Question

我有一个名单需要转换为具有相同姓氏的人都被标记的结果。例如：

原产地清单：

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
    let cell = tableView.dequeueReusableCellWithIdentifier(reuseIdentifier, forIndexPath: indexPath) as! TalesTableViewCell let tapGesture = UITapGestureRecognizer(target: self, action: "imageTapped:")

    cell.tableImage.addGestureRecognizer(tapGesture)

    cell.tableImage.userInteractionEnabled = true

    return cell
}

func imageTapped(gesture: UIGestureRecognizer) {

    //Code here I think

}

输出：

JayReese
ClaraSmith
JohnSmith

JayReese ClaraSmith1 JohnSmith2类的代码如下所示，如何比较所有Person，何时重复lastName，每个都添加唯一索引？应该添加什么方法？

我非常感谢任何输入和/或帮助。非常感谢你。

lastName

Answer 1

你可以做这样的比较方法

public compairLastName(String LastName){
if(lastName.equals(LastName){
System.out.println("These last names are the same"); 
}else{
System.out.println("These last names are not the same");
}

Answer 2

所以你问过标签（以及扩展排序），但是参考唯一标识 - 这里是两者的答案：）

排序

现实生活中的程序每天都在处理，只需确保您的可比代码默认按2个属性排序

名字
姓

如果他们按照这个顺序进行比较，你应该最终得到：

XavierAllen
JayReese
BillySmith
ClaraSmith
JohnSmith

为了比较多个属性，我将引用您的stackoverflow主题，该主题显示如何对具有相同字段名称的Person对象进行单次和多次比较; ）

How to compare objects by multiple fields

独特参与

此外，如果您担心在简单比较排序之外唯一地识别Person，那么您将添加一个int字段（或guid，或任何你的风格），它将具有唯一值

基本上与数据库PK相同 - 您永远不会在数据库中使用人名作为PK，因此理想情况下，您的Person应该具有类似的属性。

如果你想将这个PK添加到你的toString（），那么就去吧

import java.util.*;

public class Person implements Comparable<Person> {

    private int personID;
    private String firstName;
    private String lastName;

    public Person(String firstName, String lastName, int personID) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.personID = personID;
    }

    public int getID(){
        return this.personID;
    }

    public String toString() {
        return this.personID + ": " + this.lastName + ", " + this.firstName;
    }
}

@Mshnik确实发布了一个动态添加PK的方法，但你最好检查一下Person集合或一些高级变量来找到最后一个personID并从那里开始 - 否则动态值只能用于最多有限的上下文，你也可以只使用它从你正在从中检索它的Person集合中的索引

Answer 3

首先，编辑Person类或创建一个具有可设置索引字段的新类。您的用例完全没有可比性。

public class Person {

    private String firstName;
    private String lastName;

    public int index;

    public Person(String firstName, String lastName) {
        this.firstName = firstName;
        this.lastName = lastName;
    }

    public String toString() {
        return lastName + firstName + (index != 0 ? index : "");
    }
}

您可以使用HashMap<String, Integer>将姓名添加到姓氏中，如下所示：

public static void numberDuplicateLastNames(List<Person> people) {
    HashMap<String, Integer> duplicatedLastNames = new HashMap<>();
    for(Person p : people) {
        if(! duplicatedLastNames.containsKey(p.lastName)) {
            duplicatedLastNames.put(p.lastName, 1);
            p.index = 1;
        } else {
            int i = duplicatedLastNames.get(p.lastName) + 1;
            duplicatedLastNames.put(p.lastName, i);
            p.index = i;
        }
    }

    //Remove index from the people who don't have a duplicate last name
    for(Person p : people) {
        if (duplicatedLastNames.get(p.lastName) == 1) {
            p.index = 0;
        }
    }
}

Answer 4

以蛮力的方式标记Person对象并不依赖于使用Map来计算姓氏出现次数。

我还包括另一种利用Java 8流功能的方法。 IMO，蛮力方法更容易理解，但我还没有做过任何基准测试，知道哪个更有效。所以我会留下评论。

public static void main(String[] args) throws Exception {
    List<Person> persons = new ArrayList() {
        {
            add(new Person("Jay", "Reese"));
            add(new Person("Clara", "Smith"));
            add(new Person("John", "Smith"));
            add(new Person("James", "Smith"));
            add(new Person("Christie", "Mayberry"));
            add(new Person("Matthew", "Mayberry"));
        }
    };

    // Toggle calls to see results
    bruteForceLabel(persons);
    // java8StreamLabel(persons);
}

public static void bruteForceLabel(List<Person> persons) {
    for (int i = 0; i < persons.size(); i++) {
        Person currentPerson = persons.get(i);
        char lastCharacter = currentPerson.lastName.charAt(currentPerson.lastName.length() - 1);

        // Only process last names that are not labeled
        if (lastCharacter < '0' || '9' < lastCharacter) { // Not a digit
            int counter = 2;
            boolean foundDuplicateLastName = false;

            for (int j = i + 1; j < persons.size(); j++) {
                Person nextPerson = persons.get(j);
                if (nextPerson.lastName.equals(currentPerson.lastName)) {
                    foundDuplicateLastName = true;

                    // Label the next person with the counter then 
                    nextPerson.lastName = nextPerson.lastName + counter;
                    counter++;
                }
            }

            // Label the current person with the starting sequence
            if (foundDuplicateLastName) {
                currentPerson.lastName = currentPerson.lastName + 1;
            }
        }
    }

    System.out.println(persons);
}

public static void java8StreamLabel(List<Person> persons) {
    // Get a distinct count of all last names
    Map<String, Long> lastNames = persons
            .stream()
            .map(p -> p.lastName)
            .distinct()
            .collect(Collectors
                    .toMap(p -> p, p -> persons
                            .stream()
                            .filter(p2 -> p2.lastName.equals(p)).count()));

    // Apply number sequence to duplicate last names
    lastNames.keySet().stream().filter((key) -> (lastNames.get(key) > 1)).forEach((key) -> {
        int counter = 1;
        for (Person person : persons.stream().filter(p -> p.lastName.equals(key)).toArray(size -> new Person[size])) {
            person.lastName = person.lastName + counter;
            counter++;
        }
    });

    // Display the modified list
    System.out.println(persons);
}

public static class Person {

    private String firstName;
    private String lastName;

    public Person(String firstName, String lastName) {
        this.firstName = firstName;
        this.lastName = lastName;
    }

    public String toString() {
        return firstName + " " + lastName;
    }
}

结果：

[Jay Reese, Clara Smith1, John Smith2, James Smith3, Christie Mayberry1, Matthew Mayberry2]

如何标记重复的元素？

4 个答案: