Question

我有一个数据框：

data <- data.frame(
    name = c("John", "Paul", "George", "Ringo", "Stuart", "Pete"),
    instrument1 = c("guitar", "bass", "guitar", "drums", "bass","drums"),
    instrument2 = c("guitar", "bass", "guitar", "drums", "bass","drums"),stringsAsFactors=FALSE)

然后我创建一个像这样的函数：

sel <- function(df, variable) {
    var <- enquo(variable)
    df2 <- select(df, starts_with(!!var))
    return(df2)
}

如何使用以下命令运行此功能：

sel(data, instrument)

无需引用参数“仪器”。我需要不加报价，因为这是更大功能的一部分。如果我像上面那样烂了，我得到：

Error in is_string(match) : object 'instrument' not found

Answer 1

您正在寻找deparse(substitute(x))的魔力！它将返回您以字符串形式传递的参数名称。

data <- data.frame(
  name = c("John", "Paul", "George", "Ringo", "Stuart", "Pete"),
  instrument1 = c("guitar", "bass", "guitar", "drums", "bass","drums"),
  instrument2 = c("guitar", "bass", "guitar", "drums", "bass","drums"),stringsAsFactors=FALSE)

sel <- function(df, variable) {
  var <- deparse(substitute(variable))
  df2 <- select(df, starts_with(!!var))
  return(df2)
}

sel(data, instrument)

哪个返回：

  instrument1 instrument2
1      guitar      guitar
2        bass        bass
3      guitar      guitar
4       drums       drums
5        bass        bass
6       drums       drums

Answer 2

我们可以使用rlang::enexpr将variable转换为名称类，然后使用as.character将其更改为字符，并将其作为参数传递给starts_with。

sel <- function(df, variable) {
  var <- as.character(rlang::enexpr(variable))
  df2 <- dplyr::select(df, starts_with(var))
  return(df2)
}

sel(data, instrument)

#  instrument1 instrument2
#1      guitar      guitar
#2        bass        bass
#3      guitar      guitar
#4       drums       drums
#5        bass        bass
#6       drums       drums

将不带引号的参数作为字符传递给选择函数

2 个答案: