Java：按多个字段过滤收集和检索数据

Question

我有一堂课：

public class Address {
    private String country;
    private String state;
    private String city;
}

还有一个 Person 对象列表。 Person 类看起来像：

public class Person {
    private String country;
    private String state;
    private String city;
    //other fields
}

我需要过滤Person 对象并获得最合适的对象。 Address 对象至少可以有一个非空字段。 Person 对象可以没有、部分或全部初始化提到的字段。

这是可能的输入示例之一：

Three Person objects:
a. PersonA: country = 'A'
b. PersonB: country = 'A', state = 'B'
c. PersonC: country = 'A', state = 'B', city = 'C'

Address object:
a. Address: country = 'A', state = 'B'

过滤后的预期结果是PersonB。如果只有 PersonA 和 PersonC 对象，那么 PersonA 更可取。

我想展示我是如何尝试这样做的，但实际上它是纯粹的蛮力算法，我不喜欢它。算法复杂度随着新添加的字段而增加。我也想过通过谓词使用番石榴过滤器，但不知道谓词应该是什么。

如果除了蛮力之外，还有什么更好的过滤算法？

Answer 1

据我了解，暴力破解是指检查所有实体的所有字段。好吧，如果你不重构你的类，那是不可能的，但是有一个简单的技巧会有所帮助。它使用state 模式。

您可以在两个类中添加标志notNulls：

public class Address {
    private int notNulls = 0;
    private String country;
    private String state;
    private String city;
}

public class Person {
    private int notNulls = 0;
    private String country;
    private String state;
    private String city;
    //other fields
}

我将向您展示 one setter 的可能实现，因为其余部分类似：

public void setCountry(String s) {
    if (country == null {
        if (s != null) {
            country = s;
            notNulls++;
        }
    } else {
        if (s == null) {
            country == null;
            notNulls--;
        } else {
            country = s;
        }
    }
}

public boolean isValid() {
    return notNulls != 0;
}

现在您可以简单地遍历对象。

Answer 2

为避免暴力破解，您需要按地址为您的人员编制索引。为了进行良好的搜索，您肯定需要一个国家/地区（以某种方式猜测或默认它，否则结果无论如何都会太不准确）。

索引是一个数字，前 3 位代表国家，后 3 位代表州，后 4 位代表城市。在这种情况下，您可以在 int 中存储 213 个国家/地区 (only 206 as of 2016)，最多可存储 999 个州和 9999 个城市。

它使我们能够使用 hashCode 和 TreeSet 来索引您的 Person 实例，并以 O(log(n)) 的方式按地址部分查找它们，而无需触及它们的字段。字段会在 TreeSet 构造中被触及，并且您需要添加一些额外的逻辑来修改 Person 以保持索引不变。

索引是按每个部分顺序计算的，从国家开始

    import java.util.HashMap;
    import java.util.Map;

    public class PartialAddressSearch {

        private final static Map<String, AddressPartHolder> COUNTRY_MAP = new HashMap<>(200);

        private static class AddressPartHolder {
            int id;
            Map<String, AddressPartHolder> subPartMap;

            public AddressPartHolder(int id, Map<String, AddressPartHolder> subPartMap) {
                this.id = id;
                this.subPartMap = subPartMap;
            }
        }

        public static int getCountryStateCityHashCode(String country, String state, String city) {
            if (country != null && country.length() != 0) {
                int result = 0;
                AddressPartHolder countryHolder = COUNTRY_MAP.get(country);
                if (countryHolder == null) {
                    countryHolder = new AddressPartHolder(COUNTRY_MAP.size() + 1, new HashMap<>());
                    COUNTRY_MAP.put(country, countryHolder);
                }
                result += countryHolder.id * 10000000;

                if (state != null) {
                    AddressPartHolder stateHolder = countryHolder.subPartMap.get(state);
                    if (stateHolder == null) {
                        stateHolder = new AddressPartHolder(countryHolder.subPartMap.size() + 1, new HashMap<>());
                        countryHolder.subPartMap.put(state, stateHolder);
                    }
                    result += stateHolder.id * 10000;

                    if (city != null && city.length() != 0) {
                        AddressPartHolder cityHolder = stateHolder.subPartMap.get(city);
                        if (cityHolder == null) {
                            cityHolder = new AddressPartHolder(stateHolder.subPartMap.size() + 1, null);
                            stateHolder.subPartMap.put(city, cityHolder);
                        }
                        result += cityHolder.id;
                    }
                }

                return result;
            } else {
                throw new IllegalArgumentException("Non-empty country is expected");
            }
    }

对于您的 Person 和 Address 类，您可以根据 int 的自然顺序定义 hashCode 和 compareTo：

    public class Person implements Comparable {
        private String country;
        private String state;
        private String city;

        @Override
        public boolean equals(Object o) {
             //it's important but I removed it for readability
        }

        @Override
        public int hashCode() {
            return getCountryStateCityHashCode(country, state, city);
        }

        @Override
        public int compareTo(Object o) {
            //could be further improved by storing hashcode in a field to avoid re-calculation on sorting
            return hashCode() - o.hashCode();
        }

    }

    public class Address implements Comparable {
        private String country;
        private String state;
        private String city;


        @Override
        public boolean equals(Object o) {
             //removed for readability
        }

        @Override
        public int hashCode() {
            return getCountryStateCityHashCode(country, state, city);
        }

        @Override
        public int compareTo(Object o) {
            //could be further improved by storing hashcode in a field to avoid re-calculation on sorting
            return hashCode() - o.hashCode();
        }

    }

    public class AddressPersonAdapter extends Person {
        private final Address delegate;

        public AddressPersonAdapter(Address delegate) {
            this.delegate = delegate;
        }

        @Override
        public boolean equals(Object o) {
            return delegate.equals(o);
        }

        @Override
        public int hashCode() {
            return delegate.hashCode();
        }
    }

之后，您的过滤代码将缩小为填充索引并为您的部分地址计算下限：

    TreeSet<Person> personSetByAddress = new TreeSet<>();
    Person personA = new Person();
    personA.setCountry("A");
    personSetByAddress.add(personA);
    Person personB = new Person();
    personB.setCountry("A");
    personB.setState("B");
    personSetByAddress.add(personB);
    Person personC = new Person();
    personC.setCountry("A");
    personC.setState("B");
    personC.setCity("C");
    personSetByAddress.add(personC);

    Address addressAB = new Address();
    addressAB.setCountry("A");
    addressAB.setState("B");

    System.out.println(personSetByAddress.floor(new AddressPersonAdapter(addressAB)));

    Yields:
    Person{hashCode=10010000, country='A', state='B', city='null'}

如果你没有 PersonB：

    TreeSet<Person> personSetByAddress = new TreeSet<>();
    Person personA = new Person();
    personA.setCountry("A");
    personSetByAddress.add(personA);
    Person personC = new Person();
    personC.setCountry("A");
    personC.setState("B");
    personC.setCity("C");
    personSetByAddress.add(personC);

    Address addressAB = new Address();
    addressAB.setCountry("A");
    addressAB.setState("B");

    System.out.println(personSetByAddress.floor(new AddressPersonAdapter(addressAB)));

    Yields:
    Person{hashCode=10000000, country='A', state='null', city='null'}

编辑：

需要额外验证的极端情况是在同一国家/地区内没有更大（如果我们需要上限，则更小）元素。例如：

    TreeSet<Person> personSetByAddress = new TreeSet<>();
    Person personA = new Person();
    personA.setCountry("D");
    personSetByAddress.add(personA);
    Person personC = new Person();
    personC.setCountry("A");
    personC.setState("B");
    personC.setCity("C");
    personSetByAddress.add(personC);

    Address addressAB = new Address();
    addressAB.setCountry("A");
    addressAB.setState("B");

    System.out.println(personSetByAddress.floor(new AddressPersonAdapter(addressAB)));

    Yields:
    Person{hashCode=10000000, country='D', state='null', city='null'}

即我们掉到最近的国家。要解决此问题，我们需要检查国家/地区数字是​​否仍然相同。我们可以通过对 TreeSet 进行子类化并在其中添加此检查来做到这一点：

 //we need this class to allow flooring just by id
 public class IntegerPersonAdapter extends Person {
    private Integer id;
    public IntegerPersonAdapter(Integer id) {
        this.id = id;
    }

    @Override
    public boolean equals(Object o) {
        return id.equals(o);
    }

    @Override
    public int hashCode() {
        return id.hashCode();
    }

    @Override
    public int compareTo(Object o) {
        return id.hashCode() - o.hashCode();
    }

    @Override
    public String toString() {
        return id.toString();
    }
}

public class StrictCountryTreeSet extends TreeSet<Person> {

    @Override
    public Person floor(Person e) {
        Person candidate = super.floor(e);
        if (candidate != null) {
            //we check if the country is the same
            int candidateCode = candidate.hashCode();
            int eCode = e.hashCode();
            if (candidateCode == eCode) {
                return candidate;
            } else {
                int countryCandidate = candidateCode / 10000000;
                if (countryCandidate == (eCode / 10000000)) {
                    //we check if the state is the same
                    int stateCandidate = candidateCode / 10000;
                    if (stateCandidate == (eCode / 10000)) {
                        //we check if is a state
                        if (candidateCode % 10 == 0) {
                            return candidate;
                        } else { //since it's not exact match we haven't found a city - we need to get someone just from state
                            return this.floor(new IntegerPersonAdapter(stateCandidate * 10000));
                        }

                    } else if (stateCandidate % 10 == 0) { //we check if it's a country already
                        return candidate;
                    } else {
                        return this.floor(new IntegerPersonAdapter(countryCandidate * 10000000));
                    }
                }
            }
        }
        return null;
    }

现在我们的测试用例将在初始化StrictCountryTreeSet 后产生null：

    TreeSet<Person> personSetByAddress = new StrictCountryTreeSet();
    Person personA = new Person();
    personA.setCountry("D");
    personSetByAddress.add(personA);
    Person personC = new Person();
    personC.setCountry("A");
    personC.setState("B");
    personC.setCity("C");
    personSetByAddress.add(personC);

    Address addressAB = new Address();
    addressAB.setCountry("A");
    addressAB.setState("B");

    System.out.println(personSetByAddress.floor(new AddressPersonAdapter(addressAB)));

    Yields:
    null

对不同状态的测试也会产生null：

    TreeSet<Person> personSetByAddress = new StrictCountryTreeSet();
    Person personD = new Person();
    personD.setCountry("D");
    personSetByAddress.add(personD);

    Person personE = new Person();
    personE.setCountry("A");
    personE.setState("E");
    personSetByAddress.add(personE);

    Person personC = new Person();
    personC.setCountry("A");
    personC.setState("B");
    personC.setCity("C");
    personSetByAddress.add(personC);

    Address addressA = new Address();
    addressA.setCountry("A");

    Address addressAB = new Address();
    addressAB.setCountry("A");
    addressAB.setState("B");

    Address addressABC = new Address();
    addressABC.setCountry("A");
    addressABC.setState("B");
    addressABC.setCity("C");

    System.out.println(personSetByAddress.floor(new AddressPersonAdapter(addressAB)));

    Yields:
    null

请注意，在这种情况下，您需要将 hashCode 结果存储在 Address 和 Person 类中以避免重新计算。