從數組中刪除重複項，比較其對象的屬性

Question

假設我有一個類Event，它有兩個屬性：action（NSString）和date（NSDate）。

並假設我有一個Event對象的數組。問題是「日期」屬性可以匹配。

我需要刪除重複項，這意味着2個不同的對象具有相同的日期是重複的。

我可以刪除任何字符串或nsdates數組中的重複項，它們很容易比較。但是如何用複雜的對象來完成它們的屬性比較呢？

不要問我到目前爲止我做了什麼，因爲我唯一想到的就是冒泡，，但這是一個新手解決方案，並且很慢。

非常感謝任何幫助（鏈接，內容，代碼）。

在此先感謝。

編輯

感謝dasblinkenlight，我做了一個自定義的方法：

- (NSArray *)removeDuplicatesInArray:(NSArray*)arrayToFilter{ 

    NSMutableSet *seenDates = [NSMutableSet set]; 
    NSPredicate *dupDatesPred = [NSPredicate predicateWithBlock: ^BOOL(id obj, NSDictionary *bind) { 
     YourClass *e = (YourClass*)obj; 
     BOOL seen = [seenDates containsObject:e.propertyName]; 
     if (!seen) { 
      [seenDates addObject:e.when]; 
     } 
     return !seen; 
    }]; 
    return [arrayToFilter filteredArrayUsingPredicate:dupDatesPred]; 
} 

這裏YourClass是對象所屬的類的名稱，propertyName是您將要比較的對象的屬性。

假設self.arrayWithObjects包含YourClass的對象。

填充它後，使用

self.arrayWithObjects = [self removeDuplicatesInArray:self.arrayWithObjects];

和你做。

所有學分均通過dasblinkenlight。 乾杯！

Answer 1

您可以創建日期的NSMutableSet，迭代您的事件列表，並僅添加事件的日期，您沒有遇到過。

NSMutableSet *seenDates = [NSMutableSet set]; 
NSPredicate *dupDatesPred = [NSPredicate predicateWithBlock: ^BOOL(id obj, NSDictionary *bind) { 
    Event *e = (Event*)obj; 
    BOOL seen = [seenDates containsObject:e.date]; 
    if (!seen) { 
     [seenDates addObject:e.date]; 
    } 
    return !seen; 
}]; 
NSArray *events = ... // This is your array which needs to be filtered 
NSArray *filtered = [events filteredArrayUsingPredicate:dupDatesPred]; 

Answer 2

這不適用於kvc。我想下面的解決方案可以在你的情況下工作;

Event *event1 = [[Event alloc] init]; 
event1.name = @"Event1"; 
event1.date = [NSDate distantFuture]; 
Event *event2 = [[Event alloc] init]; 
event2.name = @"Event2"; 
event2.date = [NSDate distantPast]; 
Event *event3 = [[Event alloc] init]; 
event3.name = @"Event1"; 
event3.date = [NSDate distantPast]; 
NSArray *array = @[event1, event2, event3]; 

NSArray *filteredEvents = [array valueForKeyPath:@"@distinctUnionOfObjects.name"]; 

Answer 3

我認爲最有效的方法是使用NSDictionary存儲對象值和屬性值作爲重點，並添加任何對象字典，然後檢查是否存在與否是O（1 ）操作，即整個過程將花費O（n）的

下面是代碼

- (NSArray *)removeDuplicatesFromArray:(NSArray *)array onProperty:(NSString *)propertyName { 
    NSMutableDictionary *dictionary = [[NSMutableDictionary alloc] init]; 

    for (int i=0; i<array.count; i++) { 

     NSManagedObject *currentItem = array[i]; 
     NSString *propertyValue = [currentItem valueForKey:propertyName]; 

     if ([dictionary valueForKey:propertyValue] == nil) { 
      [dictionary setValue:currentItem forKey:propertyValue]; 
     } 
    } 

    NSArray *uniqueItems = [dictionary allValues]; 

    return uniqueItems; 
} 

，你可以使用它作爲以下

self.arrayWithObjects = [self removeDuplicatesFromArray:self.arrayWithObjects onProperty:@"when"]; 

Answer 4

NSMutableArray *leftObjects = [duplicateArray mutableCopy]; 
NSMutableArray *nonDuplicates = [NSMutableArray new]; 
while (leftObjects.count > 0) 
{ 
    YourClass *object = [leftObjects objectAtIndex:0]; 

    // find all objects matching your comaprison equality definition for YourClass 
    NSArray *matches = [leftObjects filteredArrayUsingPredicate: 
         [NSPredicate predicateWithBlock:^BOOL(YourClass *evaluatedObject, NSDictionary *bindings) 
         { 
          return (evaluatedObject.name == object.name); 
         }] ]; 
    [leftObjects removeObjectsInArray:matches]; 

    // add first object (arbitrary, may decide which duplicate to pick) 
    [nonDuplicates addObject:matches.firstObject]; 
} 

Answer 5

這裏是NSArray類迅捷的延伸刪除指定屬性的重複項：

extension NSArray { 
/** 
- parameter property: the name of the property to check for duplicates 

- returns: an array of objects without objects that share an identical value of the specified property 
*/ 
    func arrayWithoutObjectsOfDuplicateProperty(property : String) -> [AnyObject] { 
    var seenInstances = NSMutableSet() 

    let predicate = NSPredicate { (obj, bind) -> Bool in 
     let seen = seenInstances.containsObject(obj.valueForKey(property)!) 

     if !seen { 
     seenInstances.addObject(obj.valueForKey(property)!) 
     } 
     return !seen 
     }  
     return self.filteredArrayUsingPredicate(predicate) 
    } 
} 

Answer 6

這裏是工作斯威夫特代碼斬件刪除重複項，同時保持元素順序。

// Custom Struct. Can be also class. 
// Need to be `equitable` in order to use `contains` method below 
struct CustomStruct : Equatable { 
     let name: String 
     let lastName : String 
    } 

// conform to Equatable protocol. feel free to change the logic of "equality" 
func ==(lhs: CustomStruct, rhs: CustomStruct) -> Bool { 
    return (lhs.name == rhs.name && lhs.lastName == rhs.lastName) 
} 

let categories = [CustomStruct(name: "name1", lastName: "lastName1"), 
        CustomStruct(name: "name2", lastName: "lastName1"), 
        CustomStruct(name: "name1", lastName: "lastName1")] 
print(categories.count) // prints 3 

// remove duplicates (and keep initial order of elements) 
let uniq1 : [CustomStruct] = categories.reduce([]) { $0.contains($1) ? $0 : $0 + [$1] } 
print(uniq1.count) // prints 2 - third element has removed 

而就，如果你想知道如何減少神品 - 在這裏是完全一樣的，但使用更加擴大減少語法

let uniq2 : [CustomStruct] = categories.reduce([]) { (result, category) in 
    var newResult = result 
    if (newResult.contains(category)) {} 
    else { 
    newResult.append(category) 
    } 
    return newResult 
} 
uniq2.count // prints 2 - third element has removed 

您可以簡單地將此代碼複製粘貼到一個斯威夫特遊樂場並玩耍。