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…
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());
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);
使用列表,你可以尝试一下
List<String> strings = new ArrayList<>(Arrays.asList("First", "Second", "Third", "Fourth", "Fifth")); // An example list of Strings
strings.stream() // Turn the list into a Stream
.collect(HashMap::new, (h, o) -> h.put(h.size(), o), (h, o) -> {}) // Create a map of the index to the object
.forEach((i, o) -> { // Now we can use a BiConsumer forEach!
System.out.println(String.format("%d => %s", i, o));
});
输出:
0 => First
1 => Second
2 => Third
3 => Fourth
4 => Fifth
最简洁的方法是从一系列指数开始:
String[] names = {"Sam", "Pamela", "Dave", "Pascal", "Erik"};
IntStream.range(0, names.length)
.filter(i -> names[i].length() <= i)
.mapToObj(i -> names[i])
.collect(Collectors.toList());
结果列表只包含“Erik”。
当你习惯for循环时,另一种看起来更熟悉的方法是使用可变对象维护一个临时计数器,例如AtomicInteger:
String[] names = {"Sam", "Pamela", "Dave", "Pascal", "Erik"};
AtomicInteger index = new AtomicInteger();
List<String> list = Arrays.stream(names)
.filter(n -> n.length() <= index.incrementAndGet())
.collect(Collectors.toList());
注意,在并行流上使用后一种方法可能会中断,因为项目不一定会“按顺序”处理。