问题描述
我想要一个流的等价物:
I want the equivalent of this with a stream:
public static <T extends Number> T getSum(final Map<String, T> data) {
T sum = 0;
for (String key: data.keySet())
sum += data.get(key);
return sum;
}
这段代码实际上并没有编译,因为 0 不能分配给类型 T,但你明白了.
This code doesn't actually compile because 0 cannot be assigned to type T, but you get the idea.
推荐答案
这是另一种方法:
int sum = data.values().stream().reduce(0, Integer::sum);
(不过,对于 int
的总和,Paul 的回答减少了装箱和拆箱.)
(For a sum to just int
, however, Paul's answer does less boxing and unboxing.)
至于这样做一般,我不认为有更方便的方法.
As for doing this generically, I don't think there's a way that's much more convenient.
我们可以这样做:
static <T> T sum(Map<?, T> m, BinaryOperator<T> summer) {
return m.values().stream().reduce(summer).get();
}
int sum = MyMath.sum(data, Integer::sum);
但你总是会度过夏天.reduce
也有问题,因为它返回 Optional
.上面的 sum
方法对空映射抛出异常,但是空的 sum 应该是 0.当然,我们也可以传递 0:
But you always end up passing the summer. reduce
is also problematic because it returns Optional
. The above sum
method throws an exception for an empty map, but an empty sum should be 0. Of course, we could pass the 0 too:
static <T> T sum(Map<?, T> m, T identity, BinaryOperator<T> summer) {
return m.values().stream().reduce(identity, summer);
}
int sum = MyMath.sum(data, 0, Integer::sum);
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