如何在 anyMatch() java 8 流中应用复合 BiPredicate

Question

我有两个列表。一个是我从数据库创建的，一个是从 Csv 文件创建的。现在我想在一个包含数据库和 csv 文件的列表中收集记录。我写了如下代码

BiPredicate<Trainee, Trainee> sameTrainee = (dbTrainee, csvTrainee) -> {

    String dbTraineeFirstName = dbTrainee.getFirstName();
    String dbTraineeLastName = dbTrainee.getLastName();
    String dbTraineeEmail = dbTrainee.getEmail();
    LocalDateTime dbTraineeCompletionDate = dbTrainee.getSessionDateTime();
    String text = dbTraineeCompletionDate.format(DATE_TIME_FORMATTER); 
    LocalDateTime dbTraineeSessionDateTime = LocalDateTime.parse(text);
    String dbTraineePhoneNumber = dbTrainee.getPhoneNumber();
    String dbTraineeSsn = dbTrainee.getSocialSecurityLastFour();
    String dbTraineeStreetOne = dbTrainee.getStreetOne();
    String dbTraineeCity = dbTrainee.getCity();

    String csvTraineeFirstName = csvTrainee.getFirstName();
    String csvTraineeLastName = csvTrainee.getLastName();
    String csvTraineeEmail = csvTrainee.getEmail();
    LocalDateTime csvTraineeSessionDateTime = csvTrainee.getSessionDateTime();
    String csvTraineePhoneNumber = csvTrainee.getPhoneNumber();
    String csvTraineeSsn = csvTrainee.getSocialSecurityLastFour();
    String csvTraineeStreetOne = csvTrainee.getStreetOne();
    String csvTraineeCity = csvTrainee.getCity();

    int dbTraineeSsnLength = dbTraineeSsn.length();
    int csvTraineeSsnLength = csvTraineeSsn.length();

    if (dbTraineeSsnLength != csvTraineeSsnLength) {
        if (dbTraineeSsnLength == 4 && dbTraineeSsn.startsWith("0")) {
            String dbTraineeSsnLast3Digits = dbTraineeSsn.substring(dbTraineeSsn.length() - 3);
            if (csvTraineeSsnLength == 3 && csvTraineeSsn.endsWith(dbTraineeSsnLast3Digits)) {
                csvTraineeSsn = "0" + csvTraineeSsn;
            }
        }
    }

    return dbTraineeFirstName.equals(csvTraineeFirstName) 
            && dbTraineeLastName.equals(csvTraineeLastName)
            && dbTraineeEmail.equals(csvTraineeEmail) 
            && dbTraineeSessionDateTime.equals(csvTraineeSessionDateTime)
            && dbTraineePhoneNumber.equals(csvTraineePhoneNumber)
            && dbTraineeSsn.equals(csvTraineeSsn) 
            && dbTraineeStreetOne.equals(csvTraineeStreetOne)
            && dbTraineeCity.equals(csvTraineeCity);
};

并像这样称呼它

List<Trainee> foundInBothList = dbMonthlyTraineeList.stream()
                    .filter(dbTrainee -> csvTraineeList.stream()
                        .anyMatch(csvTrainee -> {
                            return sameTrainee.test(dbTrainee, csvTrainee);
                        })
                    ).collect(Collectors.toList());

List<Trainee> notInFileList = dbMonthlyTraineeList.stream()
                    .filter(dbTrainee -> csvTraineeList.stream()
                        .noneMatch(csvTrainee -> {
                            return sameTrainee.test(dbTrainee, csvTrainee);
                        })
                    ).collect(Collectors.toList());

它工作正常。但是由于我的 BiPredicate 变得冗长而不整洁。因此，我创建了一个类并收集了 Collection 中的所有谓词，如下所示

public class PlcbMonthlyReportStatisticsBiPredicates {

    public static BiPredicate<Trainee, Trainee> isValidFirstName() {
        return (dbTrainee, csvTrainee) -> {
            String dbTraineeFirstName = dbTrainee.getFirstName();
            String csvTraineeFirstName = csvTrainee.getFirstName();
            return dbTraineeFirstName.equals(csvTraineeFirstName);
        };
    }

    public static BiPredicate<Trainee, Trainee> isValidSsn() {
        return (dbTrainee, csvTrainee) -> {
            String dbTraineeSsn = dbTrainee.getSocialSecurityLastFour();
            String csvTraineeSsn = csvTrainee.getSocialSecurityLastFour();
            ...
            return dbTraineeSsn.equals(csvTraineeSsn);
        };
    }

    ....

    public static List<BiPredicate<Trainee, Trainee>> getAllBiPredicates() {

        List<BiPredicate<Trainee, Trainee>> allPredicates = Arrays.asList(
                isValidFirstName(),
                isValidSsn(),
                ... 
        );  
        return allPredicates;
    }
}

现在我收集了所有谓词，但是如何在我的 anyMatch() 和 noneMatch() 中应用这些谓词。我试过这个，但当然会出错

List<Trainee> foundInBothList1 = dbMonthlyTraineeList.stream()
    .filter(dbTrainee -> csvTraineeList.stream()
        .anyMatch(csvTrainee -> {
            List<BiPredicate<Trainee, Trainee>> allBiPredicates = getAllBiPredicates();
            return allBiPredicates.stream().reduce(BiPredicate::and).orElse((x,y)->true);  //error

        })
    ).collect(Collectors.toList());

我该如何应用它。我的做法对吗？

**编辑

---

@Entity
public class Trainee {

    private static final DateTimeFormatter DATE_TIME_FORMATTER = DateTimeFormatter.ofPattern("yyyy-MM-dd'T'HH:mm:ss");

    private LocalDateTime sessionDateTime;
    private String firstName;
    ....

    @Override
    public boolean equals(Object otherObject) {

        // Are the same?
        if (this == otherObject) {
            return true;
        }
        // Is otherObject a null reference?
        if (otherObject == null) {
            return false;
        }
        // Do they belong to the same class?
        if (this.getClass() != otherObject.getClass()) {
            return false;
        }

        // Get the reference of otherObject in a otherTrainee variable
        Trainee otherTrainee = (Trainee)otherObject;

        LocalDateTime dbTraineeCompletionDate = this.getSessionDateTime();
        String text = dbTraineeCompletionDate.format(DATE_TIME_FORMATTER); 
        LocalDateTime dbTraineeSessionDateTime = LocalDateTime.parse(text);

        String dbTraineeSsn = this.socialSecurityLastFour;
        String csvTraineeSsn = otherTrainee.getSocialSecurityLastFour();

        int dbTraineeSsnLength = dbTraineeSsn.length();
        int csvTraineeSsnLength = csvTraineeSsn.length();

        if (dbTraineeSsnLength != csvTraineeSsnLength) {
            if (dbTraineeSsnLength == 4 && dbTraineeSsn.startsWith("0")) {
                String dbTraineeSsnLast3Digits = dbTraineeSsn.substring(dbTraineeSsn.length() - 3);
                if (csvTraineeSsnLength == 3 && csvTraineeSsn.endsWith(dbTraineeSsnLast3Digits)) {
                    csvTraineeSsn = "0" + csvTraineeSsn;
                }
            }
        }

        boolean isEqual = (this.firstName.equals(otherTrainee.firstName)
            && this.lastName.equals(otherTrainee.lastName)
            && this.email.equals(otherTrainee.email) 
            && dbTraineeSessionDateTime.equals(otherTrainee.sessionDateTime)
            && this.phoneNumber.equals(otherTrainee.phoneNumber)
            && dbTraineeSsn.equals(csvTraineeSsn) 
            && this.streetOne.equals(otherTrainee.streetOne)
            && this.city.equals(otherTrainee.city)
        );

        return isEqual;
    }

    @Override
    public int hashCode() {

        int hash = 37;
        int code = 0;

        code = (firstName == null ? 0 : firstName.hashCode());
        hash = hash * 59 + code;

        code = (lastName == null ? 0 : lastName.hashCode());
        hash = hash * 59 + code;

        code = (email == null ? 0 : email.hashCode());
        hash = hash * 59 + code;

        code = (sessionDateTime == null ? 0 : sessionDateTime.hashCode());
        hash = hash * 59 + code;

        code = (phoneNumber == null ? 0 : phoneNumber.hashCode());
        hash = hash * 59 + code;

        code = (socialSecurityLastFour == null ? 0 : socialSecurityLastFour.hashCode());
        hash = hash * 59 + code;

        code = (streetOne == null ? 0 : streetOne.hashCode());
        hash = hash * 59 + code;

        code = (city == null ? 0 : city.hashCode());
        hash = hash * 59 + code;

        return hash;

    }
}

编辑 2（覆盖 hascode() 和 equals() 之后） --------------------------------------------------

两者都有：

List<Trainee> foundInBothList1 = dbMonthlyTraineeList.stream()
    .filter(dbTrainee -> csvTraineeList.stream()
        .anyMatch(csvTrainee -> {
            return allBiPredicates.stream().reduce(BiPredicate::and).orElse((x,y)->true).test(dbTrainee, csvTrainee);

        })
    ).collect(Collectors.toList());

    List<Trainee> foundInBothList = new ArrayList<>(dbMonthlyTraineeList);
    //foundInBothList.retainAll(new HashSet<>(csvTraineeList));
    foundInBothList.retainAll(csvTraineeList);

在数据库中找到但在 CSV 中没有

List<Trainee> notInCsvFileList1 = dbMonthlyTraineeList.stream()
    .filter(dbTrainee -> csvTraineeList.stream()
        .noneMatch(csvTrainee -> {
            return allBiPredicates.stream().reduce(BiPredicate::and).orElse((x,y)->true).test(dbTrainee, csvTrainee);
        })
    ).collect(Collectors.toList());

//find out that elements of dbMonthlyTraineeList which is not present in arraylist(csvTraineeList).
List<Trainee> notInCsvFileList = new ArrayList<>(dbMonthlyTraineeList);
notInCsvFileList.removeAll(csvTraineeList);

Answer 1

看来你想多了。为什么不用sameTrainee 双谓词代码覆盖equals？ （也不要忘记覆盖hashCode）。

完成此操作后，您可以使用以下命令保留两个列表中的 Trainees：

Set<Trainee> foundInBothList = new HashSet<>(dbMonthlyTraineeList);
foundInBothList.retainAll(new HashSet<>(csvTraineeList));

这个解决方案是 O(n)，因此它的性能会比你的解决方案 O(n²) 好很多。这是因为contains 操作在Set 上是恒定时间的。

---

但如果你真的想让你的代码编译，你只需要在你有错误的地方调用test方法：

return allBiPredicates.stream().reduce(BiPredicate::and)
                               .orElse((x,y)->true)
                               .test(dbTrainee, csvTrainee);