使用Try，Success和Failure进行错误处理时，Scala声明变量吗？

Question

The context of my problem is the following:

I'm trying to apply a linear regression model to some data but the model will throw an Exception error if the data is not presented in the required format. If successful, then I want to save the residuals of the output as a new variable, but if failure then I want to define an array of zeroes instead as the new variable.

My problem is that I'm not being able to declare this variable depending on the success or failure of the function.

More specifically, the model is called as follows:

import org.apache.commons.math3.stat.regression.OLSMultipleLinearRegression
import org.apache.commons.math3.stat.regression.AbstractMultipleLinearRegression
import scala.util.{Try,Success,Failure}

val model=new OLSMultipleLinearRegression()
val goodData=Array(Array(.1),Array(.2))
val badData=Array(Array(.1,.1),Array(.2,.2))
val y=Array(.5,.6)

The following yields the exception error MathIllegalArgumentException

model.newSampleData(y,badData)

but this yields the normal behavior

model.newSampleData(y,goodData)

If there is an exception error then I want to declare the variable val params=Array.fill(2)(0.0) but if there is no exception error then I want to declare the variable val params=model.estimateResiduals() .

My attempt (using Try , Success and Failure ) is:

Try{model.newSampleData(y,badData)} match {
case Success(_)=>val params=model.estimateResiduals()
case Failure(_)=>val params=Array.fill(2)(0.0)
}

but when I type params I get the error error: not found: value params .

How can I declare the variable params given the success or failure of the linear regression method?

Answer 1

Variables you declare are local to the block. Use

val params = Try{model.newSampleData(y,badData)} match {
  case Success(_) => model.estimateResiduals()
  case Failure(_) => Array.fill(2)(0.0)
}

or just

val params = Try(model.newSampleData(y,badData))
  .map(_ => model.estimateResiduals())
  .getOrElse(Array.fill(2)(0.0))