“数组中多个最常见的元素”

huangapple
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英文:

Multiple most frequent elements in array

问题

我想检测并在屏幕上打印出出现频率最高的项。我知道当最常见的元素只有一个时该如何做。然而，如果我们有这样一个包含以下元素的列表：

10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10

我想打印出六和十，我的意思是无论有多少个，我都想打印出所有出现频率最高的元素。

英文:

Suppose we have a list of integers. I would like to detect and print on the screen the most frequently repeated item . I know how to do it when the most common element is only one. However, if we have such a list which contains these elements:

  1. 10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10

i want to print a six and a ten so I mean I want to print all the most frequently repeated elements no matter how many there are..

答案1

得分: 1

Sure, here's the translated Python code:

  1. nums = [10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10]
  3. def most_recurrent_ints(arr):
  4.     log = {}
  5.     for i in arr:
  6.         if i in log:
  7.             log[i] += 1
  8.         else:
  9.             log[i] =  1
  10.     current_max = 0
  11.     for i in log.values():
  12.         if i > current_max:
  13.             current_max = i
  14.     results = []
  15.     for k, v in log.items():
  16.         if v == current_max:
  17.             results.append(k)
  18.     return results
  20. print(most_recurrent_ints(nums))

Please note that the code translation is provided as requested, and no additional content or answers are included.

英文: 
  1. nums = [10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10]
  3. def most_recurrent_ints(arr):
  4.     log = {}
  5.     for i in arr:
  6.         if i in log:
  7.             log[i] += 1
  8.         else:
  9.             log[i] =  1
  10.     current_max = 0
  11.     for i in log.values():
  12.         if i > current_max:
  13.             current_max = i
  14.     results = []
  15.     for k, v in log.items():
  16.         if v == current_max:
  17.             results.append(k)
  18.     return results
  19.     
  20. print(most_recurrent_ints(nums))

I'm bad at Java but this is the Python solution, maybe someone can translate. I can do it in JS if you need.

答案2

得分: 1

以下是翻译好的部分：

你考虑过使用字典类型的数据结构吗？我不懂 Java，所以可能不太优雅，但它实现了你所需的功能：

  1. final int[] array = new int[]{ 10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10 };
  3. Map<Integer, Integer> dictionary = new HashMap<Integer,Integer>(); 
  5. for(int i = 0; i < array.length; ++i){
  6.     int val = array[i];
  7.     if(dictionary.containsKey(val)){
  8.         dictionary.put(val, dictionary.get(val) + 1);
  9.     }else{
  10.         dictionary.put(val, 1);
  11.     }
  12. }
  14. for(Map.Entry<Integer,Integer> entry : dictionary.entrySet()){
  15.     System.out.println(entry.getKey() + ": " + entry.getValue());
  16. }

这将输出：

  1. 1: 1
  2. 2: 2
  3. 4: 1
  4. 5: 1
  5. 6: 4
  6. 10: 4
英文:

Have you considered using a dictionary-type data structure? I am not a java person, so this may not be pretty, but it does what you are looking for:

  1.     final int[] array = new int[]{ 10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10 };
  3.     Map&lt;Integer, Integer&gt; dictionary = new HashMap&lt;Integer,Integer&gt;(); 
  4.     
  5.     for(int i = 0; i &lt; array.length; ++i){
  6.         int val = array[i];
  7.         if(dictionary.containsKey(val)){
  8.             dictionary.put(val, dictionary.get(val) + 1);
  9.         }else{
  10.             dictionary.put(val, 1);
  11.         }
  12.     }
  14.     for(Map.Entry&lt;Integer,Integer&gt; entry : dictionary.entrySet()){
  15.         System.out.println(entry.getKey() + &quot;: &quot; + entry.getValue());
  16.     }

This would output:

  1. 1: 1
  2. 2: 2
  3. 4: 1
  4. 5: 1
  5. 6: 4
  6. 10: 4

答案3

得分: 1

以下是您提供的代码的中文翻译部分：

  1. public static void main(String[] args) {
  2.     MostFrequent(new Integer[] {10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10});
  3. }
  5. static void MostFrequent(Integer[] arr) {
  6.     Map<Integer, Integer> count = new HashMap<Integer, Integer>();
  8.     for (Integer element : arr) {
  9.         if (!count.containsKey(element)) {
  10.             count.put(element, 0);
  11.         }
  12.         count.put(element, count.get(element) + 1);
  13.     }
  15.     Map.Entry<Integer, Integer> maxEntry = null;
  16.     ArrayList<Integer> list = new ArrayList<Integer>();
  18.     for (Map.Entry<Integer, Integer> entry : count.entrySet()) {
  19.         if (maxEntry == null || entry.getValue() > maxEntry.getValue()) {
  20.             list.clear();
  21.             list.add(entry.getKey());
  22.             maxEntry = entry;
  23.         }
  24.         else if (entry.getValue() == maxEntry.getValue()) {
  25.             list.add(entry.getKey());
  26.         }
  27.     }
  29.     for (Integer item: list) {
  30.         System.out.println(item); 
  31.     }        
  32. }

输出结果：

  1. 6
  2. 10
英文: 
  1. public static void main(String[] args) {
  2. 	MostFrequent(new Integer[] {10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10});
  3. }
  5. static void MostFrequent(Integer[] arr)	{
  6. 	Map&lt;Integer, Integer&gt; count = new HashMap&lt;Integer, Integer&gt;();
  7. 	
  8. 	for (Integer element : arr) {
  9. 		if (!count.containsKey(element)) {
  10. 			count.put(element,0);
  11. 		}
  12. 		count.put(element, count.get(element) + 1);
  13. 	}
  15. 	Map.Entry&lt;Integer, Integer&gt; maxEntry = null;
  16. 	ArrayList&lt;Integer&gt; list = new ArrayList&lt;Integer&gt;();
  18. 	for (Map.Entry&lt;Integer, Integer&gt; entry : count.entrySet()) {
  19. 		if (maxEntry == null || entry.getValue() &gt; maxEntry.getValue()) {
  20. 			list.clear();
  21. 			list.add(entry.getKey());
  22. 			maxEntry = entry;
  23. 		}
  24. 		else if (entry.getValue() == maxEntry.getValue()) {
  25. 			list.add(entry.getKey());
  26. 		}
  27. 	}
  29. 	for (Integer item: list) {
  30. 		System.out.println(item); 
  31. 	}		
  32. }

Output:

  1. 6
  2. 10

答案4

得分: 1

以下是翻译好的代码部分：

  1. 你需要一个地图（也称为字典）数据结构来解决这个问题。这是我能想到的最快最简洁的解决方案。
  3. public static void main(String[] args){
  4.     int[] arr = new int[]{10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10};
  5.     printMostFrequent(arr);
  6. }
  7.  
  8. private static void printMostFrequent(int[] arr){
  9.     // 键：输入数组中的数字
  10.     // 值：数字在输入数组中出现的次数
  11.     Map<Integer, Integer> counts = new HashMap<>();
  12.     
  13.     // 同一个数字在输入数组中出现的最大次数。在这个例子中，它是4。
  14.     int highestFrequency = 0;
  15.     
  16.     // 遍历输入数组，填充映射。
  17.     for (int num : arr){
  18.         
  19.         // 如果数字在映射中不存在，它的频率是1。否则，在当前频率上加1。
  20.         int currFrequency = counts.getOrDefault(num, 0) + 1;
  21.         
  22.         // 更新当前数字的频率。
  23.         counts.put(num, currFrequency);
  24.         
  25.         // 如果当前数字的频率是目前为止最高的，存储它的频率以备后用。
  26.         highestFrequency = Math.max(currFrequency, highestFrequency);
  27.     }
  29.     // 遍历数组中的唯一数字（请记住，Java中的Map不允许重复的键）。
  30.     for (int key : counts.keySet()){
  31.         
  32.         // 如果当前数字具有最高的频率，则将其打印到控制台。
  33.         if (counts.get(key) == highestFrequency){
  34.             System.out.println(key);
  35.         }
  36.     }
  37. }

输出：

  1. 6
  2. 10
英文:

You would need a map (aka dictionary) data structure to solve this problem. This is the fastest and most concise solution I could come up with.

  1. public static void main(String[] args){
  2.     int[] arr = new int[]{10, 5, 2, 1, 2, 4, 6, 6, 6, 6, 10, 10, 10};
  3.     printMostFrequent(arr);
  4. }
  5.  
  6. private static void printMostFrequent(int[] arr){
  7.     // Key: number in input array
  8.     // Value: amount of times that number appears in the input array
  9.     Map&lt;Integer, Integer&gt; counts = new HashMap&lt;&gt;();
  11.     // The most amount of times the same number appears in the input array. In this example, it&#39;s 4.
  12.     int highestFrequency = 0;
  13.     
  14.     // Iterate through input array, populating map.
  15.     for (int num : arr){
  16.         
  17.         // If number doesn&#39;t exist in map already, its frequency is 1. Otherwise, add 1 to its current frequency.
  18.         int currFrequency = counts.getOrDefault(num, 0) + 1;
  19.         
  20.         // Update frequency of current number.
  21. 	    counts.put(num, currFrequency);
  22. 	    
  23. 	    // If the current number has the highest frequency so far, store its frequency for later use.
  24. 	    highestFrequency = Math.max(currFrequency, highestFrequency);
  25.     }
  27.     // Iterate through unique numbers in array (remember, a Map in Java allows no duplicate keys).
  28.     for (int key : counts.keySet()){
  29.         
  30.         // If the current number has the highest frequency, then print it to console.
  31. 	    if (counts.get(key) == highestFrequency){
  32. 		    System.out.println(key);
  33. 	    }
  34.     }
  35. }

Output

  1. 6
  2. 10

huangapple
  • 本文由 发表于 2020年7月31日 10:25:09
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