Question

用于验证过滤器仅返回一个结果（或该项中的特定金额）的最佳Scala惯用方法是什么？如果数量正确，则继续使用它？

例如：

String line = "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\", Delete , Hello , How are you ? , ";
String line2 = "Test asda ds asd, tesat2 . test3";
Pattern pattern = Pattern.compile("(?:\"[^\"]*(?:\"\"[^\"]*)*\"|[^,])+");
Matcher matcher = pattern.matcher(line);
if (matcher.find()){                        // if is used to get the 1st match only
    System.out.println(matcher.group(0)); 
}
Matcher matcher2 = pattern.matcher(line2); 
if (matcher2.find()){
    System.out.println(matcher2.group(0)); 
}

Answer 1

考虑这个类型T，

的列表

val myFilteredListWithDesiredOneItem = {
  val xs = unfilteredList.filter(x => x.getId.equals(something))
  if (xs.size == n) xs.toList
  else List.empty[T]
}

不是oneliner，代码仍然很简单。

Answer 2

尝试与守卫比赛，也许？

list.filter(...) match {
    case Nil => // empty
    case a if (a.size == 5) => // five items
    case b@(List(item1, item2) => // two (explicit) items
    case _ => // default
}

Answer 3

或许这样的事情：

Option(list.filter(filterFunc))
  .filter(_.size == n) 
  .getOrElse(throw new Exception("wrong size!"))

Scala过滤器仅返回一个（或特定数量）的结果

3 个答案: