如何加入 java 中的两个地图？

Question

我有以下地图：

LonM =[0, 0, 0, 0, 46, 15]
LatM =[5, 52, 35, 16, 37, 5]

并行地，我实现了一个复合模式class:

public class CompositeDataFrame implements IDataFrame, ICompositeDataFrame {

    private String name;
    private List <IDataFrame> children;

    public CompositeDataFrame(String name){
        this.name = name;
        children = new LinkedList<>();
    }

    public void addChildren(IDataFrame child){
        children.add(child);
    }

并且得到如下查询方法：

public Map<Object, List<Object>> query(String keySelector, Predicate<Object> valuePredicate) {
    Map<Object, List<Object>> result = new HashMap<>();
    for (IDataFrame child:children)
        result.putAll(child.query(keySelector, valuePredicate));
    return result;
}

那么我们有下面调用查询方法的main:

CompositeDataFrame comp = new CompositeDataFrame("CompositeCities");
        CompositeDataFrame comp1 = new CompositeDataFrame("2");

        comp.addChildren(comp1);
        comp1.addChildren(df);
        comp.addChildren(df);
        
        System.out.println("\n"+comp.query("LonM", entry -> ((Integer) entry) == 0));

问题是：如何连接每个 child 的地图？

PD：我使用了 putAll 方法，但它删除了具有相同键的映射，因此在这种情况下它没有用。

输出应该是：

LonM =[0, 0, 0, 0, 0, 0, 0, 0] LatM =[5, 52, 35, 16, 5, 52, 35, 16]

非常感谢！

Answer 1

是的，如果您使用 putAll，它会删除具有相同键的映射条目。这就是地图的工作原理。您正在寻找的是一种按键对所有条目进行分组的方法，使用 Streams.

可以很容易地做到这一点

Map<Object, List<Object>> result = children.stream()
        // Map from Stream<IDataFrame> to Stream<Map.Entry<Object, List<Object>>>
        .flatMap(child -> child.query(keySelector, predicate).entrySet().stream())
        // Collect back into a Map<Object, List<Object>> using groupinBy
        .collect(Collectors.groupingBy(
            // Use the key to group the values
            Map.Entry::getKey, 
            // Since we don't want a Map<Object, List<List<Object>>> the List needs to be flattend using Collectors.flatMapping.
            Collectors.flatMapping(entry -> entry.getValue().stream(), Collectors.toList())));

当然你不必使用Streams。

Map<Object, List<Object>> result = new HashMap<>();
for (IDataFrame child : children) {
    // Loop over each entry of the query result.
    child.query(keySelector, predicate).forEach((key, value) -> {
        // computeIfAbsent creates a new List if none exists for the given key, otherwise it returns the existing one.
        List<Object> valueList = result.computeIfAbsent(key, o -> new ArrayList<>());
        // Finally just add all the values to the list.
        valueList.addAll(value);
    });
}

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