是否有一种简洁的方法在流上迭代,同时访问流中的索引?
String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList;
Stream<Integer> indices = intRange(1, names.length).boxed();
nameList = zip(indices, stream(names), SimpleEntry::new)
.filter(e -> e.getValue().length() <= e.getKey())
.map(Entry::getValue)
.collect(toList());
与这里给出的LINQ示例相比,这似乎相当令人失望
string[] names = { "Sam", "Pamela", "Dave", "Pascal", "Erik" };
var nameList = names.Where((c, index) => c.Length <= index + 1).ToList();
有更简洁的方式吗?
此外,似乎拉链已经移动或被拆除…
String[] namesArray = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
String completeString
= IntStream.range(0,namesArray.length)
.mapToObj(i -> namesArray[i]) // Converting each array element into Object
.map(String::valueOf) // Converting object to String again
.collect(Collectors.joining(",")); // getting a Concat String of all values
System.out.println(completeString);
山姆,帕梅拉,戴夫,帕斯卡,埃里克
String[] namesArray = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
IntStream.range(0,namesArray.length)
.mapToObj(i -> namesArray[i]) // Converting each array element into Object
.map(String::valueOf) // Converting object to String again
.forEach(s -> {
//You can do various operation on each element here
System.out.println(s);
}); // getting a Concat String of all
收集清单:
String[] namesArray = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> namesList
= IntStream.range(0,namesArray.length)
.mapToObj(i -> namesArray[i]) // Converting each array element into Object
.map(String::valueOf) // Converting object to String again
.collect(Collectors.toList()); // collecting elements in List
System.out.println(listWithIndex);