比较两个字典数组并添加/删除项目

Question

我有两个这样的字典数组

  arrCurrent = [{
        id = 1;
        name = "Name1";
    },{
        id = 2;
        name = "Name2";
    },{
        id = 3;
        name = "Name3";
    }];


    arrUpdated = [{
        id = 1;
        name = "Name1 has changed";
    },{
        id = 2;
        name = "Name2 has changed";
    },{
        id = 4;
        name = "Name4 is a new item";
    }];

我的要求是根据以下条件合并这两个数组

1) 如果 arrUpdated 包含新项目（具有新 id 的项目），则应将其添加为新项目。

2) 如果 arrUpdated 包含与 arrCurrent 包含的 id 相同的项目，则应将这些项目替换为更新后的项目

所以，最终的数组应该是这样的

arrFinal = [{
        id = 1;
        name = "Name1 has changed";
    },{
        id = 2;
        name = "Name2 has changed";
    },{
        id = 3;
        name = "Name3";
    },{
        id = 4;
        name = "Name4 is a new item";
    }];

希望我的要求很明确，最好的方法是什么？

这就是我正在尝试的方式。但是这种方法会得到重复的项目。另外，遍历数组也不是最好的方法

请注意：实际代码包含与上述数组名称不同的名称。但同样的逻辑

arrSavedSectors = arrCurrent

arrAllUpdatedSectors = arrUpdated

arrFilteredSectors = arrFinal

NSMutableArray *arrSavedSectors = [[NSMutableArray alloc]initWithArray:  [dictionary objectForKey:@"ArrSectors"]];
NSMutableArray *arrFilteredSectors = [[NSMutableArray alloc]initWithArray:arrSavedSectors];

// add updated sectors to list
NSArray *arrAllUpdatedSectors = [NSArray arrayWithArray:[sectorDetails objectForKey:@"AllUpdatedSectors"]];

if([arrSavedSectors count] > 0){
    // check for updated sectors (Which are already saved in the plist, but recently updated some details)
    for(NSDictionary *dicSector in arrSavedSectors){
        for(NSDictionary *dicUpdatedSector in arrAllUpdatedSectors){
            if([[dicUpdatedSector objectForKey:@"id"] isEqualToString:[dicSector objectForKey:@"id"]]){
                [arrFilteredSectors removeObject:dicSector];
                [arrFilteredSectors addObject:dicUpdatedSector];
            }
            else{
                [arrFilteredSectors addObject:dicUpdatedSector];
            }
        }

    }
}
else{
    [arrFilteredSectors addObjectsFromArray:arrAllUpdatedSectors];
}

Answer 1

一些 Set 操作的时间。

最难的部分是我们需要让我们使用的集合理解对象是相等和相同的。

字典都不是，但我们可以引入一个包装类来使用它的实现。

@interface Wrapper : NSObject
@property NSDictionary *dictionary;
-(instancetype)initWithDictionary: (NSDictionary *)dict;
@end

@implementation Wrapper

-(instancetype)initWithDictionary: (NSDictionary *)dict
{
    self = [super init];
    if (self) {
        self.dictionary = dict;
    }
    return self;
}

-(BOOL)isEqual:(Wrapper *)object
{
    return [self.dictionary[@"id"] isEqual: object.dictionary[@"id"]];
}

-(NSUInteger)hash
{
    return [self.dictionary[@"id"] unsignedIntegerValue];
}

-(NSString *)description
{
    return [self.dictionary description];
}

@end

现在我们可以用它来包装每个字典

NSArray *arrCurrent = @[[[Wrapper alloc] initWithDictionary: @{@"id": @(1), @"name" : @"Name 1"}],
                        [[Wrapper alloc] initWithDictionary: @{@"id": @(2), @"name" : @"Name 2"}],
                        [[Wrapper alloc] initWithDictionary: @{@"id": @(3), @"name" : @"Name 3"}]];

NSArray *arrUpdated = @[[[Wrapper alloc] initWithDictionary: @{@"id": @(1), @"name" : @"Name 1 has Changed"}],
                        [[Wrapper alloc] initWithDictionary: @{@"id": @(2), @"name" : @"Name 2 has Changed"}],
                        [[Wrapper alloc] initWithDictionary: @{@"id": @(4), @"name" : @"Name 4"}]];

接下来我们从这个创建集合

NSSet *setCurrent = [NSSet setWithArray:arrCurrent];
NSSet *setUpdated = [NSSet setWithArray:arrUpdated];

现在我们做集合算术

NSMutableSet *setFinal = [setUpdated mutableCopy];
[setFinal unionSet:setCurrent];

最后我们使用键值编码来解开字典

NSArray *arrFinal = [setFinal valueForKey:@"dictionary"];

结果：

{(
    {
        id = 1;
        name = "Name 1 has Changed";
    },
        {
        id = 2;
        name = "Name 2 has Changed";
    },
        {
        id = 3;
        name = "Name 3";
    },
        {
        id = 4;
        name = "Name 4";
    }
)}

---

此代码的一个问题：包装器在其他情况下没有用，因为确定相等性的规则可能不同，因此能够在需要时定义它们会很有用。我们可以用块来做到这一点：

包装器变成

@interface Wrapper : NSObject
@property id object;
@property (copy) BOOL (^equalComparator)(id a, id b);
@property (copy) NSUInteger (^hashBlock)(id a);
@end

@implementation Wrapper

-(instancetype)initWithObject: (id)obj
              equalComparator:(BOOL (^)(id a, id b))equalComparator
                    hashBlock:(NSUInteger (^)(id a))hashBlock
{
    self = [super init];
    if (self) {
        self.equalComparator = equalComparator;
        self.hashBlock = hashBlock;
        self.object = obj;
    }
    return self;
}

-(BOOL)isEqual:(Wrapper *)object
{
    return self.equalComparator(self.object, object.object);
}

-(NSUInteger)hash
{
    return self.hashBlock(self.object);
}

-(NSString *)description
{
    return [self.object description];
}

@end

我们像这样使用它

BOOL (^eq)(NSDictionary *a, NSDictionary *b) = ^(NSDictionary *a, NSDictionary *b){
    return [a[@"id"] isEqual: b[@"id"]];
};

NSUInteger (^hash)(NSDictionary *a) = ^(NSDictionary *a){
    return [a[@"id"] unsignedIntegerValue];
};

NSArray *arrCurrent = @[[[Wrapper alloc] initWithObject: @{@"id": @(1), @"name" : @"Name 1"} equalComparator:eq hashBlock:hash],
                        [[Wrapper alloc] initWithObject: @{@"id": @(2), @"name" : @"Name 2"} equalComparator:eq hashBlock:hash],
                        [[Wrapper alloc] initWithObject: @{@"id": @(3), @"name" : @"Name 3"} equalComparator:eq hashBlock:hash]];

NSArray *arrUpdated = @[[[Wrapper alloc] initWithObject: @{@"id": @(1), @"name" : @"Name 1 has Changed"} equalComparator:eq hashBlock:hash],
                        [[Wrapper alloc] initWithObject: @{@"id": @(2), @"name" : @"Name 2 has Changed"} equalComparator:eq hashBlock:hash],
                        [[Wrapper alloc] initWithObject: @{@"id": @(4), @"name" : @"Name 4"} equalComparator:eq hashBlock:hash]];

NSSet *setCurrent = [NSSet setWithArray:arrCurrent];
NSSet *setUpdated = [NSSet setWithArray:arrUpdated];
NSMutableSet *setFinal = [setUpdated mutableCopy];
[setFinal unionSet:setCurrent];

NSArray *arrFinal = [setFinal valueForKey:@"object"];

Answer 2

我认为循环会做到这一点。

   for(int x=0;x<[arrUpdate count];x++){
       //condition 2
        if([[arrCurrent valueForKey:@"id"] containsObject:[[arrUpdate objectAtIndex:x] valueForKey:@"id"]]){

       }
     //condition 1
      else{
         [arrCurrent addObject: [arrUpdate objectAtIndex:x]]

       }
    }

虽然你也可以使用谓词。

我没有测试此代码，因为我没有机器。但你至少可以修改它。

更新：

我最终找到了一台机器来证明我的代码有效。使用提问者提供的测试数据：

NSMutableArray *arrCurrent = [NSMutableArray arrayWithArray:@[@{
    @"id":@"1", @"name":@"Name1"
},@{
    @"id":@"2", @"name":@"Name2"
},@{
    @"id":@"3", @"name":@"Name3"
}]];


NSMutableArray *arrUpdated = [NSMutableArray arrayWithArray:@[@{
    @"id":@"1", @"name":@"Name1 has changed"
},@{
    @"id":@"2", @"name":@"Name2 has changed"
},@{
    @"id":@"4", @"name":@"Name4 is a new item"
}]];


for(int x=0;x<[arrUpdated count];x++){
    //condition 2
    if([[arrCurrent valueForKey:@"id"] containsObject:[[arrUpdated objectAtIndex:x] valueForKey:@"id"]]){

        //start updating current array
        for(int y=0;y<[arrCurrent count];y++){
            if([[[arrCurrent objectAtIndex:y] valueForKey:@"id"] isEqualToString:[[arrUpdated objectAtIndex:x] valueForKey:@"id"]]){
                NSLog(@" (%@) will change to (%@)",[[arrCurrent objectAtIndex:y] valueForKey:@"name"],[[arrUpdated objectAtIndex:x] valueForKey:@"name"]);
                [arrCurrent replaceObjectAtIndex:y withObject:[arrUpdated objectAtIndex:x] ];
                break;
            }
        }


    }
    //condition 1
    else{
        [arrCurrent addObject: [arrUpdated objectAtIndex:x]];

    }
}
NSLog(@"RESULT>: %@",arrCurrent);

结果：>>

2015-06-04 23:24:30.949 dd[7523:60b]  (Name1) will change to (Name1 has changed)
    2015-06-04 23:24:30.950 dd[7523:60b]  (Name2) will change to (Name2 has changed)
    2015-06-04 23:24:30.951 dd[7523:60b] RESULT>: (
            {
            id = 1;
            name = "Name1 has changed";
        },
            {
            id = 2;
            name = "Name2 has changed";
        },
            {
            id = 3;
            name = Name3;
        },
            {
            id = 4;
            name = "Name4 is a new item";
        }
    )