Java List 根据整数排序，带有多个下划线的情况

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英文:

Java List String sort Based on Integer with multiple underscore's

问题

你好，你可以使用Java 8中的lambda表达式和Comparator来实现这个排序逻辑。以下是你的代码片段的排序实现：

  1. keyValues.sort(Comparator.comparing(s -> Arrays.asList(s.split("_"))
  2.         .stream()
  3.         .map(Integer::parseInt)
  4.         .collect(Collectors.toList())));

这将按照你描述的逻辑对keyValues进行排序。

英文:

Hi All I have arrayList of String datatype but needs to sort like integer with multiple underscore's using Java 8 only ,For your reference I have added sample data, the original data came Liferay 7.1 DDL

  1. List<String> keyValues = new ArrayList<String>();
  2. keyValues.add("100_10_20_1");
  3. keyValues.add("100001");
  4. keyValues.add("100002");
  5. keyValues.add("100002_1");
  6. keyValues.add("100003");
  7. keyValues.add("100");
  8. keyValues.add("100_1");
  9. keyValues.add("100_2");
  10. keyValues.add("100_1_1");
  11. keyValues.add("100_10_20");
  12. keyValues.add("10000001");
  13. keyValues.add("100_10_20_2");

i tried to replace the underscore with 0 but what happens sometimes 100_1 replace _ with 0 then it becomes 10001 but what if i already have 10001 value present in list ? even same thing for empty if i replace _ with "" then it becomes 1001 but what if i already have 1001 then again i am facing issue

Exact logic based on underscore was

  1.  100
  2.  100_1
  3.  100_1_2 
  4.  /after that only/
  5.  100_2
  6.  100_2_2
  7. /after that/
  8.  103
  9.  104
  10.  105
  11.  10000
  12. /similarly the pattern repeats/
  13.  10000_1
  14.  10000_2
  15.  10000_2_1
  16.  100002

i need the exact output as

  1.     100
  2.     100_1
  3.     100_1_1
  4.     100_2
  5.     100_10_20
  6.     100_10_20_1
  7.     100_10_20_2
  8.     100001
  9.     100002
  10.     100002_1
  11.     100003
  12.     10000001

i tried with replacing underscore with 0 or empty space but there is possiblity of if it had already that number present in list,if i replace with dot and storing as Double then multiple pointer exception will occur

答案1

得分: 4

我创建了一个比较器，然后根据每个字符串具有的最小标记数量进行分割和处理。

  1. static Comparator<String> stringComparator = new Comparator<String>() {
  2.     @Override
  3.     public int compare(String o1, String o2) {
  4.         String[] s1 = o1.split("_");
  5.         String[] s2 = o2.split("_");
  7.         int minLen = Math.min(s1.length, s2.length);
  9.         for (int i=0; i < minLen; i++){
  10.             if (!s1[i].equals(s2[i])){
  11.                 return Integer.parseInt(s1[i]) - Integer.parseInt(s2[i]);
  12.             }
  13.         }
  15.         return o1.compareTo(o2);
  16.     }
  17. };

调用方式如下：

  1. List<String> keyValues = new ArrayList<String>();
  2. keyValues.add("100001");
  3. keyValues.add("100002");
  4. keyValues.add("100002_1");
  5. keyValues.add("100003");
  6. keyValues.add("100");
  7. keyValues.add("100_1");
  8. keyValues.add("100_2");
  9. keyValues.add("100_1_1");
  10. keyValues.add("100_10_20");
  11. keyValues.add("10000001");
  12. keyValues.stream().sorted(stringComparator).forEach(System.out::println);

输出结果：

  1. 100
  2. 100_1
  3. 100_1_1
  4. 100_2
  5. 100_10_20
  6. 100001
  7. 100002
  8. 100002_1
  9. 100003
  10. 10000001
英文:

I made a comparator then split the string and process based on the min amount of tokens each string has.

  1. static Comparator&lt;String&gt; stringComparator = new Comparator&lt;String&gt;() {
  2.         @Override
  3.         public int compare(String o1, String o2) {
  4.             String[] s1 = o1.split(&quot;_&quot;);
  5.             String[] s2 = o2.split(&quot;_&quot;);
  7.             int minLen = Math.min(s1.length, s2.length);
  9.             for (int i=0; i &lt; minLen; i++){
  10.                 if (!s1[i].equals(s2[i])){
  11.                     return Integer.parseInt(s1[i]) - Integer.parseInt(s2[i]);
  12. //                    return s1[i].compareTo(s2[i]);  //not natural order/sort
  13.                 }
  14.             }
  16.             return o1.compareTo(o2);
  17.         }
  18.     };

Calling it like so:

  1.     List&lt;String&gt; keyValues = new ArrayList&lt;String&gt;();
  2.     keyValues.add(&quot;100001&quot;);
  3.     keyValues.add(&quot;100002&quot;);
  4.     keyValues.add(&quot;100002_1&quot;);
  5.     keyValues.add(&quot;100003&quot;);
  6.     keyValues.add(&quot;100&quot;);
  7.     keyValues.add(&quot;100_1&quot;);
  8.     keyValues.add(&quot;100_2&quot;);
  9.     keyValues.add(&quot;100_1_1&quot;);
  10.     keyValues.add(&quot;100_10_20&quot;);
  11.     keyValues.add(&quot;10000001&quot;);
  12.     keyValues.stream().sorted(stringComparator).forEach(System.out::println);

Gives output:

  1. 100
  2. 100_1
  3. 100_1_1
  4. 100_2
  5. 100_10_20
  6. 100001
  7. 100002
  8. 100002_1
  9. 100003
  10. 10000001

huangapple
  • 本文由 发表于 2023年5月10日 17:57:42
  • 转载请务必保留本文链接：https://go.coder-hub.com/76217086.html
  • arraylist
  • collections
  • java
  • java-8
  • sorting
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