungroup
 }
f()
# A tibble: 8 x 4
#   gear   cyl Count  disp
#  <dbl> <dbl> <int> <dbl>
#1     3     4     1  120.
#2     3     6     2  242.
#3     3     8    12  358.
#4     4     4     8  103.
#5     4     6     4  164.
#6     5     4     2  108.
#7     5     6     1  145 
#8     5     8     2  326
于 2020-07-15T01:43:03.520 回答