有没有办法在Java中随机获取HashMap的值？

有没有办法在Java中随机获取HashMap的值？

Answer 1

你可以使用类似的东西：

Random generator = new Random(); Map.Entry[] entries = myHashMap.entrySet().toArray(); randomValue = entries[generator.nextInt(entries.length)].getValue(); 

更新。

上面不起作用 ， Set.toArray()总是返回一个Object数组，它不能被强制转换为Map.Entry数组。

这个（更简单的版本）确实有效：

 Random generator = new Random(); Object[] values = myHashMap.values().toArray(); Object randomValue = values[generator.nextInt(values.length)]; 

如果您希望随机值是Object以外的类型，只需将转换添加到最后一行。 因此，如果myHashMap被声明为：

 Map myHashMap = new HashMap(); 

最后一行可以是：

 String randomValue = (String) values[generator.nextInt(value.length)]; 

Answer 2

由于要求只要求来自HashMap的随机值，这里的方法是：

1. HashMap有一个values方法，它返回map中值的Collection 。
2. Collection用于创建List 。
3. size方法用于查找List的大小， Random.nextInt方法使用它来获取List的随机索引。
4. 最后，使用随机索引从List get方法中检索该值。

执行：

 HashMap map = new HashMap(); map.put("Hello", 10); map.put("Answer", 42); List valuesList = new ArrayList(map.values()); int randomIndex = new Random().nextInt(valuesList.size()); Integer randomValue = valuesList.get(randomIndex); 

关于这种方法的好处是所有方法都是通用的 – 不需要进行类型转换。

Answer 3

如果您需要从地图中绘制更多的值而不重复任何元素，您可以将地图放入List中然后随机播放。

 List 

Answer 4

生成0和HashMap中键数之间的随机数。 获取随机数的密钥。 从该键获取值。

伪代码 ：

  int n = random(map.keys().length()); String key = map.keys().at(n); Object value = map.at(key); 

如果在Java中很难实现它，那么您可以使用Set的toArray()函数从此代码创建和数组。

  Object[] values = map.values().toArray(new Object[map.size()]); Object random_value = values[random(values.length)]; 

我不确定如何做随机数。

Answer 5

一个好的答案略微取决于具体情况，特别是你需要为给定地图获取随机密钥的频率（注意，无论你采用密钥还是值，技术基本相同）。

• 如果您需要来自给定地图的各种随机密钥 ， 而在获取随机密钥之间不会改变地图 ，那么在迭代密钥集时使用随机抽样方法 。 实际上，你所做的就是遍历keySet（）返回的集合，并在每个项目上计算想要获取该密钥的概率，给定总体需要的数量以及到目前为止所用的数量。 然后生成一个随机数，看看该数字是否低于概率。 （注意：即使您只需要1个密钥， 此方法也将始终有效 ;在这种情况下，它不一定是最有效的方法。）
• HashMap中的键实际上已经是伪随机顺序。 在极端情况下，您只需要一个给定可能映射的随机密钥 ，您甚至可以拔出keySet（）的第一个元素 。
• 在其他情况下 （对于给定的可能地图，您需要多个可能的随机键，或者地图将在您使用随机键之间发生变化），您基本上必须创建或维护从中选择随机的键的数组/列表键。

Answer 6

我真的不知道你为什么要这样做…但如果它有帮助，我创建了一个RandomMap，当你调用values（）时会自动随机化这些值，那么下面的可运行的演示应用程序可能会完成这项任务。 。

  package random; import java.util.ArrayList; import java.util.Collection; import java.util.Collections; import java.util.HashMap; import java.util.Iterator; import java.util.List; import java.util.Map; import java.util.TreeMap; public class Main { public static void main(String[] args) { Map hashMap = makeHashMap(); // you can make any Map random by making them a RandomMap // better if you can just create the Map as a RandomMap instead of HashMap Map randomMap = new RandomMap(hashMap); // just call values() and iterate through them, they will be random Iterator iter = randomMap.values().iterator(); while (iter.hasNext()) { String value = (String) iter.next(); System.out.println(value); } } private static Map makeHashMap() { Map retVal; // HashMap is not ordered, and not exactly random (read the javadocs) retVal = new HashMap(); // TreeMap sorts your map based on Comparable of keys retVal = new TreeMap(); // RandomMap - a map that returns stuff randomly // use this, don't have to create RandomMap after function returns // retVal = new HashMap(); for (int i = 0; i < 20; i++) { retVal.put("key" + i, "value" + i); } return retVal; } } /** * An implementation of Map that shuffles the Collection returned by values(). * Similar approach can be applied to its entrySet() and keySet() methods. */ class RandomMap extends HashMap { public RandomMap() { super(); } public RandomMap(Map map) { super(map); } /** * Randomize the values on every call to values() * * @return randomized Collection */ @Override public Collection values() { List randomList = new ArrayList(super.values()); Collections.shuffle(randomList); return randomList; } } 

Answer 7

下面是一个如何使用Peter Stuifzand描述的数组方法的示例，也是通过values() -method：

 // Populate the map // ... Object[] keys = map.keySet().toArray(); Object[] values = map.values().toArray(); Random rand = new Random(); // Get random key (and value, as an example) String randKey = keys[ rand.nextInt(keys.length) ]; String randValue = values[ rand.nextInt(values.length) ]; // Use the random key System.out.println( map.get(randKey) ); 

Answer 8

通常你并不真正想要一个随机值，而只需要任何值，然后这样做很好：

 Object selectedObj = null; for (Object obj : map.values()) { selectedObj = obj; break; } 

Answer 9

将其转换为数组，然后在热路径中获取该值太慢。

所以得到集合（键或键值集）并执行以下操作：

  public class SetUtility { public static Type getRandomElementFromSet(final Set set, Random random) { final int index = random.nextInt(set.size()); Iterator iterator = set.iterator(); for( int i = 0; i < index-1; i++ ) { iterator.next(); } return iterator.next(); } 

Answer 10

我编写了一个实用程序来从地图，条目集或迭代器中检索随机条目，键或值。

既然你不能也不应该弄清楚迭代器的大小（ Guava可以做到这一点 ），你将不得不重载randEntry()方法来接受一个大小，该大小应该是条目的长度。

 package util; import java.util.HashMap; import java.util.Iterator; import java.util.Map; import java.util.Map.Entry; import java.util.Set; public class MapUtils { public static void main(String[] args) { Map map = new HashMap() { private static final long serialVersionUID = 1L; { put("Foo", 1); put("Bar", 2); put("Baz", 3); } }; System.out.println(randEntryValue(map)); } static  Entry randEntry(Iterator> it, int count) { int index = (int) (Math.random() * count); while (index > 0 && it.hasNext()) { it.next(); index--; } return it.next(); } static  Entry randEntry(Set> entries) { return randEntry(entries.iterator(), entries.size()); } static  Entry randEntry(Map map) { return randEntry(map.entrySet()); } static  K randEntryKey(Map map) { return randEntry(map).getKey(); } static  V randEntryValue(Map map) { return randEntry(map).getValue(); } } 

Answer 11

这取决于你的密钥是什么 – hashmap的性质不允许这很容易发生。

我能想到的方法是选择一个介于1和hashmap大小之间的随机数，然后开始迭代它，保持计数 – 当count等于你的随机数时你选择，这是你的随机元素。

Answer 12

这是一本来自Objects First with Java的练习。它如下所示.pickDefaultResponse（）使用ArrayList生成一个响应并获取一个随机生成的整数并将其用作ArrayList的索引。 这是有效的但我想要克服的是填写两个列表。这里HashMap实例responseMap以及ArrayList实例defaultResponses都必须单独填写。

 import java.util.HashMap; import java.util.HashSet; import java.util.ArrayList; import java.util.Iterator; import java.util.Random; /** * The responder class represents a response generator object. * It is used to generate an automatic response, based on specified input. * Input is presented to the responder as a set of words, and based on those * words the responder will generate a String that represents the response. * * Internally, the reponder uses a HashMap to associate words with response * strings and a list of default responses. If any of the input words is found * in the HashMap, the corresponding response is returned. If none of the input * words is recognized, one of the default responses is randomly chosen. * * @version 1.0 * @author Michael Kolling and David J. Barnes */ public class Responder { // Used to map key words to responses. private HashMap responseMap; // Default responses to use if we don't recognise a word. private ArrayList defaultResponses; private Random randomGenerator; /** * Construct a Responder */ public Responder() { responseMap = new HashMap(); defaultResponses = new ArrayList(); fillResponseMap(); fillDefaultResponses(); randomGenerator = new Random(); } /** * Generate a response from a given set of input words. * * @param words A set of words entered by the user * @return A string that should be displayed as the response */ public String generateResponse(HashSet words) { Iterator it = words.iterator(); while(it.hasNext()) { String word = it.next(); String response = responseMap.get(word); if(response != null) { return response; } } // If we get here, none of the words from the input line was recognized. // In this case we pick one of our default responses (what we say when // we cannot think of anything else to say...) return **pickDefaultResponse();** } /** * Enter all the known keywords and their associated responses * into our response map. */ private void fillResponseMap() { responseMap.put("crash", "Well, it never crashes on our system. It must have something\n" + "to do with your system. Tell me more about your configuration."); responseMap.put("crashes", "Well, it never crashes on our system. It must have something\n" + "to do with your system. Tell me more about your configuration."); responseMap.put("slow", "I think this has to do with your hardware. Upgrading your processor\n" + "should solve all performance problems. Have you got a problem with\n" + "our software?"); responseMap.put("performance", "Performance was quite adequate in all our tests. Are you running\n" + "any other processes in the background?"); responseMap.put("bug", "Well, you know, all software has some bugs. But our software engineers\n" + "are working very hard to fix them. Can you describe the problem a bit\n" + "further?"); responseMap.put("buggy", "Well, you know, all software has some bugs. But our software engineers\n" + "are working very hard to fix them. Can you describe the problem a bit\n" + "further?"); responseMap.put("windows", "This is a known bug to do with the Windows operating system. Please\n" + "report it to Microsoft. There is nothing we can do about this."); responseMap.put("macintosh", "This is a known bug to do with the Mac operating system. Please\n" + "report it to Apple. There is nothing we can do about this."); responseMap.put("expensive", "The cost of our product is quite competitive. Have you looked around\n" + "and really compared our features?"); responseMap.put("installation", "The installation is really quite straight forward. We have tons of\n" + "wizards that do all the work for you. Have you read the installation\n" + "instructions?"); responseMap.put("memory", "If you read the system requirements carefully, you will see that the\n" + "specified memory requirements are 1.5 giga byte. You really should\n" + "upgrade your memory. Anything else you want to know?"); responseMap.put("linux", "We take Linux support very seriously. But there are some problems.\n" + "Most have to do with incompatible glibc versions. Can you be a bit\n" + "more precise?"); responseMap.put("bluej", "Ahhh, BlueJ, yes. We tried to buy out those guys long ago, but\n" + "they simply won't sell... Stubborn people they are. Nothing we can\n" + "do about it, I'm afraid."); } /** * Build up a list of default responses from which we can pick one * if we don't know what else to say. */ private void fillDefaultResponses() { defaultResponses.add("That sounds odd. Could you describe that problem in more detail?"); defaultResponses.add("No other customer has ever complained about this before. \n" + "What is your system configuration?"); defaultResponses.add("That sounds interesting. Tell me more..."); defaultResponses.add("I need a bit more information on that."); defaultResponses.add("Have you checked that you do not have a dll conflict?"); defaultResponses.add("That is explained in the manual. Have you read the manual?"); defaultResponses.add("Your description is a bit wishy-washy. Have you got an expert\n" + "there with you who could describe this more precisely?"); defaultResponses.add("That's not a bug, it's a feature!"); defaultResponses.add("Could you elaborate on that?"); } /** * Randomly select and return one of the default responses. * @return A random default response */ private String **pickDefaultResponse()** { // Pick a random number for the index in the default response list. // The number will be between 0 (inclusive) and the size of the list (exclusive). int index = randomGenerator.nextInt(defaultResponses.size()); return defaultResponses.get(index); } }