blocks|key|2541249|text|/**
+*+Zips+lists.+E.g.+given+[[1,2,3],[4,5,6]],+returns+[[1,4],[2,5],[3,6]].
+*+@param+listOfLists+an+N+x+M+list
+*+@returns+an+M+x+N+list
+*/
static+<T>+List<List<T>>+zip(List<List<T>>+listOfLists)+{
++++int+size+=+listOfLists.get(0).size();
++++List<List<T>>+result+=+new+ArrayList<>(size);
++++for+(int+i+=+0;+i+<+size;+%2B%2Bi)
++++++++result.add(
++++++++++++++++listOfLists.stream()
++++++++++++++++++++++++.map(list+->+list.get(i))
++++++++++++++++++++++++.collect(toList()));
++++return+result;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2541250|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>/**
 * Zips lists. E.g. given [[1,2,3],[4,5,6]], returns [[1,4],[2,5],[3,6]].
 * @param listOfLists an N x M list
 * @returns an M x N list
 */
static &lt;T&gt; List&lt;List&lt;T&gt;&gt; zip(List&lt;List&lt;T&gt;&gt; listOfLists) {
 int size = listOfLists.get(0).size();
 List&lt;List&lt;T&gt;&gt; result = new ArrayList&lt;&gt;(size);
 for (int i = 0; i &lt; size; ++i)
 result.add(
 listOfLists.stream()
 .map(list -&gt; list.get(i))
 .collect(toList()));
 return result;
}
</code></pre>

blocks|key|621634|text|Java+streams本身并不支持压缩。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|621635|如果你想手动完成，那么使用IntStream作为列表的‘迭代器’是可行的：|621636|++++List<Integer>+l1+=+Arrays.asList(1,+2,+3);
++++List<Integer>+l2+=+Arrays.asList(2,+3,+4);

++++List<Object[]>+zipped+=+IntStream.range(0,+3).mapToObj(i+->+new+Object[]{l1.get(i),+l2.get(i)}).collect(Collectors.toList());

++++zipped.stream().forEach(i+->+System.out.println(i[0]+%2B+"+"+%2B+i[1]));|code-block|syntax|javascript|621637|然而，这是丑陋的，而且不是“java”的方式(就像使用一个“properties”数组而不是一个类)。|621638|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Java streams don't natively support zipping.

If you want to do it manually, then using an IntStream as an 'iterator' over the lists is the way to go:

<pre><code> List&lt;Integer&gt; l1 = Arrays.asList(1, 2, 3);
 List&lt;Integer&gt; l2 = Arrays.asList(2, 3, 4);

 List&lt;Object[]&gt; zipped = IntStream.range(0, 3).mapToObj(i -&gt; new Object[]{l1.get(i), l2.get(i)}).collect(Collectors.toList());

 zipped.stream().forEach(i -&gt; System.out.println(i[0] + " " + i[1]));
</code></pre>

This is ugly, however, and not the 'java' way (as in using an array of 'properties' instead of a class).

blocks|key|3837974|text|如果我们认为所有的列表都有相同的大小，那么为什么是Java+8呢？你可以像这样使用一个简单的循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3837975|List<List<Integer>>+list+=+new+ArrayList<>();
for(int+i+=+0;+i<a.size();+i%2B%2B){
++++list.add(Arrays.asList(a.get(i),+b.get(i),+c.get(i),+d.get(i)));
}|code-block|syntax|javascript|3837976|输出|offset|length|style|BOLD|3837977|[[1,+1,+1,+1],+[2,+2,+2,+2],+[3,+3,+3,+3]]|3837978|3837979|我真的坚持要在这里读这篇文章+|3837980|entityMap^0|0|0|0|2|0|0|0|0|E|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@$I|W|J|X|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|S|$]]

If we consider that all the Lists have the same size, so why Java 8? you can just use a simple loop like this :

<pre><code>List&lt;List&lt;Integer&gt;&gt; list = new ArrayList&lt;&gt;();
for(int i = 0; i&lt;a.size(); i++){
 list.add(Arrays.asList(a.get(i), b.get(i), c.get(i), d.get(i)));
}
</code></pre>

Output

<pre><code>[[1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3]]
</code></pre>

<hr>

I really insist to read this post here <a href="https://softwareengineering.stackexchange.com/questions/341066/is-using-lambda-expressions-whenever-possible-in-java-good-practice">Is using Lambda expressions whenever possible in java good practice?</a>

blocks|key|3838043|text|如果我没弄错你的问题，你需要这样的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3838044|private+<T>+List<List<T>>+zip(Stream<Iterable<T>>+stream)+{
++++return+stream.collect(LinkedList::new,
++++++++++++(lists,+ts)+->+{
++++++++++++++++if+(lists.isEmpty())+{
++++++++++++++++++++ts.forEach(t+->+lists.add(new+LinkedList<T>()+{{
++++++++++++++++++++++++add(t);
++++++++++++++++++++}}));
++++++++++++++++}+else+{
++++++++++++++++++++Iterator<List<T>>+listIterator+=+lists.iterator();
++++++++++++++++++++Iterator<T>+elementIterator+=+ts.iterator();
++++++++++++++++++++while+(listIterator.hasNext()+&&+elementIterator.hasNext())+{
++++++++++++++++++++++++listIterator.next().add(elementIterator.next());
++++++++++++++++++++}
++++++++++++++++++++while+(listIterator.hasNext())+{
++++++++++++++++++++++++listIterator.next();
++++++++++++++++++++++++listIterator.remove();
++++++++++++++++++++}
++++++++++++++++}
++++++++++++},+(lists,+lists2)+->+{
++++++++++++++++Iterator<List<T>>+firstListIterator+=+lists.iterator();
++++++++++++++++Iterator<List<T>>+secondListIterator+=+lists2.iterator();
++++++++++++++++while+(firstListIterator.hasNext()+&&+secondListIterator.hasNext())+{
++++++++++++++++++++firstListIterator.next().addAll(secondListIterator.next());
++++++++++++++++}
++++++++++++++++while+(firstListIterator.hasNext())+{
++++++++++++++++++++firstListIterator.next();
++++++++++++++++++++firstListIterator.remove();
++++++++++++++++}
++++++++++++});
}|code-block|syntax|javascript|3838045|您可以使用以下命令获得所需结果|3838046|System.out.println(zip(sample.stream()));|3838047|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If i right understand your question you need something like that:

<pre><code>private &lt;T&gt; List&lt;List&lt;T&gt;&gt; zip(Stream&lt;Iterable&lt;T&gt;&gt; stream) {
 return stream.collect(LinkedList::new,
 (lists, ts) -&gt; {
 if (lists.isEmpty()) {
 ts.forEach(t -&gt; lists.add(new LinkedList&lt;T&gt;() {{
 add(t);
 }}));
 } else {
 Iterator&lt;List&lt;T&gt;&gt; listIterator = lists.iterator();
 Iterator&lt;T&gt; elementIterator = ts.iterator();
 while (listIterator.hasNext() &amp;&amp; elementIterator.hasNext()) {
 listIterator.next().add(elementIterator.next());
 }
 while (listIterator.hasNext()) {
 listIterator.next();
 listIterator.remove();
 }
 }
 }, (lists, lists2) -&gt; {
 Iterator&lt;List&lt;T&gt;&gt; firstListIterator = lists.iterator();
 Iterator&lt;List&lt;T&gt;&gt; secondListIterator = lists2.iterator();
 while (firstListIterator.hasNext() &amp;&amp; secondListIterator.hasNext()) {
 firstListIterator.next().addAll(secondListIterator.next());
 }
 while (firstListIterator.hasNext()) {
 firstListIterator.next();
 firstListIterator.remove();
 }
 });
}
</code></pre>

And you can get desired results with

<pre><code>System.out.println(zip(sample.stream()));
</code></pre>

Given :
<pre><code>List&lt;Integer&gt; a = Arrays.asList(1,2,3);
List&lt;Integer&gt; b = Arrays.asList(1,2,3);
List&lt;Integer&gt; c = Arrays.asList(1,2,3);
List&lt;Integer&gt; d = Arrays.asList(1,2,3);

List&lt;List&lt;Integer&gt;&gt; sample = Arrays.asList(a,b,c,d);
</code></pre>
How can I get this result with java 8?
<pre><code>[(1,1,1,1),(2,2,2,2),(3,3,3,3)]
</code></pre>

How to zip multiple lists using java 8?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

教程

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云智能顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

EdgeOne AI 安全实战专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

聚焦“写作效率、视觉美观与运行性能”三方面进行全面升级，为您提供更高效、稳定的创作环境

社区富文本&Markdown编辑器全新改版上线，欢迎大家体验!

诚挚邀请您参与本次调研，分享您的真实使用感受与建议。您的反馈至关重要，感谢您的支持与参与！

社区新版编辑器体验调研

给定：List<Integer> a = Arrays.asList(1,2,3);List<Integer> b = Arrays.asList(1,2,3);List<Integer> c = Arrays.asList(1,2,3);List<Integer> d = Arrays.asList(1,2,3);List<List<Integer>> sample = Arrays.asLis

问如何使用java 8压缩多个列表？
EN

回答 4

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何使用java 8压缩多个列表？EN

回答 4

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何使用java 8压缩多个列表？
EN