如何在r中的函数定义中正确使用dplyr动词？ [英] How to correctly use dplyr verbs inside a function definition in r?

问题描述

我想使用过滤器和总结从 dplyr 我的功能内没有功能，它的工作方式如下：

I want to use filter and summarise from dplyr inside my function. Without a function it works like following:

library(dplyr)
> Orange %>% 
+     filter(Tree==1) %>% 
+     summarise(age_max = max(age))
  age_max
1    1582  

我想在函数中做同样的事情，但是以下失败：

I want to do the same inside a function, but following fails:

## Function definition:

df.maker <- function(df, plant, Age){

  require(dplyr)

  dfo <- df %>% 
    filter(plant==1) %>% 
    summarise(age_max = max(Age))

  return(dfo)
}

## Use:
> df.maker(Orange, Tree, age)

 Rerun with Debug
 Error in as.lazy_dots(list(...)) : object 'Tree' not found

我知道以前有类似的问题。我还经历了一些相关链接，例如 page1 和第2页。但是我不能完全把握NSE和SE的概念。我尝试以下操作：

I know that similar questions have been asked before. I've also gone through some relevant links such as page1 and page2. But I can't fully grasp the concepts of NSE and SE. I tried following:

df.maker <- function(df, plant, Age){

  require(dplyr)

  dfo <- df %>% 
    filter_(plant==1) %>% 
    summarise_(age_max = ~max(Age))

  return(dfo)
} 

但是得到同样的错误。请帮我理解发生了什么我如何正确地创建我的功能？谢谢！

But get the same error. Please help me understand what's going on. And how can I correctly create my function? Thanks!

编辑：

我也尝试过：

I also tried following:

df.maker <- function(df, plant, Age){

  require(dplyr)

  dfo <- df %>% 
    #filter_(plant==1) %>% 
    summarise_(age_max = lazyeval::interp(~max(x),
                                          x = as.name(Age)))

  return(dfo)
}  

> df.maker(Orange, Tree, age)
 Error in as.name(Age) : object 'age' not found 

推荐答案

提供字符参数并使用 as.name ：

df.maker1 <- function(d, plant, Age){
  require(dplyr)
  dfo <- d %>% 
    filter_(lazyeval::interp(~x == 1, x = as.name(plant))) %>% 
    summarise_(age_max = lazyeval::interp(~max(x), x = as.name(Age)))
  return(dfo)
}  
df.maker1(Orange, 'Tree', 'age')

  age_max
1    1582

或者用替换 ：

df.maker2 <- function(d, plant, Age){
  require(dplyr)
  plant <- substitute(plant)
  Age <- substitute(Age)

  dfo <- d %>% 
    filter_(lazyeval::interp(~x == 1, x = plant)) %>% 
    summarise_(age_max = lazyeval::interp(~max(x), x = Age))
  return(dfo)
}  
df.maker2(Orange, Tree, age)

  age_max
1    1582

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