NSDictionary with ordered keys

IosObjective CCocoaNsdictionary

Ios Problem Overview


I have an NSDictionary (stored in a plist) that I'm basically using as an associative array (strings as keys and values). I want to use the array of keys as part of my application, but I'd like them to be in a specific order (not really an order that I can write an algorithm to sort them into). I could always store a separate array of the keys, but that seems kind of kludgey because I'd always have to update the keys of the dictionary as well as the values of the array, and make sure they always correspond. Currently I just use [myDictionary allKeys], but obviously this returns them in an arbitrary, non-guaranteed order. Is there a data structure in Objective-C that I'm missing? Does anyone have any suggestions on how to more elegantly do this?

Ios Solutions


Solution 1 - Ios

The solution of having an associated NSMutableArray of keys isn't so bad. It avoids subclassing NSDictionary, and if you are careful with writing accessors, it shouldn't be too hard to keep synchronised.

Solution 2 - Ios

I'm late to the game with an actual answer, but you might be interested to investigate CHOrderedDictionary. It's a subclass of NSMutableDictionary which encapsulates another structure for maintaining key ordering. (It's part of CHDataStructures.framework.) I find it to be more convenient than managing a dictionary and array separately.

Disclosure: This is open-source code which I wrote. Just hoping it may be useful to others facing this problem.

Solution 3 - Ios

There is no such inbuilt method from which you can acquire this. But a simple logic work for you. You can simply add few numeric text in front of each key while you prepare the dictionary. Like

NSDictionary *dict = [[NSDictionary alloc] initWithObjectsAndKeys:
                       @"01.Created",@"cre",
                       @"02.Being Assigned",@"bea",
                       @"03.Rejected",@"rej",
                       @"04.Assigned",@"ass",
                       @"05.Scheduled",@"sch",
                       @"06.En Route",@"inr",
                       @"07.On Job Site",@"ojs",
                       @"08.In Progress",@"inp",
                       @"09.On Hold",@"onh",
                       @"10.Completed",@"com",
                       @"11.Closed",@"clo",
                       @"12.Cancelled", @"can",
                       nil]; 

Now if you can use sortingArrayUsingSelector while getting all keys in the same order as you place.

NSArray *arr =  [[dict allKeys] sortedArrayUsingSelector:@selector(localizedStandardCompare:)];

At the place where you want to display keys in UIView, just chop off the front 3 character.

Solution 4 - Ios

If you're going to subclass NSDictionary you need to implement these methods as a minimum:

  • NSDictionary
    • -count
    • -objectForKey:
    • -keyEnumerator
  • NSMutableDictionary
    • -removeObjectForKey:
    • -setObject:forKey:
  • NSCopying/NSMutableCopying
    • -copyWithZone:
    • -mutableCopyWithZone:
  • NSCoding
    • -encodeWithCoder:
    • -initWithCoder:
  • NSFastEnumeration (for Leopard)
    • -countByEnumeratingWithState:objects:count:

The easiest way to do what you want is to make a subclass of NSMutableDictionary that contains its' own NSMutableDictionary that it manipulates and an NSMutableArray to store an ordered set of keys.

If you're never going to encode your objects you could conceivable skip implementing -encodeWithCoder: and -initWithCoder:

All of your method implementations in the 10 methods above would then either go directly through your hosted dictionary or your ordered key array.

Solution 5 - Ios

My little addition: sorting by numeric key (Using shorthand notations for smaller code)

// the resorted result array
NSMutableArray *result = [NSMutableArray new];
// the source dictionary - keys may be Ux timestamps (as integer, wrapped in NSNumber)
NSDictionary *dict =
@{
  @0: @"a",
  @3: @"d",
  @1: @"b",
  @2: @"c"
};

{// do the sorting to result
    NSArray *arr = [[dict allKeys] sortedArrayUsingSelector:@selector(compare:)];
    
    for (NSNumber *n in arr)
        [result addObject:dict[n]];
}

Solution 6 - Ios

Quick 'n dirty:

When you need to order your dictionary (herein called “myDict”), do this:

     NSArray *ordering = [NSArray arrayWithObjects: @"Thing",@"OtherThing",@"Last Thing",nil];

Then, when you need to order your dictionary, create an index:

    NSEnumerator *sectEnum = [ordering objectEnumerator];
    NSMutableArray *index = [[NSMutableArray alloc] init];
		id sKey;
		while((sKey = [sectEnum nextObject])) {
			if ([myDict objectForKey:sKey] != nil ) {
				[index addObject:sKey];
			}
		}

Now, the *index object will contain the appropriate keys in the correct order. Note that this solution does not require that all the keys necessarily exist, which is the usual situation we're dealing with...

Solution 7 - Ios

Minimal implementation of an ordered subclass of NSDictionary (based on https://github.com/nicklockwood/OrderedDictionary). Feel free to extend for your needs:

##Swift 3 and 4

class MutableOrderedDictionary: NSDictionary {
    let _values: NSMutableArray = []
    let _keys: NSMutableOrderedSet = []

    override var count: Int {
        return _keys.count
    }
    override func keyEnumerator() -> NSEnumerator {
        return _keys.objectEnumerator()
    }
    override func object(forKey aKey: Any) -> Any? {
        let index = _keys.index(of: aKey)
        if index != NSNotFound {
            return _values[index]
        }
        return nil
    }
    func setObject(_ anObject: Any, forKey aKey: String) {
        let index = _keys.index(of: aKey)
        if index != NSNotFound {
            _values[index] = anObject
        } else {
            _keys.add(aKey)
            _values.add(anObject)
        }
    }
}

###usage

let normalDic = ["hello": "world", "foo": "bar"]
// initializing empty ordered dictionary
let orderedDic = MutableOrderedDictionary()
// copying normalDic in orderedDic after a sort
normalDic.sorted { $0.0.compare($1.0) == .orderedAscending }
         .forEach { orderedDic.setObject($0.value, forKey: $0.key) }
// from now, looping on orderedDic will be done in the alphabetical order of the keys
orderedDic.forEach { print($0) }

##Objective-C

@interface MutableOrderedDictionary<__covariant KeyType, __covariant ObjectType> : NSDictionary<KeyType, ObjectType>
@end
@implementation MutableOrderedDictionary
{
    @protected
    NSMutableArray *_values;
    NSMutableOrderedSet *_keys;
}

- (instancetype)init
{
    if ((self = [super init]))
    {
        _values = NSMutableArray.new;
        _keys = NSMutableOrderedSet.new;
    }
    return self;
}

- (NSUInteger)count
{
    return _keys.count;
}

- (NSEnumerator *)keyEnumerator
{
    return _keys.objectEnumerator;
}

- (id)objectForKey:(id)key
{
    NSUInteger index = [_keys indexOfObject:key];
    if (index != NSNotFound)
    {
        return _values[index];
    }
    return nil;
}

- (void)setObject:(id)object forKey:(id)key
{
    NSUInteger index = [_keys indexOfObject:key];
    if (index != NSNotFound)
    {
        _values[index] = object;
    }
    else
    {
        [_keys addObject:key];
        [_values addObject:object];
    }
}
@end

###usage

NSDictionary *normalDic = @{@"hello": @"world", @"foo": @"bar"};
// initializing empty ordered dictionary
MutableOrderedDictionary *orderedDic = MutableOrderedDictionary.new;
// copying normalDic in orderedDic after a sort
for (id key in [normalDic.allKeys sortedArrayUsingSelector:@selector(compare:)]) {
    [orderedDic setObject:normalDic[key] forKey:key];
}
// from now, looping on orderedDic will be done in the alphabetical order of the keys
for (id key in orderedDic) {
    NSLog(@"%@:%@", key, orderedDic[key]);
}

Solution 8 - Ios

For, Swift 3. Please try out the following approach

        //Sample Dictionary
        let dict: [String: String] = ["01.One": "One",
                                      "02.Two": "Two",
                                      "03.Three": "Three",
                                      "04.Four": "Four",
                                      "05.Five": "Five",
                                      "06.Six": "Six",
                                      "07.Seven": "Seven",
                                      "08.Eight": "Eight",
                                      "09.Nine": "Nine",
                                      "10.Ten": "Ten"
                                     ]
        
        //Print the all keys of dictionary
        print(dict.keys)
        
        //Sort the dictionary keys array in ascending order
        let sortedKeys = dict.keys.sorted { $0.localizedCaseInsensitiveCompare($1) == ComparisonResult.orderedAscending }
        
        //Print the ordered dictionary keys
        print(sortedKeys)
        
        //Get the first ordered key
        var firstSortedKeyOfDictionary = sortedKeys[0]
        
        // Get range of all characters past the first 3.
        let c = firstSortedKeyOfDictionary.characters
        let range = c.index(c.startIndex, offsetBy: 3)..<c.endIndex
        
        // Get the dictionary key by removing first 3 chars
        let firstKey = firstSortedKeyOfDictionary[range]
        
        //Print the first key
        print(firstKey)

Solution 9 - Ios

I don’t like C++ very much, but one solution that I see myself using more and more is to use Objective-C++ and std::map from the Standard Template Library. It is a dictionary whose keys are automatically sorted on insertion. It works surprisingly well with either scalar types or Objective-C objects both as keys and as values.

If you need to include an array as a value, just use std::vector instead of NSArray.

One caveat is that you might want to provide your own insert_or_assign function, unless you can use C++17 (see this answer). Also, you need to typedef your types to prevent certain build errors. Once you figure out how to use std::map, iterators etc., it is pretty straightforward and fast.

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAndy BourassaView Question on Stackoverflow
Solution 1 - IosAbizernView Answer on Stackoverflow
Solution 2 - IosQuinn TaylorView Answer on Stackoverflow
Solution 3 - IosRajan TwanabashuView Answer on Stackoverflow
Solution 4 - IosAshley ClarkView Answer on Stackoverflow
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