java stream过滤

Java Stream filter can be very helpful when you want to do some processing only on some elements of the Stream based on certain condition.

当您希望仅基于特定条件对Stream的某些元素进行某些处理时，Java Stream过滤器可能会非常有用。

Java Stream过滤器 (Java Stream filter)

was introduced in Java 8. Java Stream filter is an intermediate operation, i.e. it returns a new Stream.

是Java 8中引入的。Java Stream过滤器是一个中间操作，即它返回一个新的Stream。

Java Stream过滤器示例 (Java Stream filter example)

Let’s look at a simple example to count the even numbers in a list of integers. We can do it like below.

让我们看一个简单的示例来计算整数列表中的偶数。 我们可以像下面这样。

List
    
      intList = generateList(); //some method to create the list of integersint evenCount = 0;for(int i : intList){	if(i%2==0){		evenCount++;		//code to do some processing	}}System.out.println("Even Number Count: "+evenCount);
    

Now let’s see how we can do the same thing using java stream filter api.

现在，让我们看看如何使用java流过滤器api来做同样的事情。

Predicate
    
      predicate = new Predicate
     
      (){	@Override	public boolean test(Integer i) {		return (i%2==0);	}	};List
      
        evenList = intList.parallelStream().filter(predicate).collect(Collectors.toList());evenCount = evenList.size();System.out.println("Even Number Count: "+evenCount);
      
     
    

Predicate is a functional interface that takes one argument and return boolean based on the test function. Above code can also be written as below using .

谓词是一个功能接口，它接受一个参数并根据测试函数返回布尔值。 上面的代码也可以使用如下编写。

evenCount = intList.parallelStream().filter(i -> {	return (i % 2 == 0);}).collect(Collectors.toList()).size();

Java流过滤器示例2 (Java Stream Filter Example 2)

Let’s look at another example where we want to filter a list of integers and keep only numbers greater than 90.

让我们看另一个示例，我们要过滤一个整数列表，并且仅保留大于90的数字。

List
    
      intList = new ArrayList<>();for(int i=50; i<100; i++) intList.add(i);Stream
     
       sequentialStream = myList.stream(); // we can create parallel stream tooStream
      
        highNumsStream = sequentialStream.filter(p -> p > 90); //filter numbers greater than 90System.out.print("High Nums greater than 90=");highNumsStream.forEach(p -> System.out.print(p+" "));//prints "High Nums greater than 90=91 92 93 94 95 96 97 98 99 "
      
     
    

带对象的Java流过滤器示例 (Java Stream filter Example with Objects)

Let’s say we have list of employees and we want to print names of all employees with age greater than 32. Below is a simple program to do this using java stream filter.

假设我们有一个雇员列表，并且想打印所有年龄大于32岁的雇员的姓名。下面是一个使用Java流过滤器执行此操作的简单程序。

package com.journaldev.examples;import java.util.ArrayList;import java.util.List;public class StreamFilterExample {	public static void main(String[] args) {		List
    
      empList = new ArrayList<>();		empList.add(new Employee("Pankaj", 35));		empList.add(new Employee("David", 25));		empList.add(new Employee("Lisa", 31));		empList.add(new Employee("Dean", 40));		// print all employees above 32 years of age		empList.stream().filter(p -> (p.getAge() > 32)).forEach(p -> System.out.println(p.getName()));	}}class Employee {	private String name;	private int age;	public Employee() {	}	public Employee(String n, int a) {		this.name = n;		this.age = a;	}	public String getName() {		return name;	}	public void setName(String name) {		this.name = name;	}	public int getAge() {		return age;	}	public void setAge(int age) {		this.age = age;	}}
    

Above program produces below output.

上面的程序产生下面的输出。

PankajDean

If we want to get the first employee only with age over 32, then we can use filter with findFirst method as shown below.

如果我们只想获取年龄在32岁以上的第一位员工，则可以将filter与findFirst方法结合使用，如下所示。

Optional
    
      empO = empList.stream().filter(p -> (p.getAge() > 32)).findFirst();if(empO.isPresent())	System.out.println(empO.get().getName()+","+empO.get().getAge());
    

That’s all for java stream filter examples.

Java流过滤器示例的全部内容。

Reference:

参考：

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java stream过滤