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…
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
如果您需要forEach中的索引,那么这提供了一种方法。
public class IndexedValue {
private final int index;
private final Object value;
public IndexedValue(final int index, final Object value) {
this.index = index;
this.value = value;
}
public int getIndex() {
return index;
}
public Object getValue() {
return value;
}
}
然后像下面这样使用它。
@Test
public void withIndex() {
final List<String> list = Arrays.asList("a", "b");
IntStream.range(0, list.size())
.mapToObj(index -> new IndexedValue(index, list.get(index)))
.forEach(indexValue -> {
System.out.println(String.format("%d, %s",
indexValue.getIndex(),
indexValue.getValue().toString()));
});
}