NSArray：删除具有重复属性的对象

我有一个NSMutableArray包含一些自定义对象。 其中两个对象具有相同的属性，如标题和作者。 我想删除重复的对象，离开另一个。

Asset *asset; NSMutableArray *items = [[[NSMutableArray alloc] init] autorelease]; // First asset = [[Asset alloc] init]; asset.title = @"Developer"; asset.author = @"John Smith"; [items addObject:asset]; [asset release]; // Second asset = [[Asset alloc] init]; asset.title = @"Writer"; asset.author = @"Steve Johnson"; [items addObject:asset]; [asset release]; // Third asset = [[Asset alloc] init]; asset.title = @"Developer"; asset.author = @"John Smith"; [items addObject:asset]; [asset release]; 

由于它们不是同一个对象，但只有重复的属性，我怎样才能删除重复？

Answer 1

你可以创build一个HashSet，当你循环，你可以添加“标题+作者”连接集到HashSet（NSMutableSet）。 当您到达每个项目时，如果HashSet包含您的密钥，请将其移除或不要复制（删除或创build没有重复的副​​本）。

这使得n（1循环）

这里是NSMutableSet类：

http://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSMutableSet_Class/Reference/NSMutableSet.html#//apple_ref/occ/cl/NSMutableSet

编辑代码：

代码的肉是一个循环。

 void print(NSMutableArray *assets) { for (Asset *asset in assets) { NSLog(@"%@/%@", [asset title], [asset author]); } } int main (int argc, const char * argv[]) { NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init]; // // Create the initial data set // Asset *asset; NSMutableArray *items = [[[NSMutableArray alloc] init] autorelease]; // First asset = [[Asset alloc] init]; asset.title = @"Developer"; asset.author = @"John Smith"; [items addObject:asset]; [asset release]; // Second asset = [[Asset alloc] init]; asset.title = @"Writer"; asset.author = @"Steve Johnson"; [items addObject:asset]; [asset release]; // Third asset = [[Asset alloc] init]; asset.title = @"Developer"; asset.author = @"John Smith"; [items addObject:asset]; [asset release]; NSLog(@"****Original****"); print(items); // // filter the data set in one pass // NSMutableSet *lookup = [[NSMutableSet alloc] init]; for (int index = 0; index < [items count]; index++) { Asset *curr = [items objectAtIndex:index]; NSString *identifier = [NSString stringWithFormat:@"%@/%@", [curr title], [curr author]]; // this is very fast constant time lookup in a hash table if ([lookup containsObject:identifier]) { NSLog(@"item already exists. removing: %@ at index %d", identifier, index); [items removeObjectAtIndex:index]; } else { NSLog(@"distinct item. keeping %@ at index %d", identifier, index); [lookup addObject:identifier]; } } NSLog(@"****Filtered****"); print(items); [pool drain]; return 0; } 

这是输出：

 Craplet[11991:707] ****Original**** Craplet[11991:707] Developer/John Smith Craplet[11991:707] Writer/Steve Johnson Craplet[11991:707] Developer/John Smith Craplet[11991:707] distinct item. keeping Developer/John Smith at index 0 Craplet[11991:707] distinct item. keeping Writer/Steve Johnson at index 1 Craplet[11991:707] item already exists. removing: Developer/John Smith at index 2 Craplet[11991:707] ****Filtered**** Craplet[11991:707] Developer/John Smith Craplet[11991:707] Writer/Steve Johnson 

Answer 2

您可以使用NSSet的唯一性从原始数组中获取不同的项目。 如果你有Assest的源代码，你需要重写Asset类的hash和isEqual:方法。

 @interface Asset : NSObject @property(copy) NSString *title, *author; @end @implementation Asset @synthesize title, author; -(NSUInteger)hash { NSUInteger prime = 31; NSUInteger result = 1; result = prime * result + [self.title hash]; result = prime * result + [self.author hash]; return result; } -(BOOL)isEqual:(id)object { return [self.title isEqualToString:[object title]] && [self.author isEqualToString:[object author]]; } - (void)dealloc { [title release]; [author release]; [super dealloc]; } @end 

然后执行：

 Asset *asset; NSMutableArray *items = [[[NSMutableArray alloc] init] autorelease]; // First asset = [[Asset alloc] init]; asset.title = @"Developer"; asset.author = @"John Smith"; [items addObject:asset]; [asset release]; // Second asset = [[Asset alloc] init]; asset.title = @"Writer"; asset.author = @"Steve Johnson"; [items addObject:asset]; [asset release]; // Third asset = [[Asset alloc] init]; asset.title = @"Developer"; asset.author = @"John Smith"; [items addObject:asset]; [asset release]; NSLog(@"Items: %@", items); NSSet *distinctItems = [NSSet setWithArray:items]; NSLog(@"Distinct: %@", distinctItems); 

如果你最后需要一个数组，你可以调用[distinctItems allObjects]

Answer 3

首先，我会重写像这样的资产的isEqual：方法：

 -(BOOL)isEqual:(Asset *)otherAsset { return [self.title isEqual:otherAsset.title] && [self.author isEqual:otherAsset.author]; } 

那么，如果你想避免在数组中放置重复：

 NSUInteger idx = [items indexOfObject:asset]; // tests objects for equality using isEqual: if (idx == NSNotFound) [items addObject:asset]; 

如果数组已经包含重复项，那么find它们的任何algorithm的运行时间已经比线性差，但是我认为创build一个新的数组并且只添加上面的独特元素是最好的algorithm。 像这样的东西：

 NSMutableArray *itemsWithUniqueElements = [NSMutableArray arrayWithCapacity:[items count]]; for (Asset *anAsset in items) { if ([items indexOfObject:anAsset] == NSNotFound) [itemsWithUniqueElements addObject:anAsset]; } [items release]; items = [itemsWithUniqueElements retain]; 

在最坏的情况下（所有元素都是唯一的）迭代的次数是：

 1 + 2 + 3 + ... + n = n * (n+1) / 2 

这仍然是O（n ^ 2），但比@Justin Meiners的algorithm稍好。 没有冒犯的意思！ 🙂

Answer 4

这是你可以做到的一个方法：

 NSMutableArray* toRemove = [NSMutableArray array]; for (Asset* asset1 in items) { for (Asset* asset2 in items) { if (asset1 != asset2) { if ([asset1.title isEqualToString:asset2.title] && [asset1.author isEqualToString:asset2.author]) { [toRemove addObject:asset2]; } } } } for (Asset* deleted in toRemove) { [items removeObject:toRemove]; } 

Answer 5

如果你希望自定义的NSObject子类在它们的名字相同时被认为是相等的，你可以实现isEqual:和hash 。 这将允许您将对象添加到NSSet / NSMutableSet （一组不同的对象）。

您可以使用NSSet的sortedArrayUsingDescriptors:方法轻松地创build一个sorting的NSArray 。

MikeAsh写了一篇关于实现自定义平等的文章： 星期五问答2010-06-18：实现平等和散列