如何在Java中实现列表折叠

本文介绍了如何在Java中实现列表折叠的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个列表，并希望将其减少为单个值（函数式编程术语fold，Ruby术语 inject ），如

  Arrays.asList（a，b，c）... fold ...a，b，c

当我感染了函数式编程思想（Scala）时，我正在寻找一种更简单的方法来编写它

  sb = new StringBuilder 
 for ... {
 append ... 
 } 
 sb.toString

解决方案

寻找是一个字符串 join（）方法，Java自8.0以来。请尝试以下方法之一。

 
  静态方法：
 集合< String& source = Arrays.asList（a，b，c）; 
 String result = String.join（，，source）; 
  
 
 
   接口支持与Scala的 foldLeft 函数非常相似的折叠操作。请查看以下连结收集器：
 集合< String> source = Arrays.asList（a，b，c）; 
 String result = source.stream（）。collect（Collectors.joining（，））; 
  
您可能希望静态导入 Collectors.joining  
 
 $ b           pre>  Collection< Integer> numbers = Arrays.asList（1，2，3）; 
 String result = numbers.stream（）
 .map（Object :: toString）
 .collect（Collectors.joining（，））;

I have a List and want to reduce it to a single value (functional programming term "fold", Ruby term inject), like

Arrays.asList("a", "b", "c") ... fold ... "a,b,c"

As I am infected with functional programming ideas (Scala), I am looking for an easier/shorter way to code it than

sb = new StringBuilder
for ... {
  append ...
}
sb.toString

解决方案

What you are looking for is a string join() method which Java has since 8.0. Try one of the methods below.

Static method String#join(delimiter, elements):

Collection<String> source = Arrays.asList("a", "b", "c");
String result = String.join(",", source);

Stream interface supports a fold operation very similar to Scala’s foldLeft function. Take a look at the following concatenating Collector:
```
Collection<String> source = Arrays.asList("a", "b", "c");
String result = source.stream().collect(Collectors.joining(","));
```
You may want to statically import Collectors.joining to make your code clearer.
By the way this collector can be applied to collections of any particular objects:
```
Collection<Integer> numbers = Arrays.asList(1, 2, 3);
String result = numbers.stream()
        .map(Object::toString)
        .collect(Collectors.joining(","));
```

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如何在Java中实现列表折叠

问题描述