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如何遍历Hashmap中的对象

[英]How to iterate through objects in Hashmap

I am having 我有

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

I want to iterate through each values in list By doing this I am able to get the key and its entire values.But how to get each single value for a key. 我想遍历列表中的每个值,这样就能得到键及其整个值。但是,如何获得键的每个值。

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());
    }

OUTPUT 输出值

->>>>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}]

How to iterate through values 如何遍历值

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

You have a Map that contains values that are instances of List<Attribute> . 您有一个Map ,其中包含作为List<Attribute>实例的值。

If you want to iterate though those lists and display their contents... you'd need to iterate through those lists; 如果要遍历这些列表并显示其内容...,则需要遍历这些列表。

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);
    }
}

Edit to add: You show an Iterator in your example, and you're using raw types instead of generics. 编辑添加:在示例中显示一个Iterator ,并且您使用的是原始类型而不是泛型。 The above does away with the Iterator but presumably this is the correct version of what you were trying to do using Iterator s. 上面的方法消除了Iterator但是大概这是您尝试使用Iterator进行操作的正确版本。 The "for-each" loops shown above are equivalent: 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);
    }
}

try with foreach style in java 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
    }
}

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