[英]Composing streams with flatmap in Java 8
Let's consider I have the following class: 我们考虑一下我有以下课程:
class A {
int i, j, k;
public A(int i, int j, int k) {
this.i = i; this.j = j; this.k = k;
}
}
where i
, j
, k
have a known range: r_i
, r_j
, r_k
. 其中i
, j
, k
具有已知范围: r_i
, r_j
, r_k
。 Now I want to to generate all possible instances of A
in this range. 现在我想在此范围内生成A
所有可能实例。 I could come up with something like: 我可以想出类似的东西:
Stream.iterate(0, n -> ++n).limit(r_i)
.flatMap(i -> Stream.iterate(0, n -> ++n).limit(r_j)
.flatMap(j -> Stream.iterate(0, n -> ++n).limit(r_k)
.map(k -> new A(i, j, k)))).collect(Collectors.toList())
First, it's too verbose. 首先,它太冗长了。 Is there a way to shorten it? 有没有办法缩短它? In particular I couldn't find a range
on Stream
. 特别是我在Stream
上找不到range
。 Second, the compiler cannot determine the type of the returned type. 其次,编译器无法确定返回类型的类型。 It considers it as List<Object>
instead of the expected List<A>
. 它将其视为List<Object>
而不是预期的List<A>
。 How can I fix that? 我该如何解决这个问题?
One way of using range
is to perform a boxing conversion right afterwards: 使用range
一种方法是在之后执行拳击转换:
List<A> list=IntStream.range(0, r_i).boxed()
.flatMap(i -> IntStream.range(0, r_j).boxed()
.flatMap(j -> IntStream.range(0, r_k)
.mapToObj(k -> new A(i, j, k)))).collect(Collectors.toList());
It's not the most beautiful code but IntStream.range(0, max).boxed()
is still better than Stream.iterate(0, n -> n+1).limit(max)
… 它不是最漂亮的代码,但IntStream.range(0, max).boxed()
仍然比Stream.iterate(0, n -> n+1).limit(max)
...
One alternative is to use a real flattening operation rather than nested operations: 一种替代方法是使用实际展平操作而不是嵌套操作:
List<A> list=IntStream.range(0, r_i).boxed()
.flatMap(i -> IntStream.range(0, r_j).mapToObj(j -> new int[]{i,j}))
.flatMap(ij -> IntStream.range(0, r_k).mapToObj(k -> new A(ij[0], ij[1], k)))
.collect(Collectors.toList());
The main drawback I see is that it suffers from the absence of an IntPair
or Tuple<int,int>
type. 我看到的主要缺点是它缺少IntPair
或Tuple<int,int>
类型。 So it uses an array as a work-around. 所以它使用数组作为解决方法。
if it's ok to have mutable variable, you could get the List of class A like this.... 如果有可变变量,你可以像这样获得A类的List ....
List<A> newCollect = new ArrayList<>();
IntStream.range(0, r_i).forEach(
i -> IntStream.range(0, r_j).forEach(
j -> IntStream.range(0, r_k).forEach(
k -> newCollect.add(new A(i, j, k))
)
)
);
or you can make List of List of List of A, then flatMap twice like this... 或者你可以制作A列表列表,然后像这样两次flatMap ...
List<A> newCollect2 = IntStream.range(0, r_i).mapToObj(
i -> IntStream.range(0, r_j).mapToObj(
j -> IntStream.range(0, r_k).mapToObj(
k -> new A(i, j, k)
).collect(Collectors.toList())
).collect(Collectors.toList())
)
.flatMap(a -> a.stream())
.flatMap(a -> a.stream())
.collect(Collectors.toList());
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.