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…
下面是标准Java的解决方案:
在线解决方案:
Arrays.stream("zero,one,two,three,four".split(","))
.map(new Function<String, Map.Entry<Integer, String>>() {
int index;
@Override
public Map.Entry<Integer, String> apply(String s) {
return Map.entry(index++, s);
}
})
.forEach(System.out::println);
更可读的解决方案与实用方法:
static <T> Function<T, Map.Entry<Integer, T>> mapWithIntIndex() {
return new Function<T, Map.Entry<Integer, T>>() {
int index;
@Override
public Map.Entry<Integer, T> apply(T t) {
return Map.entry(index++, t);
}
};
}
...
Arrays.stream("zero,one,two,three,four".split(","))
.map(mapWithIntIndex())
.forEach(System.out::println);
你可以使用IntStream.iterate()来获取索引:
String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i < names.length, i -> i + 1)
.filter(i -> names[i].length() <= i)
.mapToObj(i -> names[i])
.collect(Collectors.toList());
这只适用于Java 9以上的Java 8，你可以使用这个:
String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i + 1)
.limit(names.length)
.filter(i -> names[i].length() <= i)
.mapToObj(i -> names[i])
.collect(Collectors.toList());
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);
下面是标准Java的解决方案:
在线解决方案:
Arrays.stream("zero,one,two,three,four".split(","))
.map(new Function<String, Map.Entry<Integer, String>>() {
int index;
@Override
public Map.Entry<Integer, String> apply(String s) {
return Map.entry(index++, s);
}
})
.forEach(System.out::println);
更可读的解决方案与实用方法:
static <T> Function<T, Map.Entry<Integer, T>> mapWithIntIndex() {
return new Function<T, Map.Entry<Integer, T>>() {
int index;
@Override
public Map.Entry<Integer, T> apply(T t) {
return Map.entry(index++, t);
}
};
}
...
Arrays.stream("zero,one,two,three,four".split(","))
.map(mapWithIntIndex())
.forEach(System.out::println);
