如何在Java中找到多个集合的所有交集的列表？
How to find the list of all intersections of multiple sets in Java?
我有一个集合列表：
setlist = [s1,s2,s3...sn]
我想要一个集合的全方位比较，即 2^n 个集合：
setIntersection = [s1 ∩ s2, s1 ∩ s2 ∩ s3, ....., s2 ∩ s3 ∩ s4, ...., sn-1 ∩ sn]
在 Java 中执行此操作的最佳方法是什么？
例如，如果我只使用 5 套。我希望能够填充 5 circle venn diagram.
的所有重叠
我正在尝试使用集合列表来执行此操作：
List<Set<People>> lListSets = new ArrayList<Set<People>>();
for (DaysObject day : listOfDaysInJanuary) {
lListSets.add(day.peopleOneInternet());
}
findPowerSetsAndCompare(lListSets, listOfDaysInJanuary);
我想找到类似这样的结果：
January 1 (Bob, Sally, Tommy)
January 1, January 2 (Sally, Tommy)
...
so on for all possible combination of days.
基本上我要问的是如何应用 powerset algorithm 与集合并集的组合。
What would be the best way to do this in Java?
您描述的第一部分是 powerset（因为我上周编辑了您的问题以包含在内）。然后，您将获得幂集中每组集合的交集。
因为您正在做集合的幂集，而不是像整数这样的简单幂集，所以实现会涉及更多一些。
额外学分
我写了一个基本的实现你的要求，作为你如何做的一个例子。此示例中的所有方法和类型都是 Example class.
的成员
Example class 仅使用其 main 方法来演示工作代码。我相信您会原谅我在演示中使用已弃用的 Date 构造函数。
import java.text.*;
import java.util.*;
public class Example
{
public static void main(String[] args) {
// create simple test harness
Set<PeopleByDays> peepsByDay = new HashSet<PeopleByDays>();
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 1),
Person.BOB, Person.FRANK, Person.JIM, Person.JUDY, Person.SALLY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 2),
Person.BOB, Person.FRANK, Person.JIM, Person.JUDY, Person.TOMMY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 3),
Person.BOB, Person.FRANK, Person.JIM, Person.SALLY, Person.TOMMY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 4),
Person.BOB, Person.FRANK, Person.JUDY, Person.SALLY, Person.TOMMY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 5),
Person.BOB, Person.JIM, Person.JUDY, Person.SALLY, Person.TOMMY));
// make powerSet, then intersect, then sort
Set<Set<PeopleByDays>> powerPeeps = powerSet(peepsByDay);
List<PeopleByDays> powerPeepsIntersected = intersect(powerPeeps);
sort(powerPeepsIntersected);
// print out results
for (PeopleByDays peeps: powerPeepsIntersected) {
String daysFormatted = format(peeps.getDays());
System.out.print(daysFormatted);
System.out.println(peeps);
}
}
// all other Example members as defined in this answer
}
Person 是人名的简单 enum 类型。在这里使用枚举的好处是它会为所需的 HashSet 行为处理适当的 equals() 和 hashCode() 实现。
static enum Person {
BOB, FRANK, JIM, JUDY, SALLY, TOMMY;
}
PeopleByDays 扩展了 HashSet<Person> 以收集一组额外的 Date 对象来表示日期。覆盖 retainAll()（相交）以合并日期；覆盖 equals() 和 hashSet() 以获得外部集中的正确行为。
static class PeopleByDays extends HashSet<Person> {
private final Set<Date> days = new HashSet<Date>();
public PeopleByDays() {
super();
}
public PeopleByDays(Date day, Person... people) {
super(Arrays.asList(people));
this.days.add(day);
}
public PeopleByDays(PeopleByDays other) {
super(other);
this.days.addAll(other.days);
}
public List<Date> getDays() {
return new ArrayList<Date>(this.days);
}
@Override
public boolean retainAll(Collection<?> c) {
if (c instanceof PeopleByDays) {
this.days.addAll(((PeopleByDays)c).days);
}
return super.retainAll(c);
}
@Override
public boolean equals(Object o) {
return super.equals(o) && this.days.equals(((PeopleByDays) o).days);
}
@Override
public int hashCode() {
return super.hashCode() + this.days.hashCode();
}
}
powerSet() 方法，逐字取自 this answer.
public static <T> Set<Set<T>> powerSet(Set<T> originalSet) {
Set<Set<T>> sets = new HashSet<Set<T>>();
if (originalSet.isEmpty()) {
sets.add(new HashSet<T>());
return sets;
}
List<T> list = new ArrayList<T>(originalSet);
T head = list.get(0);
Set<T> rest = new HashSet<T>(list.subList(1, list.size()));
for (Set<T> set: powerSet(rest)) {
Set<T> newSet = new HashSet<T>();
newSet.add(head);
newSet.addAll(set);
sets.add(newSet);
sets.add(set);
}
return sets;
}
intersect() 方法为幂集中的每组集合创建交集。
static List<PeopleByDays> intersect(Set<Set<PeopleByDays>> powerSet) {
List<PeopleByDays> intersected = new ArrayList<PeopleByDays>();
for (Set<PeopleByDays> powerElement: powerSet) {
PeopleByDays intersection = null;
if (powerElement.isEmpty()) {
intersection = new PeopleByDays();
} else for (PeopleByDays peeps: powerElement) {
if (intersection == null) {
intersection = new PeopleByDays(peeps);
} else {
intersection.retainAll(peeps);
}
}
intersected.add(intersection);
}
return intersected;
}
sort() 方法按日期对结果相交集进行排序。
static void sort(List<PeopleByDays> peeps) {
Collections.sort(peeps, new Comparator<PeopleByDays>() {
@Override
public int compare(PeopleByDays p1, PeopleByDays p2) {
List<Date> days1 = p1.getDays();
List<Date> days2 = p2.getDays();
Collections.sort(days1);
Collections.sort(days2);
for (int i = 0; i < days1.size() && i < days2.size(); i++) {
int compare = days1.get(i).compareTo(days2.get(i));
if (compare != 0) {
return compare;
}
}
return days1.size() - days2.size();
}
});
}
format() 方法来格式化每个路口的天数列表。
static String format(List<Date> days) {
if (days.isEmpty()) {
return "";
}
StringBuilder sb = new StringBuilder();
DateFormat format = new SimpleDateFormat("MMM d");
Collections.sort(days);
String separator = "";
for (Date day: days) {
sb.append(separator);
sb.append(format.format(day));
separator = ", ";
}
sb.append(" ");
return sb.toString();
}
最后，输出。
[]
Jan 1 [BOB, JUDY, FRANK, JIM, SALLY]
Jan 1, Jan 2 [BOB, JUDY, FRANK, JIM]
Jan 1, Jan 2, Jan 3 [BOB, FRANK, JIM]
Jan 1, Jan 2, Jan 3, Jan 4 [BOB, FRANK]
Jan 1, Jan 2, Jan 3, Jan 4, Jan 5 [BOB]
Jan 1, Jan 2, Jan 3, Jan 5 [BOB, JIM]
Jan 1, Jan 2, Jan 4 [BOB, JUDY, FRANK]
Jan 1, Jan 2, Jan 4, Jan 5 [BOB, JUDY]
Jan 1, Jan 2, Jan 5 [BOB, JUDY, JIM]
Jan 1, Jan 3 [BOB, FRANK, JIM, SALLY]
Jan 1, Jan 3, Jan 4 [BOB, FRANK, SALLY]
Jan 1, Jan 3, Jan 4, Jan 5 [BOB, SALLY]
Jan 1, Jan 3, Jan 5 [BOB, JIM, SALLY]
Jan 1, Jan 4 [BOB, JUDY, FRANK, SALLY]
Jan 1, Jan 4, Jan 5 [BOB, JUDY, SALLY]
Jan 1, Jan 5 [BOB, JUDY, JIM, SALLY]
Jan 2 [BOB, JUDY, TOMMY, FRANK, JIM]
Jan 2, Jan 3 [BOB, TOMMY, FRANK, JIM]
Jan 2, Jan 3, Jan 4 [BOB, TOMMY, FRANK]
Jan 2, Jan 3, Jan 4, Jan 5 [BOB, TOMMY]
Jan 2, Jan 3, Jan 5 [BOB, TOMMY, JIM]
Jan 2, Jan 4 [BOB, JUDY, TOMMY, FRANK]
Jan 2, Jan 4, Jan 5 [BOB, JUDY, TOMMY]
Jan 2, Jan 5 [BOB, JUDY, TOMMY, JIM]
Jan 3 [BOB, TOMMY, FRANK, JIM, SALLY]
Jan 3, Jan 4 [BOB, TOMMY, FRANK, SALLY]
Jan 3, Jan 4, Jan 5 [BOB, TOMMY, SALLY]
Jan 3, Jan 5 [BOB, TOMMY, JIM, SALLY]
Jan 4 [BOB, JUDY, TOMMY, FRANK, SALLY]
Jan 4, Jan 5 [BOB, JUDY, TOMMY, SALLY]
Jan 5 [BOB, JUDY, TOMMY, JIM, SALLY]
希望对您有所帮助。我修改它的时间比我预期的要长得多 ;) 仍然没有对输出中的人名进行排序。
我有一个集合列表：
setlist = [s1,s2,s3...sn]
我想要一个集合的全方位比较，即 2^n 个集合：
setIntersection = [s1 ∩ s2, s1 ∩ s2 ∩ s3, ....., s2 ∩ s3 ∩ s4, ...., sn-1 ∩ sn]
在 Java 中执行此操作的最佳方法是什么？
例如，如果我只使用 5 套。我希望能够填充 5 circle venn diagram.
的所有重叠
我正在尝试使用集合列表来执行此操作：
List<Set<People>> lListSets = new ArrayList<Set<People>>();
for (DaysObject day : listOfDaysInJanuary) {
lListSets.add(day.peopleOneInternet());
}
findPowerSetsAndCompare(lListSets, listOfDaysInJanuary);
我想找到类似这样的结果：
January 1 (Bob, Sally, Tommy)
January 1, January 2 (Sally, Tommy)
...
so on for all possible combination of days.
基本上我要问的是如何应用 powerset algorithm 与集合并集的组合。
What would be the best way to do this in Java?
您描述的第一部分是 powerset（因为我上周编辑了您的问题以包含在内）。然后，您将获得幂集中每组集合的交集。
因为您正在做集合的幂集，而不是像整数这样的简单幂集，所以实现会涉及更多一些。
额外学分
我写了一个基本的实现你的要求，作为你如何做的一个例子。此示例中的所有方法和类型都是 Example class.
Example class 仅使用其 main 方法来演示工作代码。我相信您会原谅我在演示中使用已弃用的 Date 构造函数。
import java.text.*;
import java.util.*;
public class Example
{
public static void main(String[] args) {
// create simple test harness
Set<PeopleByDays> peepsByDay = new HashSet<PeopleByDays>();
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 1),
Person.BOB, Person.FRANK, Person.JIM, Person.JUDY, Person.SALLY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 2),
Person.BOB, Person.FRANK, Person.JIM, Person.JUDY, Person.TOMMY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 3),
Person.BOB, Person.FRANK, Person.JIM, Person.SALLY, Person.TOMMY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 4),
Person.BOB, Person.FRANK, Person.JUDY, Person.SALLY, Person.TOMMY));
peepsByDay.add(new PeopleByDays(new Date(2015 - 1900, Calendar.JANUARY, 5),
Person.BOB, Person.JIM, Person.JUDY, Person.SALLY, Person.TOMMY));
// make powerSet, then intersect, then sort
Set<Set<PeopleByDays>> powerPeeps = powerSet(peepsByDay);
List<PeopleByDays> powerPeepsIntersected = intersect(powerPeeps);
sort(powerPeepsIntersected);
// print out results
for (PeopleByDays peeps: powerPeepsIntersected) {
String daysFormatted = format(peeps.getDays());
System.out.print(daysFormatted);
System.out.println(peeps);
}
}
// all other Example members as defined in this answer
}
Person 是人名的简单 enum 类型。在这里使用枚举的好处是它会为所需的 HashSet 行为处理适当的 equals() 和 hashCode() 实现。
static enum Person {
BOB, FRANK, JIM, JUDY, SALLY, TOMMY;
}
PeopleByDays 扩展了 HashSet<Person> 以收集一组额外的 Date 对象来表示日期。覆盖 retainAll()（相交）以合并日期；覆盖 equals() 和 hashSet() 以获得外部集中的正确行为。
static class PeopleByDays extends HashSet<Person> {
private final Set<Date> days = new HashSet<Date>();
public PeopleByDays() {
super();
}
public PeopleByDays(Date day, Person... people) {
super(Arrays.asList(people));
this.days.add(day);
}
public PeopleByDays(PeopleByDays other) {
super(other);
this.days.addAll(other.days);
}
public List<Date> getDays() {
return new ArrayList<Date>(this.days);
}
@Override
public boolean retainAll(Collection<?> c) {
if (c instanceof PeopleByDays) {
this.days.addAll(((PeopleByDays)c).days);
}
return super.retainAll(c);
}
@Override
public boolean equals(Object o) {
return super.equals(o) && this.days.equals(((PeopleByDays) o).days);
}
@Override
public int hashCode() {
return super.hashCode() + this.days.hashCode();
}
}
powerSet() 方法，逐字取自 this answer.
public static <T> Set<Set<T>> powerSet(Set<T> originalSet) {
Set<Set<T>> sets = new HashSet<Set<T>>();
if (originalSet.isEmpty()) {
sets.add(new HashSet<T>());
return sets;
}
List<T> list = new ArrayList<T>(originalSet);
T head = list.get(0);
Set<T> rest = new HashSet<T>(list.subList(1, list.size()));
for (Set<T> set: powerSet(rest)) {
Set<T> newSet = new HashSet<T>();
newSet.add(head);
newSet.addAll(set);
sets.add(newSet);
sets.add(set);
}
return sets;
}
intersect() 方法为幂集中的每组集合创建交集。
static List<PeopleByDays> intersect(Set<Set<PeopleByDays>> powerSet) {
List<PeopleByDays> intersected = new ArrayList<PeopleByDays>();
for (Set<PeopleByDays> powerElement: powerSet) {
PeopleByDays intersection = null;
if (powerElement.isEmpty()) {
intersection = new PeopleByDays();
} else for (PeopleByDays peeps: powerElement) {
if (intersection == null) {
intersection = new PeopleByDays(peeps);
} else {
intersection.retainAll(peeps);
}
}
intersected.add(intersection);
}
return intersected;
}
sort() 方法按日期对结果相交集进行排序。
static void sort(List<PeopleByDays> peeps) {
Collections.sort(peeps, new Comparator<PeopleByDays>() {
@Override
public int compare(PeopleByDays p1, PeopleByDays p2) {
List<Date> days1 = p1.getDays();
List<Date> days2 = p2.getDays();
Collections.sort(days1);
Collections.sort(days2);
for (int i = 0; i < days1.size() && i < days2.size(); i++) {
int compare = days1.get(i).compareTo(days2.get(i));
if (compare != 0) {
return compare;
}
}
return days1.size() - days2.size();
}
});
}
format() 方法来格式化每个路口的天数列表。
static String format(List<Date> days) {
if (days.isEmpty()) {
return "";
}
StringBuilder sb = new StringBuilder();
DateFormat format = new SimpleDateFormat("MMM d");
Collections.sort(days);
String separator = "";
for (Date day: days) {
sb.append(separator);
sb.append(format.format(day));
separator = ", ";
}
sb.append(" ");
return sb.toString();
}
最后，输出。
[]
Jan 1 [BOB, JUDY, FRANK, JIM, SALLY]
Jan 1, Jan 2 [BOB, JUDY, FRANK, JIM]
Jan 1, Jan 2, Jan 3 [BOB, FRANK, JIM]
Jan 1, Jan 2, Jan 3, Jan 4 [BOB, FRANK]
Jan 1, Jan 2, Jan 3, Jan 4, Jan 5 [BOB]
Jan 1, Jan 2, Jan 3, Jan 5 [BOB, JIM]
Jan 1, Jan 2, Jan 4 [BOB, JUDY, FRANK]
Jan 1, Jan 2, Jan 4, Jan 5 [BOB, JUDY]
Jan 1, Jan 2, Jan 5 [BOB, JUDY, JIM]
Jan 1, Jan 3 [BOB, FRANK, JIM, SALLY]
Jan 1, Jan 3, Jan 4 [BOB, FRANK, SALLY]
Jan 1, Jan 3, Jan 4, Jan 5 [BOB, SALLY]
Jan 1, Jan 3, Jan 5 [BOB, JIM, SALLY]
Jan 1, Jan 4 [BOB, JUDY, FRANK, SALLY]
Jan 1, Jan 4, Jan 5 [BOB, JUDY, SALLY]
Jan 1, Jan 5 [BOB, JUDY, JIM, SALLY]
Jan 2 [BOB, JUDY, TOMMY, FRANK, JIM]
Jan 2, Jan 3 [BOB, TOMMY, FRANK, JIM]
Jan 2, Jan 3, Jan 4 [BOB, TOMMY, FRANK]
Jan 2, Jan 3, Jan 4, Jan 5 [BOB, TOMMY]
Jan 2, Jan 3, Jan 5 [BOB, TOMMY, JIM]
Jan 2, Jan 4 [BOB, JUDY, TOMMY, FRANK]
Jan 2, Jan 4, Jan 5 [BOB, JUDY, TOMMY]
Jan 2, Jan 5 [BOB, JUDY, TOMMY, JIM]
Jan 3 [BOB, TOMMY, FRANK, JIM, SALLY]
Jan 3, Jan 4 [BOB, TOMMY, FRANK, SALLY]
Jan 3, Jan 4, Jan 5 [BOB, TOMMY, SALLY]
Jan 3, Jan 5 [BOB, TOMMY, JIM, SALLY]
Jan 4 [BOB, JUDY, TOMMY, FRANK, SALLY]
Jan 4, Jan 5 [BOB, JUDY, TOMMY, SALLY]
Jan 5 [BOB, JUDY, TOMMY, JIM, SALLY]
希望对您有所帮助。我修改它的时间比我预期的要长得多 ;) 仍然没有对输出中的人名进行排序。