将变量名称向量传递给 dplyr 中的排列() [英] Pass a vector of variable names to arrange() in dplyr

问题描述

我想传递 arrange() {dplyr} 一个变量名称向量来排序.通常我只是输入我想要的变量，但我正在尝试创建一个函数，其中排序变量可以作为函数参数输入.

I want to pass arrange() {dplyr} a vector of variable names to sort on. Usually I just type in the variables I want, but I'm trying to make a function where the sorting variables can be input as a function parameter.

df <- structure(list(var1 = c(1L, 2L, 2L, 3L, 1L, 1L, 3L, 2L, 4L, 4L
  ), var2 = structure(c(10L, 1L, 8L, 3L, 5L, 4L, 7L, 9L, 2L, 6L
  ), .Label = c("b", "c", "f", "h", "i", "o", "s", "t", "w", "x"
  ), class = "factor"), var3 = c(7L, 5L, 5L, 8L, 5L, 8L, 6L, 7L, 
  5L, 8L), var4 = structure(c(8L, 5L, 1L, 4L, 7L, 4L, 3L, 6L, 9L, 
  2L), .Label = c("b", "c", "d", "e", "f", "h", "i", "w", "y"), 
  class = "factor")), .Names = c("var1", "var2", "var3", "var4"), 
  row.names = c(NA, -10L), class = "data.frame")

# this is the normal way to arrange df with dplyr
df %>% arrange(var3, var4)

# but none of these (below) work for passing a vector of variables
vector_of_vars <- c("var3", "var4")
df %>% arrange(vector_of_vars)
df %>% arrange(get(vector_of_vars))
df %>% arrange(eval(parse(text = paste(vector_of_vars, collapse = ", "))))

推荐答案

Hadley 没有在帮助文件中明确说明这一点——仅在他的 NSE 小插图中.下划线后的函数版本使用标准评估，因此您可以将字符串向量等传递给它们.

Hadley hasn't made this obvious in the help file--only in his NSE vignette. The versions of the functions followed by underscores use standard evaluation, so you pass them vectors of strings and the like.

如果我正确理解您的问题，您只需将 arrange() 替换为 arrange_() 即可.

If I understand your problem correctly, you can just replace arrange() with arrange_() and it will work.

具体来说，在执行时将字符串向量作为 .dots 参数传递.

Specifically, pass the vector of strings as the .dots argument when you do it.

> df %>% arrange_(.dots=c("var1","var3"))
   var1 var2 var3 var4
1     1    i    5    i
2     1    x    7    w
3     1    h    8    e
4     2    b    5    f
5     2    t    5    b
6     2    w    7    h
7     3    s    6    d
8     3    f    8    e
9     4    c    5    y
10    4    o    8    c

========== 2018 年 3 月更新 ==============

========== Update March 2018 ==============

在 dplyr 中使用标准评估版本，正如我在此处展示的那样现在被认为已弃用.您可以阅读Hadley 的编程小插图以了解新方法.基本上，您将使用 !! 取消引用一个变量或使用 !!! 取消引用 arrange() 内的变量向量.

Using the standard evaluation versions in dplyr as I have shown here is now considered deprecated. You can read Hadley's programming vignette for the new way. Basically you will use !! to unquote one variable or !!! to unquote a vector of variables inside of arrange().

当您传递这些列时，如果它们是空的，请使用 quo() 引用它们作为一个变量或使用 quos() 引用它们作为向量.不要使用引号.请参阅 Akrun 的回答.

When you pass those columns, if they are bare, quote them using quo() for one variable or quos() for a vector. Don't use quotation marks. See the answer by Akrun.

如果您的列已经是字符串，则使用 rlang::sym() 为单个列或rlang::syms() 为向量命名.请参阅克里斯托斯的答案.您还可以将 as.name() 用于单个列.不幸的是，在撰写本文时，有关如何使用 rlang::sym() 的信息尚未包含在我上面链接的小插图中(最终它将在可变参数准引用"部分根据他的草稿).

If your columns are already strings, then make them names using rlang::sym() for a single column or rlang::syms() for a vector. See the answer by Christos. You can also use as.name() for a single column. Unfortunately as of this writing, the information on how to use rlang::sym() has not yet made it into the vignette I link to above (eventually it will be in the section on "variadic quasiquotation" according to his draft).

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