我有一个List<String>,其中包含一些名称,我想将该列表转换为其他唯一名称列表,我发现一种有效的方式是:

public class Main {

  public Main() {
    List<String> nameList = new ArrayList<>();

    nameList.add("Ali");
    nameList.add("Al");
    nameList.add("Ali");
    nameList.add("Al");
    nameList.add("Ai");
    nameList.add("li");
    nameList.add("li");
    nameList.add("Ai");
    System.out.print("All names are: ");
    for (String string : nameList) {
      System.out.print(string + ", ");
    }
    System.out.println("");

    System.out.print("Unique names are: ");

    for (String string : convertToUniqueList(nameList)) {
      System.out.print(string + ", ");
    }

  }

  public List<String> convertToUniqueList(List<String> listInts) {
    List<String> listDistinctInts = new ArrayList<>(listInts.size());
    for (String i : listInts) {
      if (!listDistinctInts.contains(i)) {
        listDistinctInts.add(i);
      }
    }
    return listDistinctInts;
  }

  public static void main(String[] args) {
    new Main();
  }

}


一切工作都很好,并且我得到了预期的结果,但是我想对下面的场景实现与上面相同的逻辑:

  @Data // lombok annotation
  public class Foo {
    String name;
    int age;
    List<Bar> barList = new ArrayList<>();
  }

  @Data // lombok annotation
  public class Bar {
    String bloodGroup;
    String dayOfPassing;
  }


我想从尝试的Main.java中的杂项列表中获取具有唯一日期名称的列表:

public class Main {

  public Main() {
    Foo foo1 = new Foo();

    foo1.setName("Foo One");
    foo1.setAge(21);

    List<Bar> barList = new ArrayList<>();

    Bar bar1 = new Bar();
    bar1.setBloodGroup("O +ve");
    bar1.setDayOfPassing("Wednesday");

    Bar bar2 = new Bar();
    bar2.setBloodGroup("O +ve");
    bar2.setDayOfPassing("Wednesday");

    Bar bar3 = new Bar();
    bar3.setBloodGroup("O +ve");
    bar3.setDayOfPassing("Thursday");

    Bar bar4 = new Bar();
    bar4.setBloodGroup("O -ve");
    bar4.setDayOfPassing("Friday");

    barList.add(bar1);
    barList.add(bar2);
    barList.add(bar3);
    barList.add(bar4);

    System.out.print("All Bars with all days are: ");
    for (Bar bar : barList) {
      System.out.print(bar.dayOfPassing + ", ");
    }
    System.out.println("");

    System.out.print("Bars with Unique days are: ");
    int count = 0;
    for (Bar bar : getBarsWithUniqueDays(barList, count)) {
      System.out.print(bar.dayOfPassing + ", ");
      count++;
    }

  }

  public List<Bar> getBarsWithUniqueDays(List<Bar> barList, int count) {
    List<Bar> listDistinctBars = new ArrayList<>(barList.size());

    for (Bar bar : barList) {
      if (!listDistinctBars.get(count).getDayOfPassing().contains(bar.dayOfPassing)) {
        listDistinctBars.add(bar);
      }
      count++;
    }
    return listDistinctBars;
  }

  public static void main(String[] args) {
    new Main();
  }
}


与上面的代码,我得到异常:

All Bars with all days are: Wednesday, Wednesday, Thursday, Friday,
Bars with Unique days are: Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 0, Size: 0
    at java.util.ArrayList.rangeCheck(Unknown Source)
    at java.util.ArrayList.get(Unknown Source)
    at Main.getBarsWithUniqueDays(Main.java:124)
    at Main.<init>(Main.java:113)
    at Main.main(Main.java:133)


我已经搜索并尝试了很多,但徒劳无功地解决了这个问题。

最佳答案

    Map<String, Bar> map = new LinkedHashMap<>();
    for (Bar ays : barList) {
        map.put(ays.dayOfPassing, ays);
    }
    barList.clear();
    barList.addAll(map.values());


    for (Bar ays : barList) {

        System.out.println("Unique names are: "+ays.dayOfPassing);
    }

关于java - 如何将重复值列表转换为相对于T类字段的不同值列表,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/58975032/

10-12 00:10
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