Java 如何计算ArrayList中的重复元素？
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How to count duplicate elements in ArrayList?
提问by Piotr Szczepanik
I need to separate and count how many values in arraylist are the same and print them according to the number of occurrences.
我需要分离并计算arraylist中有多少个值是相同的，并根据出现的次数打印它们。
I've got an arraylist called digits :
我有一个名为 numbers 的数组列表：
[1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]
I created a method which separates each value and saves it to a new array.
我创建了一个方法来分隔每个值并将其保存到一个新数组中。
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;
}
After this I've got a new array called numbers. I'm using sort on this array
在此之后，我有了一个名为 numbers 的新数组。我在这个数组上使用排序
Collections.sort(numbers);
and my ArrayList looks like this:
我的 ArrayList 看起来像这样：
[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]
It has:
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;
I need to print out the string of numbers depend on how many are they So it suppose to look like this : 1354678290
我需要打印出一串数字取决于它们有多少所以它看起来像这样：1354678290
采纳答案by pruthwiraj.kadam
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");
回答by PrabaharanKathiresan
use Collections.frequency method to count the duplicates
使用 Collections.frequency 方法计算重复项
回答by Andrew Petryk
Well, for that you can try to use Map
好吧，为此你可以尝试使用 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);
}
After end of forEach loop you will have a filled map with your items against it count
在 forEach 循环结束后，您将有一个填充的地图，其中包含您的项目计数
回答by Florian Salihovic
By using the Stream API for example.
例如，通过使用 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));
}
}
回答by Srivastava
You can count the number of duplicate elements in a list by adding all the elements of the list and storing it in a hashset, once that is done, all you need to know is get the difference in the size of the hashset and the list.
您可以通过添加列表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，您需要知道的就是获取哈希集和列表的大小差异。
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);
Let me know if this needs more clarification
让我知道这是否需要更多说明
回答by Soudipta Dutta
The question is to count how many ones twos and threes are there in an array. In Java 7 solution is:
问题是计算数组中有多少个三三两两。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
}}
In Java 8, the solution to the problem is :
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 ) );
}}
One another method is to use Collections.frequency. The solution is:
另一种方法是使用 Collections.frequency。解决办法是：
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));
}
}}
Another method is to change the int array to Integer List using method => Arrays.stream(array).boxed().collect(Collectors.toList()) and then get the integer using for loop.
另一种方法是使用 method => Arrays.stream(array).boxed().collect(Collectors.toList()) 将 int 数组更改为整数列表，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
}
回答by Madina Mountaniol
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");
});
}
回答by Ramij
Use below function for count duplicate elements :
使用以下函数计算重复元素：
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();
}
}
回答by Bhushan Uniyal
Java 8 can handle this problem with 3 lines of code.
Java 8 可以用 3 行代码处理这个问题。
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));
