Question

我需要分开并计算arraylist中的多少个值是相同的，并根据出现次数打印它们。

我有一个名为digits的arraylist：

 [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]

我创建了一个方法，用于分隔每个值并将其保存到新数组中。

public static ArrayList<Integer> myNumbers(int z) {

    ArrayList<Integer> digits = new ArrayList<Integer>();
    String number = String.valueOf(z);
    for (int a = 0; a < number.length(); a++) {
        int j = Character.digit(number.charAt(a), 10);
        digits.add(j);
    }
    return digits;

}

在此之后，我得到了一个名为numbers的新数组。我在这个阵列上使用了排序

Collections.sort(numbers);

我的ArrayList如下所示：

[0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9]

它有：

2 times 0; 
9 times 1;
4 times 2;
6 times 3;
5 times 4;
6 times 5;
5 times 6;
5 times 7;
5 times 8;
3 times 9;

我需要打印出一串数字取决于它们的数量所以它假设看起来像这样： 1354678290

Answer 1

使用Collections.frequency方法计算重复项

Answer 2

那么，你可以尝试使用Map

Map<Integer, Integer> countMap = new HashMap<>();

  for (Integer item: yourArrayList) {

      if (countMap.containsKey(item))
          countMap.put(item, countMap.get(item) + 1);
      else
          countMap.put(item, 1);
  }

在forEach循环结束后，您将获得一个填充的地图，其中包含您的项目计数

Answer 3

例如，使用Stream API。

package tests;

import org.junit.Assert;
import org.junit.Test;

import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;

public class Duplicates {

    @Test
    public void duplicates() throws Exception {
        List<Integer> items = Arrays.asList(1, 1, 2, 2, 2, 2);

        Map<Integer, Long> result = items.stream()
                .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

        Assert.assertEquals(Long.valueOf(2), result.get(1));
        Assert.assertEquals(Long.valueOf(4), result.get(2));
    }
}

Answer 4

问题是计算阵列中有多少个三三分。在Java 7中，解决方案是：

import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

public class howMany1 {
public static void main(String[] args) {

    List<Integer> list = Arrays.asList(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765);

    Map<Integer ,Integer> map = new HashMap<>();

      for(  Integer r  : list) {
          if(  map.containsKey(r)   ) {
                 map.put(r, map.get(r) + 1);
          }//if
          else {
              map.put(r, 1);
          }
      }//for

      //iterate

      Set< Map.Entry<Integer ,Integer> > entrySet = map.entrySet();
      for(    Map.Entry<Integer ,Integer>  entry : entrySet     ) {
          System.out.printf(   "%s : %d %n "    , entry.getKey(),entry.getValue()  );
      }//for

}}

在Java 8中，问题的解决方案是：

import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.stream.Collectors;

public class howMany2 {
public static void main(String[] args) {

    List<Integer> list = Arrays.asList(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765);
     // we can also use Function.identity() instead of c->c
    Map<Integer ,Long > map = list.stream()
            .collect(  Collectors.groupingBy(c ->c , Collectors.counting())         ) ;


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

}}

另一种方法是使用Collections.frequency。解决方案是：

import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Duplicates1 {
public static void main(String[] args) {

    List<Integer> list = Arrays.asList(1, 1, 2, 3, 5, 8, 13,13, 21, 34, 55, 89, 144, 233);

    System.out.println("Count all with frequency");
    Set<Integer> set = new HashSet<Integer>(list);
    for (Integer r : set) {
        System.out.println(r + ": " + Collections.frequency(list, r));
    }

}}

另一种方法是使用method =＆gt;将int数组更改为Integer List。 Arrays.stream（array）.boxed（）。collect（Collectors.toList（））然后使用for循环获取整数。

public class t7 {
    public static void main(String[] args) {
        int[] a = { 1, 1, 2, 3, 5, 8, 13, 13 };
        List<Integer> list = Arrays.stream(a).boxed().collect(Collectors.toList());

        for (Integer ch : list) {
            System.out.println(ch + " :  " + Collections.frequency(list, ch));
        }

    }// main
}

Answer 5

List<String> list = new ArrayList<String>();
    list.add("a");
    list.add("b");
    list.add("c");
    list.add("a");
    list.add("a");
    list.add("a");

int countA=Collections.frequency(list, "a");
int countB=Collections.frequency(list, "b");
int countC=Collections.frequency(list, "c");

Answer 6

您可以通过添加列表中的所有元素并将其存储在哈希集中来计算列表中重复元素的数量，一旦完成，您需要知道的是获得哈希值的大小和清单。

ArrayList<String> al = new ArrayList<String>();
al.add("Santosh");
al.add("Saket");
al.add("Saket");
al.add("Shyam");
al.add("Santosh");
al.add("Shyam");
al.add("Santosh");
al.add("Santosh");
HashSet<String> hs = new HashSet<String>();
hs.addAll(al);
int totalDuplicates =al.size() - hs.size();
System.out.println(totalDuplicates);

如果需要更多说明，请告诉我

Answer 7

Java 8, the solution: 1. Create Map when the Key is the Value of Array and Value is counter. 
2. Check if Map contains the Key increase counter or add a new set.

 private static void calculateDublicateValues(int[] array) {
      //key is value of array, value is counter
      Map<Integer, Integer> map = new HashMap<Integer, Integer>();

      for (Integer element : array) {
        if (map.containsKey(element)) {
          map.put(element, map.get(element) + 1); // increase counter if contains
        } else
          map.put(element, 1);
      }

      map.forEach((k, v) -> {
        if (v > 1)
          System.out.println("The element " + k + " duplicated " + v + " times");
      });

    }

Answer 8

将以下功能用于计数重复元素：

public void countDuplicate(){
        try {
            Set<String> set = new HashSet<>(original_array);
            ArrayList<String> temp_array = new ArrayList<>();
            temp_array.addAll(set);
            for (int i = 0 ; i < temp_array.size() ; i++){
                Log.e(temp_array.get(i),"=>"+Collections.frequency(original_array,temp_array.get(i)));
            }
        }catch (Exception e){
            e.printStackTrace();
        }
    }

Answer 9

Java 8可以用3行代码来解决这个问题。

    Map<Integer, Integer>  duplicatedCount = new LinkedHashMap<>();
    list.forEach(a -> duplicatedCount.put(a, duplicatedCount.getOrDefault(a, 0) +1));
    duplicatedCount.forEach((k,v) -> System.out.println(v+" times "+k));

Answer 10

还可以使用

Map＃merge 方法：

 List<Integer> input = Arrays.asList(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584,
            4181, 6765);
 final Map<Integer, Integer> result = new HashMap<>();
 input.forEach(in -> result.merge(in, 1, Integer::sum));

Answer 11

import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;

public class A_List_Find_Duplicate_Count {

    public static void main(String[] args) {
        
        List<String> list = Arrays.asList(
                "Chennai","Bangalore","Pune","Hyderabad",
                "Chennai","Pune","Mysore","Delhi","Hyderabad",
                "Pune"
                );
        String elementToFound = "Chennai";
        
        //JAVA 8 
        long count = list.stream().filter(i->elementToFound.equals(i)).count();
        System.out.println("elementToFound : "+count);
        
        
        //using Collections
        int frequency = Collections.frequency(list, elementToFound);
        System.out.println("frequency : "+frequency);
        
        
        //using Map
        HashMap<String, Integer> map = new HashMap<>();
        for(String s : list)
        {
            map.put(s, map.get(s)!=null ? map.get(s)+1 : 1);
        }
        
        System.out.println("map : "+map.get(elementToFound));
        
        //JAVA 8 using groupingBy
        Map<String, Long> collect = list.stream().collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
        System.out.println("collect groupingBy : "+collect.get(elementToFound));
        

    }

}

输出

<块引用>

elementToFound : 2 频率 : 2 地图 : 2 收集 groupingBy : 2

如何计算ArrayList中的重复元素？

11 个答案: