Java Streams：将列表分组为地图映射
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Java Streams: group a List into a Map of Maps
提问by mrod
How could I do the following with Java Streams?
我如何使用 Java Streams 执行以下操作？
Let's say I have the following classes:
假设我有以下课程：
class Foo {
Bar b;
}
class Bar {
String id;
String date;
}
I have a List<Foo>and I want to convert it to a Map <Foo.b.id, Map<Foo.b.date, Foo>. I.e: group first by the Foo.b.idand then by Foo.b.date.
我有一个List<Foo>，我想将它转换为一个Map <Foo.b.id, Map<Foo.b.date, Foo>. 即：首先分组Foo.b.id，然后分组Foo.b.date。
I'm struggling with the following 2-step approach, but the second one doesn't even compile:
我正在努力使用以下两步方法，但第二个方法甚至无法编译：
Map<String, List<Foo>> groupById =
myList
.stream()
.collect(
Collectors.groupingBy(
foo -> foo.getBar().getId()
)
);
Map<String, Map<String, Foo>> output = groupById.entrySet()
.stream()
.map(
entry -> entry.getKey(),
entry -> entry.getValue()
.stream()
.collect(
Collectors.groupingBy(
bar -> bar.getDate()
)
)
);
Thanks in advance.
提前致谢。
采纳答案by Flown
You can group your data in one go assuming there are only distinct Foo:
假设只有不同的数据，您可以一次性对数据进行分组Foo：
Map<String, Map<String, Foo>> map = list.stream()
.collect(Collectors.groupingBy(f -> f.b.id,
Collectors.toMap(f -> f.b.date, Function.identity())));
Saving some characters by using static imports:
使用静态导入保存一些字符：
Map<String, Map<String, Foo>> map = list.stream()
.collect(groupingBy(f -> f.b.id, toMap(f -> f.b.date, identity())));
回答by Alex Salauyou
Suppose (b.id, b.date)pairs are distinct. If so,
in second step you don't need grouping, just collecting to Mapwhere key is foo.b.dateand value is fooitself:
假设(b.id, b.date)对是不同的。如果是这样，在第二步中，您不需要分组，只需收集到Mapkey 所在的位置foo.b.date， valuefoo本身即可：
Map<String, Map<String, Foo>> map =
myList.stream()
.collect(Collectors.groupingBy(f -> f.b.id)) // map {Foo.b.id -> List<Foo>}
.entrySet().stream()
.collect(Collectors.toMap(e -> e.getKey(), // id
e -> e.getValue().stream() // stream of foos
.collect(Collectors.toMap(f -> f.b.date,
f -> f))));
Or even more simple:
或者更简单：
Map<String, Map<String, Foo>> map =
myList.stream()
.collect(Collectors.groupingBy(f -> f.b.id,
Collectors.toMap(f -> f.b.date,
f -> f)));
回答by weston
An alternative is to support the equality contract on your key, Bar:
另一种选择是支持您的密钥上的平等合同，Bar：
class Bar {
String id;
String date;
public boolean equals(Object o){
if (o == null) return false;
if (!o.getClass().equals(getClass())) return false;
Bar other = (Bar)o;
return Objects.equals(o.id, id) && Objects.equals(o.date, date);
}
public int hashCode(){
return id.hashCode*31 + date.hashCode;
}
}
Now you can just have a Map<Bar, Foo>.
现在你可以只拥有一个Map<Bar, Foo>.
