如何通过迭代器分组而不将其转换为Scala中的列表? [英] How to groupBy an iterator without converting it to list in scala?

问题描述

假设我要在迭代器上使用groupBy，编译器将询问"value groupBy is not a member of Iterator[Int]".一种方法是将迭代器转换为我要避免的列表.我想做groupBy，使输入为Iterator[A]，输出为Map[B, Iterator[A]].这样，仅当访问元素的该部分时才加载迭代器的该部分，而不将整个列表加载到内存中.我也知道可能的一组键，所以我可以说是否存在特定的键.

Suppose I want to groupBy on a iterator, compiler asks to "value groupBy is not a member of Iterator[Int]". One way would be to convert iterator to list which I want to avoid. I want to do the groupBy such that the input is Iterator[A] and output is Map[B, Iterator[A]]. Such that the part of the iterator is loaded only when that part of element is accessed and not loading the whole list into memory. I also know the possible set of keys, so I can say whether a particular key exists.

def groupBy(iter: Iterator[A], f: fun(A)->B): Map[B, Iterator[A]] = {
    .........
}

推荐答案

一种可能性是，您可以将Iterator转换为view，然后将groupBy as转换为

One possibility is, you can convert Iterator to view and then groupBy as,

iter.toTraversable.view.groupBy(_.whatever)

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