使用 R 解析函数中的多个命名参数答案

【问题标题】：Parsing Multiple Naming Arguments in a Function using R使用 R 解析函数中的多个命名参数
【发布时间】：2021-01-23 15:00:11
【问题描述】：

功能

Daily <- function(DF,Roles){

DF <- select(OutputData, (Store.No:Cluster),(Roles:"Total Trading Hours"))

DF$'Roles (Monday)' <- 0
DF$'Roles (Tuesday)' <- 0
DF$'Roles (Wednesday)' <- 0
DF$'Roles (Thursday)' <- 0
DF$'Roles (Friday)' <- 0
DF$'Roles (Saturday)' <- 0
DF$'Roles (Sunday)'<- 0

DF$`Roles (Monday)` <- (DF$`Monday Trading Hours`/DF$`Total Trading 
Hours`)*DF$Roles
DF$`Roles (Tuesday)` <- (DF$`Tuesday Trading Hours`/DF$`Total Trading 
Hours`)*DF$Roles
DF$`Roles (Wednesday)` <- (DF$`Wednesday Trading Hours`/DF$`Total Trading 
Hours`)*DF$Roles
DF$`Roles (Thursday)` <- (DF$`Thursday Trading Hours`/DF$`Total Trading 
Hours`)*DF$Roles
DF$`Roles (Friday)` <- (DF$`Friday Trading Hours`/DF$`Total Trading 
Hours`)*DF$Roles
DF$`Roles (Saturday)` <- (DF$`Saturday Trading Hours`/DF$`Total Trading 
Hours`)*DF$Roles
DF$`Roles (Sunday)` <- (DF$`Sunday Trading Hours`/DF$`Total Trading 
Hours`)*DF$Roles

DF <- select(DF,(Store.No:Cluster),Roles,"Roles (Monday)":"Roles (Sunday)")

return(DF)

}

通话

RoleDailyOutput <- Daily(RoleDailyOutput, "Programmer")

错误

 Error in `$<-.data.frame`(`*tmp*`, "Roles (Monday)", value = numeric(0)) : 
 replacement has 0 rows, data has 1432

问题

我正在尝试解析“程序员”以替换函数中的“角色”。我认为问题在于我试图在引用中引用某些内容但它不喜欢它......我该如何让它工作？如果这甚至是问题。对不起，如果它很明显我对 R 很陌生

非常感谢

【问题讨论】：

Roles（一个包含字符串的变量）不能与$一起使用。使用*DF[[Roles]]。
对不起，我不明白这个...所以新行应该是：*DF[[Roles]]' (Monday)' <- 0
不，您对DF$Roles 的使用无法正常工作，请替换它。看我的回答。

标签： r function datatable

【解决方案1】：

Roles 是一个包含 字符串 的变量。它似乎适用于select（实际上我很惊讶），但它不适用于$。为此，请使用[[。

colname <- "disp"
mtcars$colname
# NULL
mtcars[[colname]]
#  [1] 160.0 160.0 108.0 258.0 360.0 225.0 360.0 146.7 140.8 167.6 167.6 275.8 275.8 275.8 472.0 460.0 440.0  78.7  75.7
# [20]  71.1 120.1 318.0 304.0 350.0 400.0  79.0 120.3  95.1 351.0 145.0 301.0 121.0

您的代码，将所有*DF$Roles 更改为*DF[[Roles]]：

Daily <- function(DF,Roles){

DF <- select(OutputData, (Store.No:Cluster),(Roles:"Total Trading Hours"))

DF$'Roles (Monday)' <- 0
DF$'Roles (Tuesday)' <- 0
DF$'Roles (Wednesday)' <- 0
DF$'Roles (Thursday)' <- 0
DF$'Roles (Friday)' <- 0
DF$'Roles (Saturday)' <- 0
DF$'Roles (Sunday)'<- 0

DF$`Roles (Monday)` <- (DF$`Monday Trading Hours`/DF$`Total Trading 
Hours`)*DF[[Roles]]
DF$`Roles (Tuesday)` <- (DF$`Tuesday Trading Hours`/DF$`Total Trading 
Hours`)*DF[[Roles]]
DF$`Roles (Wednesday)` <- (DF$`Wednesday Trading Hours`/DF$`Total Trading 
Hours`)*DF[[Roles]]
DF$`Roles (Thursday)` <- (DF$`Thursday Trading Hours`/DF$`Total Trading 
Hours`)*DF[[Roles]]
DF$`Roles (Friday)` <- (DF$`Friday Trading Hours`/DF$`Total Trading 
Hours`)*DF[[Roles]]
DF$`Roles (Saturday)` <- (DF$`Saturday Trading Hours`/DF$`Total Trading 
Hours`)*DF[[Roles]]
DF$`Roles (Sunday)` <- (DF$`Sunday Trading Hours`/DF$`Total Trading 
Hours`)*DF[[Roles]]

DF <- select(DF,(Store.No:Cluster),Roles,"Roles (Monday)":"Roles (Sunday)")

return(DF)

}

我注意到您的函数存在另一个问题：参数是 DF，但您立即从 OutputData 覆盖它（未定义，不好的做法）。试试这个：

Daily <- function(DF, Roles) {

  # DF <- select(OutputData, (Store.No:Cluster), (Roles:"Total Trading Hours"))
  for (wd in c("Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun")) {
    nm1 <- sprintf("%s (%s)", Roles, wd)
    nm2 <- sprintf("%s Trading Hours", wd)
    DF[[ nm1 ]] <- (DF[[ nm2 ]] / DF$`Total Trading Hours`) * DF[[ Roles ]]
  }

  DF <- select(DF, (Store.No:Cluster), Roles,
               sprintf("%s (Monday)", Roles):sprintf("%s (Sunday)", Roles))
  return(DF)
}

【讨论】：

行得通！你是对的，它在“选择”语句中工作有多奇怪......有一件事......有没有办法让列标题动态化？目前它显示“角色（星期一）”，但我希望它更改为“程序员（星期一）”......取决于正在解析的内容......R中是否有像DF$Roles + ' (Monday)'这样的连接字符串函数
将 DF$'Roles (Monday)' <- 0 更改为 DF[[paste(Roles, "Monday")]] <- 0。 $-vs-[[ 为 Roles 的前提会影响到所有这些（由于某种原因，select 除外）。
查看我编辑的答案，它修复了另一个问题并缩短了它。
我在这里遇到了麻烦...如果我们暂时忘记了您的新代码... select 语句不断导致错误：` 错误：无法对不存在的列进行子集化. x 列 Roles (Monday) 不存在。运行rlang::last_error() 以查看错误发生的位置。`
DF <- select(DF,(Store.No:Cluster),Roles,"Roles (Monday)":"Roles (Sunday)")