如何根据对象的属性比较两个NSSets？

Question

我有两个nssets.

nsset1: person.id = 1, person.id = 2, person.id = 3
nsset2: person.id = 1, person.id = 2

结果应该是:

nsset1 - nsset2: person (with id 3)
nsset2 - nsset1: null

这两个集中具有相同id的对象是不同的对象,所以我不能简单地做minusSet.

我想做的事情如下:

nsset1: person.id = 1, person.id = 2, person.id = 3
nsset2: person.id = 4, person.id = 5

结果应该是这样的:

nsset1 - nsset2: person (id 1), person (id 2), person (id 3)
nsset2 - nsset1: person (id 4), person (id 5)

做这个的最好方式是什么？

Answer 1

@AliSoftware的答案是一个有趣的方法.NSPredicate在Core Data之外很慢,但通常都很好.如果性能有问题,您可以使用循环实现相同的算法,这是一些代码行,但通常更快.

另一种方法是询问具有相同身份证的两个人是否应始终被视为等同.如果这是真的,那么你可以覆盖isEqual:和hash你的人类这样(假设identifier是一个NSUInteger):

- (BOOL)isEqual:(id)other {
  if ([other isMemberOfClass:[self class]) {
    return ([other identifier] == [self identifier]);
  }
  return NO;
}

- (NSUInteger)hash {
  return [self identifier];
}

这样做,所有NSSet操作都会将具有相同标识符的对象视为相等,因此您可以使用minusSet.也NSMutableSet addObject:将自动为独特的你标识符.

实施isEqual:并hash具有广泛影响,因此您需要确保遇到具有相同标识符的两个人对象的每个场所,它们应被视为相等.但如果是这种情况,这会大大简化并加快代码速度.

Answer 2

你应该尝试这样的事情

NSSet* nsset1 = [NSSet setWithObjects:person_with_id_1, person_with_id_2, person_with_id_3, nil];
NSSet* nsset2 = [NSSet setWithObjects:person_with_id_2, person_with_id_4, nil];

// retrieve the IDs of the objects in nsset2
NSSet* nsset2_ids = [nsset2 valueForKey:@"objectID"]; 
// only keep the objects of nsset1 whose 'id' are not in nsset2_ids
NSSet* nsset1_minus_nsset2 = [nsset1 filteredSetUsingPredicate:
    [NSPredicate predicateWithFormat:@"NOT objectID IN %@",nsset2_ids]];