如何在Java 8中动态过滤嵌套列表对象

huangapple
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英文:

How to filter dynamically nested list object java 8

问题

如何在Java 8中动态筛选嵌套列表对象

示例：

  1. class Items {
  2.     List<Mobile> mobiles;
  3. }
  5. class Mobile {
  6.     String mName;
  7.     List<Plans> plans;
  8. }
  10. class Plans {
  11.     String planId;
  12.     String planName;
  13. }

所以，我有3个移动设备（移动设备将是动态的，3或4等等），每个移动设备上有多个计划。如何动态筛选每个移动设备的共同计划？

示例（P1-planId）：

  1. 项目：
  2.     M1 - P1,P2,P3,P4
  3.     M2 - P4,P5,P6,P1,P8,P2
  4.     M3 - P7,P2,P4,P1,P8,P9,P10

结果：

  1. 项目：
  2.     M1 - P1,P2,P4
  3.     M2 - P1,P2,P4
  4.     M3 - P1,P2,P4
英文:

How to filter dynamically nested list object java 8

Example:

  1. class Items {
  2.     List&lt;Mobile&gt; mobiles;
  3. }
  5. class Mobile{
  6.     String mName;
  7.     List&lt;Plans&gt; plans;
  8. }
  10. class Plans{
  11.     String planId;
  12.     String planName;
  13. }

So, I have 3 mobiles (mobiles will be dynamic 3 or 4..etc) with multiple plans on each mobile device. How to dynamically filter common plan for each mobile device ?

Example(P1-planId) :

  1. Items:
  2.     M1 - P1,P2,P3,P4
  3.     M2 - P4,P5,P6,P1,P8,P2
  4.     M3 - P7,P2,P4,P1,P8,P9,P10

Result:

  1. Items:
  2.     M1 - P1,P2,P4
  3.     M2 - P1,P2,P4
  4.     M3 - P1,P2,P4

答案1

得分: 2

以下是已翻译的代码部分：

  1. 获取所有移动设备共有的计划的`Items`内部方法可能如下所示：
  3. public List<Plan> getCommonPlans() {
  4.     return mobiles.stream().flatMap(Mobile::streamPlans).distinct()
  5.         .filter(p -> mobiles.stream().allMatch(m -> m.hasPlan(p)))
  6.         .collect(Collectors.toList());
  7. }
  9. 这假设了`Mobile.streamPlans`和`Mobile.hasPlan`方法相当简单。
  11. 稍微不同的方法，更高效但可能不太直观，是计算计划并过滤出计数等于移动设备数量的计划：
  13. return mobiles.stream().flatMap(Mobile::streamPlans)
  14.     .collect(Collectors.groupingBy(m -> m, Collectors.counting())
  15.     .entrySet().stream()
  16.     .filter(e -> e.getValue() == mobiles.size())
  17.     .map(Map.Entry::getKey)
  18.     .collect(Collectors.toList());
英文:

A method inside Items to get all plans common to all mobiles might look like:

  1. public List&lt;Plan&gt; getCommonPlans() {
  2.     return mobiles.stream().flatMap(Mobile::streamPlans).distinct()
  3.         .filter(p -&gt; mobiles.stream().allMatch(m -&gt; m.hasPlan(p)))
  4.         .collect(Collectors.toList());
  5. }

this assumes Mobile.streamPlans and Mobile.hasPlan methods which are pretty trivial.

A slightly different method, more efficient but perhaps not so intuitive, is to count the plans and filter for ones that have counts equal to number of mobiles:

  1.     return mobiles.stream().flatMap(Mobile::streamPlans)
  2.         .collect(Collectors.groupingBy(m -&gt; m, Collectors.counting())
  3.         .entrySet().stream()
  4.         .filter(e -&gt; e.getValue() == mobiles.size())
  5.         .map(Map.Entry::getKey)
  6.         .collect(Collectors.toList());

答案2

得分: 1

首先，获取第一个手机的计划，并且使用 retainAll 方法筛选出所有手机的共同计划。

  1. List<Plans> commonPlans = new ArrayList<>(mobiles.get(0).getPlans());
  2. for (int i = 1; i < mobiles.size(); i++) {
  3.   commonPlans.retainAll(mobiles.get(i).getPlans());
  4. }

注意： 确保你为 Plans 类重写了 equals 和 hashCode 方法，并且检查手机列表是否为空。

英文:

First, take plans of the first mobile and retainAll plans of mobiles from that list.

  1. List&lt;Plans&gt; commonPlans = new ArrayList&lt;&gt;(mobiles.get(0).getPlans());
  2. for (int i = 1; i &lt; mobiles.size(); i++) {
  3.   commonPlans.retainAll(mobiles.get(i).getPlans());
  4. }

Note: Make sure you override equals and hashCode for Plans and check for empty mobiles list

答案3

得分: 0

  • stream all Mobiles,
  • map each Mobile to its List&lt;Plan&gt;s
  • create the union of all plans.

In code, this might look something like this:

  1. HashSet&lt;Plan&gt; initialSet = new HashSet&lt;&gt;(mobiles.get(0).getPlans());
  2. return mobiles.stream()
  3.     .map(Mobile::getPlans)
  4.     .map(HashSet&lt;Plan&gt;::new)
  5.     .reduce(initialSet, (plan1, plan2) -&gt; { 
  6.         plan1.retainAll(plan2);
  7.         return plan1;
  8.     });

<kbd>Ideone demo</kbd>

英文:

A slightly different approach would be to:

  • stream all Mobiles,
  • map each Mobile to its List&lt;Plan&gt;s
  • create the union of all plans.

In code, this might look something like this:

  1. HashSet&lt;Plan&gt; initialSet = new HashSet&lt;&gt;(mobiles.get(0).getPlans());
  2. return mobiles.stream()
  3. 	.map(Mobile::getPlans)
  4. 	.map(HashSet&lt;Plan&gt;::new)
  5. 	.reduce(initialSet, (plan1, plan2) -&gt; { 
  6. 		plan1.retainAll(plan2);
  7. 		return plan1;
  8. 	});

<kbd>Ideone demo</kbd>

huangapple
  • 本文由 发表于 2020年8月8日 13:12:06
  • 转载请务必保留本文链接：https://go.coder-hub.com/63312068.html
  • java
  • java-8
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