在 Java 8 流 API 中按计数分组
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/25441088/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Group by counting in Java 8 stream API
提问by Muhammad Hewedy
I try to find a simple way in Java 8 stream API to do the grouping, I come out with this complex way!
我尝试在 Java 8 流 API 中找到一种简单的方法来进行分组，我提出了这种复杂的方法！
List<String> list = new ArrayList<>();
list.add("Hello");
list.add("Hello");
list.add("World");
Map<String, List<String>> collect = list.stream().collect(
Collectors.groupingBy(o -> o));
System.out.println(collect);
List<String[]> collect2 = collect
.entrySet()
.stream()
.map(e -> new String[] { e.getKey(),
String.valueOf(e.getValue().size()) })
.collect(Collectors.toList());
collect2.forEach(o -> System.out.println(o[0] + " >> " + o[1]));
I appreciate your input.
我感谢您的意见。
采纳答案by Jon Skeet
I think you're just looking for the overloadwhich takes another Collectorto specify what to do with each group... and then Collectors.counting()to do the counting:
我认为你只是在寻找需要另一个来指定每个组做什么的重载Collector......然后Collectors.counting()进行计数：
import java.util.*;
import java.util.stream.*;
class Test {
public static void main(String[] args) {
List<String> list = new ArrayList<>();
list.add("Hello");
list.add("Hello");
list.add("World");
Map<String, Long> counted = list.stream()
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
System.out.println(counted);
}
}
Result:
结果：
{Hello=2, World=1}
(There's also the possibility of using groupingByConcurrentfor more efficiency. Something to bear in mind for your real code, if it would be safe in your context.)
（还有可能groupingByConcurrent用于提高效率。如果在您的上下文中是安全的，请记住您的真实代码。）
回答by Sivakumar
List<String> list = new ArrayList<>();
list.add("Hello");
list.add("Hello");
list.add("World");
Map<String, List<String>> collect = list.stream()
.collect(Collectors.groupingBy(o -> o));
collect.entrySet()
.forEach(e -> System.out.println(e.getKey() + " - " + e.getValue().size()));
回答by user_3380739
回答by fjkjava
Here is example for list of Objects
这是对象列表的示例
Map<String, Long> requirementCountMap = requirements.stream().collect(Collectors.groupingBy(Requirement::getRequirementType, Collectors.counting()));
回答by Ousmane D.
Here are slightly different options to accomplish the task at hand.
以下是完成手头任务的略有不同的选项。
using toMap:
使用toMap：
list.stream()
.collect(Collectors.toMap(Function.identity(), e -> 1, Math::addExact));
using Map::merge:
使用Map::merge：
Map<String, Integer> accumulator = new HashMap<>();
list.forEach(s -> accumulator.merge(s, 1, Math::addExact));
回答by Donald Raab
If you're open to using a third-party library, you can use the Collectors2class in Eclipse Collectionsto convert the Listto a Bagusing a Stream. A Bagis a data structure that is built for counting.
如果你打开使用第三方库，你可以使用Collectors2类的Eclipse收藏到转换List至Bag使用Stream。ABag是为计数而构建的数据结构。
Bag<String> counted =
list.stream().collect(Collectors2.countBy(each -> each));
Assert.assertEquals(1, counted.occurrencesOf("World"));
Assert.assertEquals(2, counted.occurrencesOf("Hello"));
System.out.println(counted.toStringOfItemToCount());
Output:
输出：
{World=1, Hello=2}
In this particular case, you can simply collectthe Listdirectly into a Bag.
在这种特殊情况下，你可以简单的collect将List直接进入Bag。
Bag<String> counted =
list.stream().collect(Collectors2.toBag());
You can also create the Bagwithout using a Streamby adapting the Listwith the Eclipse Collections protocols.
您还可以创建Bag不使用Stream通过调整List与Eclipse集合协议。
Bag<String> counted = Lists.adapt(list).countBy(each -> each);
or in this particular case:
或者在这种特殊情况下：
Bag<String> counted = Lists.adapt(list).toBag();
You could also just create the Bag directly.
您也可以直接创建 Bag。
Bag<String> counted = Bags.mutable.with("Hello", "Hello", "World");
A Bag<String>is like a Map<String, Integer>in that it internally keeps track of keys and their counts. But, if you ask a Mapfor a key it doesn't contain, it will return null. If you ask a Bagfor a key it doesn't contain using occurrencesOf, it will return 0.
ABag<String>类似于 a Map<String, Integer>，因为它在内部跟踪键及其计数。但是，如果您向 a 请求Map它不包含的密钥，它将返回null。如果您向 a 请求Bag它不包含 using 的密钥occurrencesOf，它将返回 0。
Note:I am a committer for Eclipse Collections.
注意：我是 Eclipse Collections 的提交者。
