Java 8 Lambda：我可以从 IntStream 生成新的对象 ArrayList 吗？
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Java 8 Lambda: Can I generate a new ArrayList of objects from an IntStream?
提问by Brennan Hoeting
I have a Card class
我有一个卡片类
public class Card {
private int value, suit;
public Card(int value, int suit) {
this.value = value;
this.suit = suit;
}
//gets, sets, toString
}
This is how I would normally fill the ArrayList of Card
这就是我通常填充 Card ArrayList 的方式
for(int suit = 1; suit <= 4; ++suit)
for(int value = 1; value <= 13; ++value)
Cards.add(new Card(value, suit));
But I want to use a Lambda expression to initialize it
但我想使用 Lambda 表达式来初始化它
ArrayList<Card> Cards = IntStream.range(1, 4)
.map(value -> IntStream.range(1, 13)
.map(suit -> new Card(value, suit)));
Intellij is giving me an error under .map(suit -> new Card(suit, value))
Intellij 在下面给了我一个错误 .map(suit -> new Card(suit, value))
It says "Incompatible return type Card in lambda expression"
它说“不兼容的 lambda 表达式中的返回类型卡”
采纳答案by Maurice Naftalin
This is what you want:
这就是你想要的：
List<Card> cards = IntStream.rangeClosed(1, 4)
.boxed()
.flatMap(value ->
IntStream.rangeClosed(1, 13)
.mapToObj(suit -> new Card(value, suit))
)
.collect(Collectors.toList());
Points to note:
注意事项：
you have to box the ints because
flatMapon primitives doesn't have any type-conversion overloads likeflatMapToObj(doesn't exist);assign to
Listrather thanArrayListas theCollectorsmethods make no guarantee as to the specific type they return (as it happens it currently isArrayList, but you can't rely on that);use
rangeClosedfor this kind of situation.
您必须将整数装箱，因为
flatMap在原语上没有任何类型转换重载，例如flatMapToObj（不存在）；分配给
List而不是ArrayList因为这些Collectors方法不能保证它们返回的特定类型（因为它目前是这样ArrayList，但你不能依赖它）；使用
rangeClosed了这种情况。
回答by srborlongan
Another way to get what you want (based on Maurice Naftalin's answer):
另一种获得您想要的东西的方法（基于 Maurice Naftalin 的回答）：
List<Card> cards = IntStream.rangeClosed(1, 4)
.mapToObj(value -> IntStream.rangeClosed(1, 13)
.mapToObj(suit -> new Card(value, suit))
)
.flatMap(Function.identity())
.collect(Collectors.toList())
;
Additional points to note:
补充注意事项：
- you have to map the int values to Stream streams, then flatMap said streams via Function.identity(), since flatMapToObj does not exists, yet said operation is translatable to a map to Stream, then an identity flatMap.
- 您必须将 int 值映射到 Stream 流，然后通过 Function.identity() 将所述流映射到 flatMap，因为 flatMapToObj 不存在，但所述操作可转换为映射到 Stream，然后是身份 flatMap。
