Java 如何使用流对 Map 中的值求和？
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How to sum values in a Map with a stream?
提问by Jay
I want the equivalent of this with a stream:
我想要一个流的等价物：
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;
}
This code doesn't actually compile because 0 cannot be assigned to type T, but you get the idea.
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。
采纳答案by Radiodef
Here's another way to do this:
这是执行此操作的另一种方法：
int sum = data.values().stream().reduce(0, Integer::sum);
(For a sum to just int, however, Paul's answer does less boxing and unboxing.)
（int然而，对于一个总和，保罗的回答没有装箱和拆箱。）
As for doing this generically, I don't think there's a way that's much more convenient.
至于一般地这样做，我认为没有更方便的方法。
We could do something like this:
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);
But you always end up passing the summer. reduceis also problematic because it returns Optional. The above summethod throws an exception for an empty map, but an empty sum should be 0. Of course, we could pass the 0 too:
但你总是以夏天过去告终。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，但空和应该是0。当然，我们也可以传递0：
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);
回答by Paul Boddington
You can do this:
你可以这样做：
int sum = data.values().stream().mapToInt(Integer::parseInt).sum();
回答by Saeed Zarinfam
You can do it like this:
你可以这样做：
int creditAmountSum = result.stream().map(e -> e.getCreditAmount()).reduce(0, (x, y) -> x + y);
