在 `dplyr` 中使用动态变量名 [英] Use dynamic variable names in `dplyr`

问题描述

我想使用 dplyr::mutate() 在数据框中创建多个新列.列名及其内容应动态生成.

I want to use dplyr::mutate() to create multiple new columns in a data frame. The column names and their contents should be dynamically generated.

来自虹膜的示例数据:

library(dplyr)
iris <- as_tibble(iris)

我创建了一个函数来从 Petal.Width 变量改变我的新列:

I've created a function to mutate my new columns from the Petal.Width variable:

multipetal <- function(df, n) {
    varname <- paste("petal", n , sep=".")
    df <- mutate(df, varname = Petal.Width * n)  ## problem arises here
    df
}

现在我创建一个循环来构建我的列:

Now I create a loop to build my columns:

for(i in 2:5) {
    iris <- multipetal(df=iris, n=i)
}

然而，由于 mutate 认为 varname 是一个字面变量名，所以循环只创建一个新变量(称为 varname)而不是四个(称为 petal.2 - petal.5).

However, since mutate thinks varname is a literal variable name, the loop only creates one new variable (called varname) instead of four (called petal.2 - petal.5).

如何让 mutate() 使用我的动态名称作为变量名称?

How can I get mutate() to use my dynamic name as variable name?

推荐答案

由于您将变量名动态构建为字符值，因此使用允许列的字符值的标准 data.frame 索引进行赋值更有意义名称.例如:

Since you are dynamically building a variable name as a character value, it makes more sense to do assignment using standard data.frame indexing which allows for character values for column names. For example:

multipetal <- function(df, n) {
    varname <- paste("petal", n , sep=".")
    df[[varname]] <- with(df, Petal.Width * n)
    df
}

mutate 函数使通过命名参数命名新列变得非常容易.但这假设您在键入命令时知道名称.如果要动态指定列名，则还需要构建命名参数.

The mutate function makes it very easy to name new columns via named parameters. But that assumes you know the name when you type the command. If you want to dynamically specify the column name, then you need to also build the named argument.

使用最新的 dplyr 版本，您可以在使用 := 命名参数时使用 glue 包中的语法.所以这里名称中的 {} 通过评估里面的表达式来获取值.

With the latest dplyr version you can use the syntax from the glue package when naming parameters when using :=. So here the {} in the name grab the value by evaluating the expression inside.

multipetal <- function(df, n) {
  mutate(df, "petal.{n}" := Petal.Width * n)
}

如果您将列名传递给函数，则可以在字符串和列名中使用 {{}}

If you are passing a column name to your function, you can use {{}} in the string as well as for the column name

meanofcol <- function(df, col) {
  mutate(df, "Mean of {{col}}" := mean({{col}}))
}
meanofcol(iris, Petal.Width)

---

dplyr 版本 >= 0.7

dplyr 从 0.7 版开始允许您使用 := 动态分配参数名称.您可以将函数编写为:

---

dplyr version >= 0.7

dplyr starting with version 0.7 allows you to use := to dynamically assign parameter names. You can write your function as:

# --- dplyr version 0.7+---
multipetal <- function(df, n) {
    varname <- paste("petal", n , sep=".")
    mutate(df, !!varname := Petal.Width * n)
}

有关更多信息，请参阅vignette(programming", dplyr") 中可用的文档.

For more information, see the documentation available form vignette("programming", "dplyr").

dplyr 的稍早版本(>=0.3 <0.7)，鼓励使用标准评估"；许多功能的替代品.有关详细信息，请参阅非标准评估小插图 (vignette("nse")).

Slightly earlier version of dplyr (>=0.3 <0.7), encouraged the use of "standard evaluation" alternatives to many of the functions. See the Non-standard evaluation vignette for more information (vignette("nse")).

所以在这里，答案是使用 mutate_() 而不是 mutate() 并执行:

So here, the answer is to use mutate_() rather than mutate() and do:

# --- dplyr version 0.3-0.5---
multipetal <- function(df, n) {
    varname <- paste("petal", n , sep=".")
    varval <- lazyeval::interp(~Petal.Width * n, n=n)
    mutate_(df, .dots= setNames(list(varval), varname))
}

---

dplyr <0.3

请注意，这在最初提出问题时存在的旧版本 dplyr 中也是可能的.它需要小心使用quote 和setName:

---

dplyr < 0.3

Note this is also possible in older versions of dplyr that existed when the question was originally posed. It requires careful use of quote and setName:

# --- dplyr versions < 0.3 ---
multipetal <- function(df, n) {
    varname <- paste("petal", n , sep=".")
    pp <- c(quote(df), setNames(list(quote(Petal.Width * n)), varname))
    do.call("mutate", pp)
}

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