如何有效地迭代Java Map中的每个条目？

If I have an object implementing the Map interface in Java and I wish to iterate over every pair contained within it, what is the most efficient way of going through the map?

如果我有一个用Java实现Map接口的对象，并希望迭代其中包含的每一对，那么通过地图的最有效方法是什么？

Will the ordering of elements depend on the specific map implementation that I have for the interface?

元素的排序是否取决于我对界面的具体映射实现？

38 个解决方案

#1

Map<String, String> map = ...
for (Map.Entry<String, String> entry : map.entrySet())
{
    System.out.println(entry.getKey() + "/" + entry.getValue());
}

#2

To summarize the other answers and combine them with what I know, I found 10 main ways to do this (see below). Also, I wrote some performance tests (see results below). For example, if we want to find the sum of all of the keys and values of a map, we can write:

总结其他答案并将它们与我所知道的结合起来，我找到了10种主要方法（见下文）。另外，我写了一些性能测试（见下面的结果）。例如，如果我们想要找到地图的所有键和值的总和，我们可以写：

1. Using iterator and Map.Entry

   使用iterator和Map.Entry

   long i = 0;
   Iterator<Map.Entry<Integer, Integer>> it = map.entrySet().iterator();
   while (it.hasNext()) {
       Map.Entry<Integer, Integer> pair = it.next();
       i += pair.getKey() + pair.getValue();
   }
   

2. Using foreach and Map.Entry

   使用foreach和Map.Entry

   long i = 0;
   for (Map.Entry<Integer, Integer> pair : map.entrySet()) {
       i += pair.getKey() + pair.getValue();
   }
   

3. Using forEach from Java 8

   使用Java 8中的forEach

   final long[] i = {0};
   map.forEach((k, v) -> i[0] += k + v);
   

4. Using keySet and foreach

   使用keySet和foreach

   long i = 0;
   for (Integer key : map.keySet()) {
       i += key + map.get(key);
   }
   

5. Using keySet and iterator

   使用keySet和iterator

   long i = 0;
   Iterator<Integer> itr2 = map.keySet().iterator();
   while (itr2.hasNext()) {
       Integer key = itr2.next();
       i += key + map.get(key);
   }
   

6. Using for and Map.Entry

   使用for和Map.Entry

   long i = 0;
   for (Iterator<Map.Entry<Integer, Integer>> entries = map.entrySet().iterator(); entries.hasNext(); ) {
       Map.Entry<Integer, Integer> entry = entries.next();
       i += entry.getKey() + entry.getValue();
   }
   

7. Using the Java 8 Stream API

   使用Java 8 Stream API

   final long[] i = {0};
   map.entrySet().stream().forEach(e -> i[0] += e.getKey() + e.getValue());
   

8. Using the Java 8 Stream API parallel

   使用Java 8 Stream API并行

   final long[] i = {0};
   map.entrySet().stream().parallel().forEach(e -> i[0] += e.getKey() + e.getValue());
   

9. Using IterableMap of Apache Collections

   使用Apache Collections的IterableMap

   long i = 0;
   MapIterator<Integer, Integer> it = iterableMap.mapIterator();
   while (it.hasNext()) {
       i += it.next() + it.getValue();
   }
   

10. Using MutableMap of Eclipse (CS) collections

    使用Eclipse（CS）集合的MutableMap

    final long[] i = {0};
    mutableMap.forEachKeyValue((key, value) -> {
        i[0] += key + value;
    });
    

Perfomance tests (mode = AverageTime, system = Windows 8.1 64-bit, Intel i7-4790 3.60 GHz, 16 GB)

性能测试（模式= AverageTime，system = Windows 8.1 64位，Intel i7-4790 3.60 GHz，16 GB）

1. For small map (100 elements), score 0.308 is the best

   对于小地图（100个元素），得分0.308是最好的

   Benchmark                          Mode  Cnt  Score    Error  Units
   test3_UsingForEachAndJava8         avgt  10   0.308 ±  0.021  µs/op
   test10_UsingEclipseMap             avgt  10   0.309 ±  0.009  µs/op
   test1_UsingWhileAndMapEntry        avgt  10   0.380 ±  0.014  µs/op
   test6_UsingForAndIterator          avgt  10   0.387 ±  0.016  µs/op
   test2_UsingForEachAndMapEntry      avgt  10   0.391 ±  0.023  µs/op
   test7_UsingJava8StreamApi          avgt  10   0.510 ±  0.014  µs/op
   test9_UsingApacheIterableMap       avgt  10   0.524 ±  0.008  µs/op
   test4_UsingKeySetAndForEach        avgt  10   0.816 ±  0.026  µs/op
   test5_UsingKeySetAndIterator       avgt  10   0.863 ±  0.025  µs/op
   test8_UsingJava8StreamApiParallel  avgt  10   5.552 ±  0.185  µs/op
   

2. For map with 10000 elements, score 37.606 is the best

   对于10000个元素的地图，得分37.606是最好的

   Benchmark                           Mode   Cnt  Score      Error   Units
   test10_UsingEclipseMap              avgt   10    37.606 ±   0.790  µs/op
   test3_UsingForEachAndJava8          avgt   10    50.368 ±   0.887  µs/op
   test6_UsingForAndIterator           avgt   10    50.332 ±   0.507  µs/op
   test2_UsingForEachAndMapEntry       avgt   10    51.406 ±   1.032  µs/op
   test1_UsingWhileAndMapEntry         avgt   10    52.538 ±   2.431  µs/op
   test7_UsingJava8StreamApi           avgt   10    54.464 ±   0.712  µs/op
   test4_UsingKeySetAndForEach         avgt   10    79.016 ±  25.345  µs/op
   test5_UsingKeySetAndIterator        avgt   10    91.105 ±  10.220  µs/op
   test8_UsingJava8StreamApiParallel   avgt   10   112.511 ±   0.365  µs/op
   test9_UsingApacheIterableMap        avgt   10   125.714 ±   1.935  µs/op
   

3. For map with 100000 elements, score 1184.767 is the best

   对于具有100000个元素的地图，得分11​​84.767是最好的

   Benchmark                          Mode   Cnt  Score        Error    Units
   test1_UsingWhileAndMapEntry        avgt   10   1184.767 ±   332.968  µs/op
   test10_UsingEclipseMap             avgt   10   1191.735 ±   304.273  µs/op
   test2_UsingForEachAndMapEntry      avgt   10   1205.815 ±   366.043  µs/op
   test6_UsingForAndIterator          avgt   10   1206.873 ±   367.272  µs/op
   test8_UsingJava8StreamApiParallel  avgt   10   1485.895 ±   233.143  µs/op
   test5_UsingKeySetAndIterator       avgt   10   1540.281 ±   357.497  µs/op
   test4_UsingKeySetAndForEach        avgt   10   1593.342 ±   294.417  µs/op
   test3_UsingForEachAndJava8         avgt   10   1666.296 ±   126.443  µs/op
   test7_UsingJava8StreamApi          avgt   10   1706.676 ±   436.867  µs/op
   test9_UsingApacheIterableMap       avgt   10   3289.866 ±  1445.564  µs/op
   

Graphs (perfomance tests depending on map size)

图表（性能测试取决于地图大小）

Table (perfomance tests depending on map size)

表（性能测试取决于地图大小）

          100     600      1100     1600     2100
test10    0.333    1.631    2.752    5.937    8.024
test3     0.309    1.971    4.147    8.147   10.473
test6     0.372    2.190    4.470    8.322   10.531
test1     0.405    2.237    4.616    8.645   10.707
test2     0.376    2.267    4.809    8.403   10.910
test7     0.473    2.448    5.668    9.790   12.125
test9     0.565    2.830    5.952   13.220   16.965
test4     0.808    5.012    8.813   13.939   17.407
test5     0.810    5.104    8.533   14.064   17.422
test8     5.173   12.499   17.351   24.671   30.403

All tests are on GitHub.

所有测试都在GitHub上。

#3

In Java 8 you can do it clean and fast using the new lambdas features:

在Java 8中，您可以使用新的lambdas功能清洁和快速地执行此操作：

 Map<String,String> map = new HashMap<>();
 map.put("SomeKey", "SomeValue");
 map.forEach( (k,v) -> [do something with key and value] );

 // such as
 map.forEach( (k,v) -> System.out.println("Key: " + k + ": Value: " + v));

The type of k and v will be inferred by the compiler and there is no need to use Map.Entry anymore.

k和v的类型将由编译器推断，不再需要使用Map.Entry。

Easy-peasy!

十分简单！

#4

Yes, the order depends on the specific Map implementation.

是的，订单取决于具体的Map实施。

@ScArcher2 has the more elegant Java 1.5 syntax. In 1.4, I would do something like this:

@ ScArcher2具有更优雅的Java 1.5语法。在1.4中，我会做这样的事情：

Iterator entries = myMap.entrySet().iterator();
while (entries.hasNext()) {
  Entry thisEntry = (Entry) entries.next();
  Object key = thisEntry.getKey();
  Object value = thisEntry.getValue();
  // ...
}

#5

Typical code for iterating over a map is:

迭代地图的典型代码是：

Map<String,Thing> map = ...;
for (Map.Entry<String,Thing> entry : map.entrySet()) {
    String key = entry.getKey();
    Thing thing = entry.getValue();
    ...
}

HashMap is the canonical map implementation and doesn't make guarantees (or though it should not change order if no mutating operation are performed on it). SortedMap will return entries based on the natural ordering of the keys, or a Comparator, if provided. LinkedHashMap will either return entries in insertion-order or access-order depending upon how it has been constructed. EnumMap returns entries in natural order of keys.

HashMap是规范的地图实现，不提供保证（或者如果不对其执行变异操作则不应更改顺序）。 SortedMap将根据键的自然顺序或比较器（如果提供）返回条目。 LinkedHashMap将以插入顺序或访问顺序返回条目，具体取决于它的构造方式。 EnumMap按键的自然顺序返回条目。

(Update: I think this is no longer true.) Note, IdentityHashMap entrySet iterator currently has a peculiar implementation which returns the same Map.Entry instance for every item in the entrySet! However, every time a new the iterator advances the Map.Entry is updated.

（更新：我认为这不再是真的。）注意，IdentityHashMap entrySet迭代器当前有一个特殊的实现，它为entrySet中的每个项返回相同的Map.Entry实例！但是，每当新的迭代器推进时，Map.Entry都会更新。

#6

Example of using iterator and generics:

使用迭代器和泛型的示例：

Iterator<Map.Entry<String, String>> entries = myMap.entrySet().iterator();
while (entries.hasNext()) {
  Map.Entry<String, String> entry = entries.next();
  String key = entry.getKey();
  String value = entry.getValue();
  // ...
}

#7

This is a two part question:

这是一个两部分问题：

How to iterate over the entries of a Map - @ScArcher2 has answered that perfectly.

如何迭代Map的条目 - @ ScArcher2已经完美地回答了这个问题。

What is the order of iteration - if you are just using Map, then strictly speaking, there are no ordering guarantees. So you shouldn't really rely on the ordering given by any implementation. However, the SortedMap interface extends Map and provides exactly what you are looking for - implementations will aways give a consistent sort order.

迭代的顺序是什么 - 如果你只是使用Map，那么严格来说，没有排序保证。所以你不应该真正依赖任何实现给出的顺序。但是，SortedMap接口扩展了Map并提供了您正在寻找的内容 - 实现将提供一致的排序顺序。

NavigableMap is another useful extension - this is a SortedMap with additional methods for finding entries by their ordered position in the key set. So potentially this can remove the need for iterating in the first place - you might be able to find the specific entry you are after using the higherEntry, lowerEntry, ceilingEntry, or floorEntry methods. The descendingMap method even gives you an explicit method of reversing the traversal order.

NavigableMap是另一个有用的扩展 - 这是一个SortedMap，带有其他方法，用于按键集中的有序位置查找条目。因此，这可能会消除首先进行迭代的需要 - 您可能能够在使用higherEntry，lowerEntry，ceilingEntry或floorEntry方法之后找到您所使用的特定条目。 descendingMap方法甚至为您提供了一种反转遍历顺序的显式方法。

#8

There are several ways to iterate over map.

迭代地图有几种方法。

Here is comparison of their performances for a common data set stored in map by storing a million key value pairs in map and will iterate over map.

这里是通过在地图中存储一百万个键值对来比较它们在地图中存储的公共数据集的性能，并将迭代在地图上。

1) Using entrySet() in for each loop

1）为每个循环使用entrySet（）

for (Map.Entry<String,Integer> entry : testMap.entrySet()) {
    entry.getKey();
    entry.getValue();
}

50 milliseconds

50毫秒

2) Using keySet() in for each loop

2）对每个循环使用keySet（）

for (String key : testMap.keySet()) {
    testMap.get(key);
}

76 milliseconds

76毫秒

3) Using entrySet() and iterator

3）使用entrySet（）和迭代器

Iterator<Map.Entry<String,Integer>> itr1 = testMap.entrySet().iterator();
while(itr1.hasNext()) {
    Map.Entry<String,Integer> entry = itr1.next();
    entry.getKey();
    entry.getValue();
}

50 milliseconds

50毫秒

4) Using keySet() and iterator

4）使用keySet（）和迭代器

Iterator itr2 = testMap.keySet().iterator();
while(itr2.hasNext()) {
    String key = itr2.next();
    testMap.get(key);
}

75 milliseconds

75毫秒

I have referred this link.

我已经提到了这个链接。

#9

FYI, you can also use map.keySet() and map.values() if you're only interested in keys/values of the map and not the other.

仅供参考，如果您只对地图的键/值感兴趣而不感兴趣，则还可以使用map.keySet（）和map.values（）。

#10

The correct way to do this is to use the accepted answer as it is the most efficient. I find the following code looks a bit cleaner.

正确的方法是使用接受的答案，因为它是最有效的。我发现以下代码看起来更清晰。

for (String key: map.keySet()) {
   System.out.println(key + "/" + map.get(key));
}

#11

With Eclipse Collections (formerly GS Collections), you would use the forEachKeyValue method on the MapIterable interface, which is inherited by the MutableMap and ImmutableMap interfaces and their implementations.

使用Eclipse Collections（以前称为GS Collections），您可以在MapIterable接口上使用forEachKeyValue方法，该方法由MutableMap和ImmutableMap接口及其实现继承。

final MutableBag<String> result = Bags.mutable.empty();
MutableMap<Integer, String> map = Maps.mutable.of(1, "One", 2, "Two", 3, "Three");
map.forEachKeyValue(new Procedure2<Integer, String>()
{
    public void value(Integer key, String value)
    {
        result.add(key + value);
    }
});
Assert.assertEquals(Bags.mutable.of("1One", "2Two", "3Three"), result);

With Java 8 lambda syntax, you can write the code as follows:

使用Java 8 lambda语法，您可以按如下方式编写代码：

MutableBag<String> result = Bags.mutable.empty();
MutableMap<Integer, String> map = Maps.mutable.of(1, "One", 2, "Two", 3, "Three");
map.forEachKeyValue((key, value) -> result.add(key + value));
Assert.assertEquals(Bags.mutable.of("1One", "2Two", "3Three"), result);

Note: I am a committer for Eclipse Collections.

注意：我是Eclipse Collections的提交者。

#12

Try this with Java 1.4:

尝试使用Java 1.4：

for( Iterator entries = myMap.entrySet().iterator(); entries.hasNext();){

  Entry entry = (Entry) entries.next();

  System.out.println(entry.getKey() + "/" + entry.getValue());

  //...
}

#13

Java 8:

Java 8：

You can use lambda expressions:

您可以使用lambda表达式：

myMap.entrySet().stream().forEach((entry) -> {
    Object currentKey = entry.getKey();
    Object currentValue = entry.getValue();
});

For more information, follow this.

有关更多信息，请按照此操作。

#14

In theory, the most efficient way will depend on which implementation of Map. The official way to do this is to call map.entrySet(), which returns a set of Map.Entry, each of which contains a key and a value (entry.getKey() and entry.getValue()).

从理论上讲，最有效的方法取决于Map的实现。执行此操作的官方方法是调用map.entrySet（），它返回一组Map.Entry，每个Map.Entry包含一个键和一个值（entry.getKey（）和entry.getValue（））。

In an idiosyncratic implementation, it might make some difference whether you use map.keySet(), map.entrySet() or something else. But I can't think of a reason why anyone would write it like that. Most likely it makes no difference to performance what you do.

在一个特殊的实现中，无论是使用map.keySet（），map.entrySet（）还是其他东西，它都可能会有所不同。但我想不出有人会这样写的原因。很可能它对你的表现没有任何影响。

And yes, the order will depend on the implementation - as well as (possibly) the order of insertion and other hard-to-control factors.

是的，订单将取决于实施 - 以及（可能）插入顺序和其他难以控制的因素。

[edit] I wrote valueSet() originally but of course entrySet() is actually the answer.

[编辑]我最初写了valueSet（）但当然entrySet（）实际上是答案。

#15

Lambda Expression Java 8

Lambda Expression Java 8

In Java 1.8 (Java 8) this has become lot easier by using forEach method from Aggregate operations(Stream operations) that looks similar to iterators from Iterable Interface.

在Java 1.8（Java 8）中，通过使用类似于来自Iterable Interface的迭代器的聚合操作（流操作）的forEach方法，这变得更加容易。

Just copy paste below statement to your code and rename the HashMap variable from hm to your HashMap variable to print out key-value pair.

只需将下面的语句粘贴到代码中，然后将HashMap变量从hm重命名为HashMap变量，以打印出键值对。

HashMap<Integer,Integer> hm = new HashMap<Integer, Integer>();
/*
 *     Logic to put the Key,Value pair in your HashMap hm
 */

// Print the key value pair in one line.

hm.forEach((k,v) -> System.out.println("key: "+k+" value:"+v));

// Just copy and paste above line to your code.

Below is the sample code that i tried using Lambda Expression. This stuff is so cool. Must try.

下面是我尝试使用Lambda Expression的示例代码。这东西很酷。一定要试。

HashMap<Integer,Integer> hm = new HashMap<Integer, Integer>();
    Random rand = new Random(47);
    int i=0;
    while(i<5){
        i++;
        int key = rand.nextInt(20);
        int value = rand.nextInt(50);
        System.out.println("Inserting key: "+key+" Value: "+value);
        Integer imap =hm.put(key,value);
        if( imap == null){
            System.out.println("Inserted");
        }
        else{
            System.out.println("Replaced with "+imap);
        }               
    }

    hm.forEach((k,v) -> System.out.println("key: "+k+" value:"+v));

Output:

Inserting key: 18 Value: 5
Inserted
Inserting key: 13 Value: 11
Inserted
Inserting key: 1 Value: 29
Inserted
Inserting key: 8 Value: 0
Inserted
Inserting key: 2 Value: 7
Inserted
key: 1 value:29
key: 18 value:5
key: 2 value:7
key: 8 value:0
key: 13 value:11

Also one can use Spliterator for the same.

也可以使用Spliterator。

Spliterator sit = hm.entrySet().spliterator();

UPDATE

UPDATE

---

Including documentation links to Oracle Docs. For more on Lambda go to this link and must read Aggregate Operations and for Spliterator go to this link.

包括Oracle Docs的文档链接。有关Lambda的更多信息，请转到此链接，并且必须阅读Aggregate Operations，对于Spliterator，请转到此链接。

#16

In Map one can Iteration over keys and/or values and/or both (e.g., entrySet) depends on one's interested in_ Like:

在Map中，可以对键和/或值和/或两者进行迭代（例如，entrySet）取决于一个人的兴趣_喜欢：

1.) Iterate through the keys -> keySet() of the map:

1.）遍历地图的键 - > keySet（）：

Map<String, Object> map = ...;

for (String key : map.keySet()) {
    //your Business logic...
}

2.) Iterate through the values -> values() of the map:

2.）迭代地图的值 - > values（）：

for (Object value : map.values()) {
    //your Business logic...
}

3.) Iterate through the both -> entrySet() of the map:

3.）遍历地图的两个 - > entrySet（）：

for (Map.Entry<String, Object> entry : map.entrySet()) {
    String key = entry.getKey();
    Object value = entry.getValue();
    //your Business logic...
}

Moreover, there are 3 difference ways to Iterate Through a HashMap. They are as below_

此外，通过HashMap迭代有3种不同的方法。它们如下_

//1.
for (Map.Entry entry : hm.entrySet()) {
    System.out.print("key,val: ");
    System.out.println(entry.getKey() + "," + entry.getValue());
}

//2.
Iterator iter = hm.keySet().iterator();
while(iter.hasNext()) {
    Integer key = (Integer)iter.next();
    String val = (String)hm.get(key);
    System.out.println("key,val: " + key + "," + val);
}

//3.
Iterator it = hm.entrySet().iterator();
while (it.hasNext()) {
    Map.Entry entry = (Map.Entry) it.next();
    Integer key = (Integer)entry.getKey();
    String val = (String)entry.getValue();
    System.out.println("key,val: " + key + "," + val);
}

#17

public class abcd{
    public static void main(String[] args)
    {
       Map<Integer, String> testMap = new HashMap<Integer, String>();
        testMap.put(10, "a");
        testMap.put(20, "b");
        testMap.put(30, "c");
        testMap.put(40, "d");
        for (Integer key:testMap.keySet()) {
            String value=testMap.get(key);
            System.out.println(value);
        }
    }
}

OR

要么

public class abcd {
    public static void main(String[] args)
    {
       Map<Integer, String> testMap = new HashMap<Integer, String>();
        testMap.put(10, "a");
        testMap.put(20, "b");
        testMap.put(30, "c");
        testMap.put(40, "d");
        for (Entry<Integer, String> entry : testMap.entrySet()) {
            Integer key=entry.getKey();
            String value=entry.getValue();
        }
    }
}

#18

If you have a generic untyped Map you can use:

如果您有一个通用的无类型地图，您可以使用：

Map map = new HashMap();
for (Map.Entry entry : ((Set<Map.Entry>) map.entrySet())) {
    System.out.println(entry.getKey() + "/" + entry.getValue());
}

#19

    Iterator iterator = map.entrySet().iterator();
    while (iterator.hasNext()) {
        Map.Entry element = (Map.Entry)it.next();
        LOGGER.debug("Key: " + element.getKey());
        LOGGER.debug("value: " + element.getValue());    
    }

#20

With Java 8

使用Java 8

map.forEach((k, v) -> System.out.println((k + ":" + v)));

#21

Most compact with Java 8:

最紧凑的Java 8：

map.entrySet().forEach(System.out::println);

#22

You can do it using generics:

你可以使用泛型来做到这一点：

Map<Integer, Integer> map = new HashMap<Integer, Integer>();
Iterator<Map.Entry<Integer, Integer>> entries = map.entrySet().iterator();
while (entries.hasNext()) {
    Map.Entry<Integer, Integer> entry = entries.next();
    System.out.println("Key = " + entry.getKey() + ", Value = " + entry.getValue());
}

#23

Iterator itr2 = testMap.keySet().iterator();
while (itr2.hasNext()) {
    String key = itr2.next();
    testMap.get(key);
}

for (String key: map.keySet()) {    
    System.out.println(key + "/" + map.get(key)); 
}

The best way is entrySet() though.

最好的方法是entrySet（）。

#24

           //Functional Oprations
            Map<String, String> mapString = new HashMap<>();
            mapString.entrySet().stream().map((entry) -> {
                String mapKey = entry.getKey();
                return entry;
            }).forEach((entry) -> {
                String mapValue = entry.getValue();
            });

            //Intrator
            Map<String, String> mapString = new HashMap<>();
            for (Iterator<Map.Entry<String, String>> it = mapString.entrySet().iterator(); it.hasNext();) {
                Map.Entry<String, String> entry = it.next();
                String mapKey = entry.getKey();
                String mapValue = entry.getValue();
            }

            //Simple for loop
            Map<String, String> mapString = new HashMap<>();
            for (Map.Entry<String, String> entry : mapString.entrySet()) {
                String mapKey = entry.getKey();
                String mapValue = entry.getValue();

            }

#25

package com.test;

import java.util.Collection;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;

public class Test {

    public static void main(String[] args) {
        Map<String, String> map = new HashMap<String, String>();
        map.put("ram", "ayodhya");
        map.put("krishan", "mathura");
        map.put("shiv", "kailash");

        System.out.println("********* Keys *********");
        Set<String> keys = map.keySet();
        for (String key : keys) {
            System.out.println(key);
        }

        System.out.println("********* Values *********");
        Collection<String> values = map.values();
        for (String value : values) {
            System.out.println(value);
        }

        System.out.println("***** Keys and Values (Using for each loop) *****");
        for (Map.Entry<String, String> entry : map.entrySet()) {
            System.out.println("Key: " + entry.getKey() + "\t Value: "
                    + entry.getValue());
        }

        System.out.println("***** Keys and Values (Using while loop) *****");
        Iterator<Entry<String, String>> entries = map.entrySet().iterator();
        while (entries.hasNext()) {
            Map.Entry<String, String> entry = (Map.Entry<String, String>) entries
                    .next();
            System.out.println("Key: " + entry.getKey() + "\t Value: "
                    + entry.getValue());
        }

        System.out
                .println("** Keys and Values (Using java 8 using lambdas )***");
        map.forEach((k, v) -> System.out
                .println("Key: " + k + "\t value: " + v));
    }
}

#26

Yes, as many people agreed this is the best way to iterate over a Map.

是的，因为很多人都认为这是迭代Map的最佳方式。

But there are chances to throw nullpointerexception if the map is null. Don't forget to put null .check in.

但是如果map是null，则有机会抛出nullpointerexception。不要忘记将null .check放入。

                                                 |
                                                 |
                                         - - - -
                                       |
                                       |
for (Map.Entry<String, Object> entry : map.entrySet()) {
    String key = entry.getKey();
    Object value = entry.getValue();
}

#27

There are a lot of ways to do this. Below is a few simple steps:

有很多方法可以做到这一点。以下是几个简单的步骤：

Suppose you have one Map like:

假设您有一个地图，如：

Map<String, Integer> m = new HashMap<String, Integer>();

Then you can do something like the below to iterate over map elements.

然后你可以做类似下面的事情迭代地图元素。

// ********** Using an iterator ****************
Iterator<Entry<String, Integer>> me = m.entrySet().iterator();
while(me.hasNext()){
    Entry<String, Integer> pair = me.next();
    System.out.println(pair.getKey() + ":" + pair.getValue());
}

// *********** Using foreach ************************
for(Entry<String, Integer> me : m.entrySet()){
    System.out.println(me.getKey() + " : " + me.getValue());
}

// *********** Using keySet *****************************
for(String s : m.keySet()){
    System.out.println(s + " : " + m.get(s));
}

// *********** Using keySet and iterator *****************
Iterator<String> me = m.keySet().iterator();
while(me.hasNext()){
    String key = me.next();
    System.out.println(key + " : " + m.get(key));
}

#28

Use Java 8:

使用Java 8：

map.entrySet().forEach(entry -> System.out.println(entry.getValue()));

#29

The ordering will always depend on the specific map implementation. Using Java 8 you can use either of these:

排序将始终取决于具体的地图实施。使用Java 8，您可以使用以下任一方法：

map.forEach((k,v) -> { System.out.println(k + ":" + v); });

Or:

要么：

map.entrySet().forEach((e) -> {
            System.out.println(e.getKey() + " : " + e.getValue());
        });

The result will be the same (same order). The entrySet backed by the map so you are getting the same order. The second one is handy as it allows you to use lambdas, e.g. if you want only to print only Integer objects that are greater than 5:

结果将是相同的（相同的顺序）。由映射支持的entrySet，以便您获得相同的顺序。第二个是方便的，因为它允许你使用lambda，例如如果您只想打印大于5的Integer对象：

map.entrySet()
    .stream()
    .filter(e-> e.getValue() > 5)
    .forEach(System.out::println);

The code below shows iteration through LinkedHashMap and normal HashMap (example). You will see difference in the order:

下面的代码显示了LinkedHashMap和普通HashMap的迭代（示例）。你会看到顺序的不同：

public class HMIteration {


    public static void main(String[] args) {
        Map<Object, Object> linkedHashMap = new LinkedHashMap<>();
        Map<Object, Object> hashMap = new HashMap<>();

        for (int i=10; i>=0; i--) {
            linkedHashMap.put(i, i);
            hashMap.put(i, i);
        }

        System.out.println("LinkedHashMap (1): ");
        linkedHashMap.forEach((k,v) -> { System.out.print(k + " (#="+k.hashCode() + "):" + v + ", "); });

        System.out.println("\nLinkedHashMap (2): ");

        linkedHashMap.entrySet().forEach((e) -> {
            System.out.print(e.getKey() + " : " + e.getValue() + ", ");
        });


        System.out.println("\n\nHashMap (1): ");
        hashMap.forEach((k,v) -> { System.out.print(k + " (#:"+k.hashCode() + "):" + v + ", "); });

        System.out.println("\nHashMap (2): ");

        hashMap.entrySet().forEach((e) -> {
            System.out.print(e.getKey() + " : " + e.getValue() + ", ");
        });
    }
}

LinkedHashMap (1):

LinkedHashMap（1）：

10 (#=10):10, 9 (#=9):9, 8 (#=8):8, 7 (#=7):7, 6 (#=6):6, 5 (#=5):5, 4 (#=4):4, 3 (#=3):3, 2 (#=2):2, 1 (#=1):1, 0 (#=0):0,

10（＃= 10）：10,9（＃= 9）：9,8（＃= 8）：8,7（＃= 7）：7,6（＃= 6）：6,5（＃= 5 ）：5,4（＃= 4）：4,3（＃= 3）：3,2（＃= 2）：2,1（＃= 1）：1,0（＃= 0）：0，

LinkedHashMap (2):

LinkedHashMap（2）：

10 : 10, 9 : 9, 8 : 8, 7 : 7, 6 : 6, 5 : 5, 4 : 4, 3 : 3, 2 : 2, 1 : 1, 0 : 0,

10：10,9：9,8：8,7：7,6：6,5：5,4：4,3：3,2：2,1：1,0：0，

HashMap (1):

HashMap（1）：

0 (#:0):0, 1 (#:1):1, 2 (#:2):2, 3 (#:3):3, 4 (#:4):4, 5 (#:5):5, 6 (#:6):6, 7 (#:7):7, 8 (#:8):8, 9 (#:9):9, 10 (#:10):10,

0（＃：0）：0,1（＃：1）：1,2（＃：2）：2,3（＃：3）：3,4（＃：4）：4,5（＃：5 ）：5,6（＃：6）：6,7（＃：7）：7,8（＃：8）：8,9（＃：9）：9,10（＃：10）：10，

HashMap (2):

HashMap（2）：

0 : 0, 1 : 1, 2 : 2, 3 : 3, 4 : 4, 5 : 5, 6 : 6, 7 : 7, 8 : 8, 9 : 9, 10 : 10,

0：0,1：1,2：2,3：3,4：4,5：5,6：6,7：7,8：8,9：9,10：10，

#30

If your reason for iterating trough the Map, is to do an operation on the value and write to a resulting Map. I recommend using the transform-methods in the Google Guava Maps class.

如果您通过Map迭代的原因是对值进行操作并写入生成的Map。我建议在Google Guava Maps类中使用transform-methods。

import com.google.common.collect.Maps;

After you have added the Maps to your imports, you can use Maps.transformValues and Maps.transformEntries on your maps, like this:

将地图添加到导入后，可以在地图上使用Maps.transformValues和Maps.transformEntries，如下所示：

public void transformMap(){
    Map<String, Integer> map = new HashMap<>();
    map.put("a", 2);
    map.put("b", 4);

    Map<String, Integer> result = Maps.transformValues(map, num -> num * 2);
    result.forEach((key, val) -> print(key, Integer.toString(val)));
    // key=a,value=4
    // key=b,value=8

    Map<String, String> result2 = Maps.transformEntries(map, (key, value) -> value + "[" + key + "]");
    result2.forEach(this::print);
    // key=a,value=2[a]
    // key=b,value=4[b]
}

private void print(String key, String val){
    System.out.println("key=" + key + ",value=" + val);
}