Question

例如：

require(RevoScaleR)

# Create a data frame
set.seed(100)
myData = data.frame(x = 1:100, y = rep(c("a", "b", "c", "d"), 25),
                     z = rnorm(100), w = runif(100))

# Create a multi-block .xdf file from the data frame
inputFile = file.path(tempdir(), "testInput.xdf")
rxDataStep(inData = myData, outFile = inputFile, rowsPerRead = 50, 
           overwrite = TRUE)

# Square the values in the column "z"; this works fine
rxDataStep(inData = inputFile, outFile = inputFile, overwrite = TRUE,
           transforms = list(z = z^2))

# Define a squaring function and try to use it to repeat the previous step:
myFun = function(x) x^2
rxDataStep(inData = inputFile, outFile = inputFile, overwrite = TRUE,
           transforms = list(z = myFun(z)))

最后一步因错误

而崩溃

Error in transformation function: Error in eval(expr, envir, enclos) : could not find function "myFun"

rxDataStep的文档指出“与所有表达式一样，变换...可以使用expression函数在函数调用之外定义。”但我不知道如何实现这个建议，也找不到一个例子。例如，以下内容不起作用：

myFun = expression(function(x) x^2)
rxDataStep(inData = inputFile, outFile = inputFile, overwrite = TRUE,
           transforms = list(z = myFun(z)))

Answer 1

您当然可以将表达式传递给在函数调用之外创建的transform。

它看起来像这样：

myFun <- expression(
  list(x2 = x^2,
       z2 = z^2))
rxDataStep(inData = inputFile, outFile = inputFile, overwrite = TRUE,
           transforms = myFun)

如果你想在第一个例子中传递一个函数，它看起来像这样：

myFun2 <- function(dataList){
  dataList$x2 <- dataList$x^2
  dataList$z2 <- dataList$z^2
  dataList
}
rxDataStep(inData = inputFile, outFile = inputFile, overwrite = TRUE,
           transformFunc = myFun2)

Answer 2

不知道为什么会这样有效！

env <- new.env()
env$myFun <- function(x) x^2
rxDataStep(inData = inputFile, outFile = inputFile, overwrite = TRUE,
           transforms = list(z = myFun(z)), transformEnvir=env)

rxDataStep转换参数在用户定义的函数上失败

2 个答案: