根据条件干净地生成和替换值

Question

我正在尝试在 R 中做一些非常简单的事情，但我做错了。

让我们从 ggplot 2 中获取“钻石”数据集

glimpse(diamonds)

$ carat   <dbl> 0.23, 0.21, 0.23, 0.29, 0.31, 0.24, 0.24, 0.26, 0.22, 0.23, 0.30, 0.23, 0.22, 0.31, 0.20, 0.32, 0.30, 0.30, 0.30, 0.30, 0.30, 0.23, 0.23, 0.31, 0.31, 0.23, ...
$ cut     <ord> Ideal, Premium, Good, Premium, Good, Very Good, Very Good, Very Good, Fair, Very Good, Good, Ideal, Premium, Ideal, Premium, Premium, Ideal, Good, Good, Ver...
$ color   <ord> E, E, E, I, J, J, I, H, E, H, J, J, F, J, E, E, I, J, J, J, I, E, H, J, J, G, I, J, D, F, F, F, E, E, D, F, E, H, D, I, I, J, D, D, H, F, H, H, E, H, F, G, ...
$ clarity <ord> SI2, SI1, VS1, VS2, SI2, VVS2, VVS1, SI1, VS2, VS1, SI1, VS1, SI1, SI2, SI2, I1, SI2, SI1, SI1, SI1, SI2, VS2, VS1, SI1, SI1, VVS2, VS1, VS2, VS2, VS1, VS1,...
$ depth   <dbl> 61.5, 59.8, 56.9, 62.4, 63.3, 62.8, 62.3, 61.9, 65.1, 59.4, 64.0, 62.8, 60.4, 62.2, 60.2, 60.9, 62.0, 63.4, 63.8, 62.7, 63.3, 63.8, 61.0, 59.4, 58.1, 60.4, ...
$ table   <dbl> 55, 61, 65, 58, 58, 57, 57, 55, 61, 61, 55, 56, 61, 54, 62, 58, 54, 54, 56, 59, 56, 55, 57, 62, 62, 58, 57, 57, 61, 57, 57, 57, 59, 58, 58, 59, 59, 54, 59, ...
$ price   <int> 326, 326, 327, 334, 335, 336, 336, 337, 337, 338, 339, 340, 342, 344, 345, 345, 348, 351, 351, 351, 351, 352, 353, 353, 353, 354, 355, 357, 357, 357, 402, 4...
$ x       <dbl> 3.95, 3.89, 4.05, 4.20, 4.34, 3.94, 3.95, 4.07, 3.87, 4.00, 4.25, 3.93, 3.88, 4.35, 3.79, 4.38, 4.31, 4.23, 4.23, 4.21, 4.26, 3.85, 3.94, 4.39, 4.44, 3.97, ...
$ y       <dbl> 3.98, 3.84, 4.07, 4.23, 4.35, 3.96, 3.98, 4.11, 3.78, 4.05, 4.28, 3.90, 3.84, 4.37, 3.75, 4.42, 4.34, 4.29, 4.26, 4.27, 4.30, 3.92, 3.96, 4.43, 4.47, 4.01, ...
$ z       <dbl> 2.43, 2.31, 2.31, 2.63, 2.75, 2.48, 2.47, 2.53, 2.49, 2.39, 2.73, 2.46, 2.33, 2.71, 2.27, 2.68, 2.68, 2.70, 2.71, 2.66, 2.71, 2.48, 2.41, 2.62, 2.59, 2.41, ...

假设我们要计算一个新价格，其中包括“公平”钻石的 10% 折扣。我想在 R 中实现的是在 Stata 中：

generate price_cut = .
replace price_cut = price if cut != "Fair"
replace price_cut = (0.90 * price) if cut =="Fair"

但我做不到。我试过了

    diamonds["price_cut"] <- 0
    diamonds[diamonds$cut == "Ideal", "price_cut"] <- diamonds$price
    Error in `[<-.data.frame`(`*tmp*`, diamonds$cut == "Ideal", "price_cut",  : 
      replacement has 53940 rows, data has 21551

我也试过了

diamonds["price_cut"] <- 0
diamonds[diamonds$cut == "Ideal", "price_cut"] <- diamonds$price
Error in `[<-.data.frame`(`*tmp*`, diamonds$cut == "Ideal", "price_cut",  : 
  replacement has 53940 rows, data has 21551
diamonds$price_cut[diamonds$cut !="Ideal"] <- diamonds$price * 0.9
Warning message:
In diamonds$price_cut[diamonds$cut != "Ideal"] <- diamonds$price :
  number of items to replace is not a multiple of replacement length

它在我的玩具示例中有些工作，但在缺少值等的更复杂的数据集中不起作用。

我做错了什么？

Answer 1

从您的 Stata 代码直接翻译为

diamonds$price_cut <- NA
diamonds$price_cut[diamonds$cut != "Fair"] <- diamonds$price[diamonds$cut != "Fair"]
diamonds$price_cut[diamonds$cut == "Fair"] <- (0.90 * diamonds$price[diamonds$cut == "Fair"])

这可以使用矢量化参数在一行中实现，例如

diamonds$price_cut <- c(1, .9)[(diamonds$cut == "Fair") + 1] * diamonds$price

或者，更常用的是ifelse：

diamonds$price_cut <- ifelse(diamonds$cut == "Fair", diamonds$price, 0.9 * diamonds$price)

---

旁注：具有相同精神的 Stata 单线

generate price_cut = price - ((cut == "Fair") * 0.1)

在R中

diamonds$price_cut <- diamonds$price - ((diamonds$cut == "Fair") * 0.1)

Answer 2

 stata.gen<-function(data, v,value=1) {
  namesd=names(data)
  if (is.na(match(v,namesd)) ) {
  x=parse(text=value)
   data[v]<-as.numeric(eval(x,data,parent.frame()))
   } else {
  print(paste0(v, " already exists"))
}  
  return(data)
}

example of calling this function:
d=stata.gen(roster,"under20","age<20")
d=stata.gen(roster,"under20",0)