Question

我有

Map<String, List<Attribute>> binList = new HashMap<String, List<Attribute>>();

我想遍历列表中的每个值通过这样做，我能够获得密钥及其全部值。但是如何获取密钥的每个单值。

BinList {3=[index=0 {from=1.3,to=2.42}, index=1 {from=2.42,to=3.54}, index=2 {from=3.54,to=4.66}, index=3 {from=4.66,to=5.78}, index=4 {from=5.78,to=6.9}], 2=[index=0 {from=2.3,to=2.76}, index=1 {from=2.76,to=3.2199999999999998}, index=2 {from=3.2199999999999998,to=3.6799999999999997}, index=3 {from=3.6799999999999997,to=4.14}, index=4 {from=4.14,to=4.6}], 1=[index=0 {from=4.3,to=5.02}, index=1 {from=5.02,to=5.739999999999999}, index=2 {from=5.739999999999999,to=6.459999999999999}, index=3 {from=6.459999999999999,to=7.179999999999999}, index=4 {from=7.179999999999999,to=7.899999999999999}], 4=[index=0 {from=0.3,to=0.76}, index=1 {from=0.76,to=1.2200000000000002}, index=2 {from=1.2200000000000002,to=1.6800000000000002}, index=3 {from=1.6800000000000002,to=2.14}, index=4 {from=2.14,to=2.6}]}



Map<String, List<Attribute>> binList = new HashMap<String, List<Attribute>>();
System.out.println("BinList "+binList);
//Iterating binList
while (it.hasNext()) {
    Map.Entry pairs = (Map.Entry)it.next();
    System.out.println("->>>>"+pairs.getKey() + " = " + pairs.getValue());
    }

输出

->>>>3 = [index=0 {from=1.3,to=2.42}, index=1 {from=2.42,to=3.54}, index=2 {from=3.54,to=4.66}, index=3 {from=4.66,to=5.78}, index=4 {from=5.78,to=6.9}]
->>>>2 = [index=0 {from=2.3,to=2.76}, index=1 {from=2.76,to=3.2199999999999998}, index=2 {from=3.2199999999999998,to=3.6799999999999997}, index=3 {from=3.6799999999999997,to=4.14}, index=4 {from=4.14,to=4.6}]
->>>>1 = [index=0 {from=4.3,to=5.02}, index=1 {from=5.02,to=5.739999999999999}, index=2 {from=5.739999999999999,to=6.459999999999999}, index=3 {from=6.459999999999999,to=7.179999999999999}, index=4 {from=7.179999999999999,to=7.899999999999999}]
->>>>4 = [index=0 {from=0.3,to=0.76}, index=1 {from=0.76,to=1.2200000000000002}, index=2 {from=1.2200000000000002,to=1.6800000000000002}, index=3 {from=1.6800000000000002,to=2.14}, index=4 {from=2.14,to=2.6}]

如何迭代值

id =3
   index=0 {from=1.3,to=2.42}
   index=1 {from=2.42,to=3.54}
.
.

Answer 1

您的Map包含List<Attribute>实例的值。

如果您想迭代这些列表并显示其内容......您需要遍历这些列表;

for (Map.Entry<String, List<Attribute>> entry : binList.entrySet())
{
    System.out.println("Key: " + entry.getKey());

    // Each value is a List<Attribute>, so you can iterate though that as well
    for (Attribute a : entry.getValue())
    {
        // This assumes Attribute.toString() prints something useful 
        System.out.println("Attribute: " + a);
    }
}

编辑以添加：您在示例中显示Iterator，而您使用的是原始类型而不是泛型。以上内容消除了Iterator，但可能这是您使用Iterator尝试执行的操作的正确版本。上面显示的“for-each”循环是等效的：

Iterator<Map.Entry<String, List<Attribute>>> it = binList.entrySet().iterator();
while (it.hasNext())
{
    Map.Entry<String, List<Attribute>> entry = it.next();
    System.out.println("Key: " + entry.getKey());

    // Each value is a List<Attribute>, so you can iterate though that as well
    Iterator<Attribute> it2 = entry.getValue().iterator();

    while (it2.hasNext())
    {
        Attribute a = it2.next();
        // This assumes Attribute.toString() prints something useful 
        System.out.println("Attribute: " + a);
    }
}

Answer 2

尝试使用java中的foreach样式

for (String entryKey:binList.keySet()){
    for (List<Attribute> attribute:binList.get(entryKey)){
        attribute.from
        attribute.to
    }
}

如何遍历Hashmap中的对象

2 个答案: