在 scala 中使用可变长度参数
我知道如何定义具有可变长度参数的方法：
case class taxonomy(vocabularies:(String,Set[String])*)
并且客户端代码非常干净：
val terms=taxonomy("topics"->Set("economic","politic")
,"tag"->Set("Libya","evolution")
)
但我想知道当我有一个像这样的变量（而不是变量序列）时如何使用这个案例类：
val notFormattedTerms = Map("topics"->Set("economic","politic")
,"tag"->Set("Libya","evolution"))
I know how to define a method with variable length argument:
case class taxonomy(vocabularies:(String,Set[String])*)
and client code is very clean:
val terms=taxonomy("topics"->Set("economic","politic")
,"tag"->Set("Libya","evolution")
)
but I want to Know how can I use this case class when I have a variable (instead of a Sequence of variable) like this:
val notFormattedTerms = Map("topics"->Set("economic","politic")
,"tag"->Set("Libya","evolution"))
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使用
: _*您实际上可以转换序列参数，使其看起来好像多个参数已传递给可变长度方法。然而，这种转换仅适用于（有序？）简单序列类型，并且在本例中不适用于Map。因此，之前必须使用显式的toSeq。With
: _*you virtually transform a sequence argument so that it looks as if a several arguments had been passed to the variable length method. This transformation, however, only works for (ordered?) simple sequence types and, as in this case, not for aMap. Therefore, one will have to use an explicittoSeqbefore.