NSArray：删除具有重复属性的对象

Question

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

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的？

Answer 1

您可以创建一个 HashSet，并在循环时将“title+author”连接集添加到 HashSet (NSMutableSet)。当您到达每个项目时，如果 HashSet 包含您的密钥，请将其删除或不复制（删除或创建不重复的副本）。

这使它排序为 n（1 个循环）

这是 NSMutableSet 类：

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

使用代码编辑：

代码的核心是一个循环。

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。如果您有 Assest 的源代码，则需要覆盖 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

然后去实现：

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，你可以打电话给[distinctItems allObjects]

Answer 3

首先，我会像这样覆盖 Asset 的 isEqual: 方法：

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

那么，如果你想首先避免在数组中放置重复项：

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，那么任何找到它们的算法的运行时间都比线性差，但我认为创建一个新数组并只添加像上面这样的唯一元素是最好的算法。像这样的：

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（所有元素已经是唯一的）迭代次数是：

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

这仍然是 O(n^2) 但比@Justin Meiners 的算法略好。没有冒犯的意思！ :)

Answer 4

这是您可以做到的一种方式 ：

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

如果您希望自定义 NSObject 子类在名称相同时被视为相同，您可以实现 isEqual: 和 hash。这将允许您将对象添加到 NSSet/NSMutableSet（一组不同的对象）。

然后，您可以使用NSSet 的sortedArrayUsingDescriptors: 方法轻松创建已排序的NSArray。

MikeAsh 写了一篇关于实现自定义相等的非常可靠的文章：Friday Q&A 2010-06-18: Implementing Equality and Hashing