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POGrouping of Custom Objects in Objective-C
 

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    <p>I have and array of custom objects of Person Class</p> <pre><code>Person : NSObject{ NSString *firstName; NSString *lastName; NSString *age; } NSMutableArray *personsArray = [NSMutableArray array]; Person *personObj1 = [[[Person alloc] init] autorelease]; personObj1.firstName = @"John"; personObj1.lastName = @"Smith"; personObj1.age = @"25"; [personsArray addObject: personObj1]; Person *personObj2 = [[[Person alloc] init] autorelease]; personObj2.firstName = @"John"; personObj2.lastName = @"Paul"; personObj2.age = @"26"; [personsArray addObject: personObj2]; Person *personObj3 = [[[Person alloc] init] autorelease]; personObj3.firstName = @"David"; personObj3.lastName = @"Antony"; personObj3.age = @"30"; [personsArray addObject: personObj3]; </code></pre> <p>Now personsArray contains 3 objects of Person objects.</p> <p>Can we group this objects by attribute like age or firstName ?</p> <p>My expected result is </p> <pre><code>NSDictionary { "John" = &gt;{ personObj1, //(Its because personObj1 firstName is John ) personObj2 //(Its because personObj2 firstName is John ) }, "David" = &gt;{ personObj3, //(Its because personObj3 firstName is David ) }, } </code></pre> <p>I know I can get this result by creating an NSDictionary and then Iterate through personsArray, then check for each first</p> <pre><code>NSMutableDictionary *myDict = [NSMutableDictionary dictionary]; for (Person *person in personsArray){ if([myDict objectForKey: person.firstName]){ NSMutableArray *array = [myDict objectForKey: person.firstName]; [array addObject:person]; [myDict setObject: array forKey:person.firstName]; }else{ NSMutableArray *array = [NSMutableArray arrayWithObject:person]; [myDict setObject: array forKey:person.firstName]; } } NSLog(@"myDict %@", myDict); //Result myDict will give the desired output. </code></pre> <p>But is there any better way to do ? </p> <p>If I am using <code>@distinctUnionOfObjects</code> , I can group by string objects only (Not such custom objects. Am I right ?).</p> <p>Thanks for the answer in advance.</p>
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