減去數組，關於字典中的鍵

Question

我說兩個數組

NSArray *array1=@[ @{@"key1":@"A",@"key2":@"AA"},@{@"key1":@"C",@"key2":@"CC"},@{@"key1":@"E",@"key2":@"EE"},@{@"key1":@"G",@"key2":@"GG"}];
NSArray *array2=@[ @{@"key1":@"A",@"key2":@"AAA"},@{@"key1":@"Z",@"key2":@"ZZZ"}];

我想減去數組，這應該是預期的重用，

NSArray *resultArray=@[ @{@"key1":@"C",@"key2":@"CC"},@{@"key1":@"E",@"key2":@"EE"},@{@"key1":@"G",@"key2":@"GG"}];

我嘗試了以下代碼，但沒有工作

NSArray *extracted = [array1 valueForKey:@"key1"];
NSMutableSet *pressieContactsSet = [NSMutableSet setWithArray:extracted];
NSMutableSet *allContactSet = [NSMutableSet setWithArray:array2];

[allContactSet minusSet:pressieContactsSet];

NSLog(@"%@",allContactSet);

Answer 1

請嘗試以下代碼

首先獲取臨時數組中的所有key1對象。 然后在array1上應用過濾器，並檢查array1對象是否包含arrayKey1對象。

確保它只會檢查key1鍵。

NSArray *arrKey1 = [array2 valueForKey:@"key1"];
NSPredicate *pred = [NSPredicate predicateWithBlock:
                            ^BOOL(id evaluatedObject, NSDictionary *bindings)
                            {
                                if ([arrKey1 containsObject:evaluatedObject[@"key1"]])
                                {
                                    NSLog(@"found : %@",evaluatedObject);
                                    return NO;
                                }
                                else
                                {
                                    NSLog(@"Not found : %@",evaluatedObject);
                                    return YES;
                                }

                            }];
NSArray *arrSubtracted = [array1 filteredArrayUsingPredicate:pred];

NSLog(@"%@", arrSubtracted);

或者您可以使用enumerateObjectsUsingBlock

NSMutableArray *resultArray = [NSMutableArray new];

[array1 enumerateObjectsUsingBlock:^(id  _Nonnull obj, NSUInteger idx, BOOL * _Nonnull stop)
{
    if (![arrKey1 containsObject:obj[@"key1"]]) {
        [resultArray addObject:obj];
    }
}];
NSLog(@"%@",resultArray);

希望這會幫助你。

Answer 2

這應該工作。

NSArray *array1=@[ @{@"key1":@"A"},@{@"key1":@"C"},@{@"key1":@"E"},@{@"key1":@"G"}];
NSArray *array2=@[ @{@"key1":@"A"},@{@"key1":@"Z"}];

NSMutableSet *pressieContactsSet = [NSMutableSet setWithArray:array1];
NSSet *allContactSet = [NSSet setWithArray:array2];

[pressieContactsSet minusSet:allContactSet];

NSArray *result = [pressieContactsSet allObjects];

NSLog(@"%@",result);

Answer 3

請享用！

NSArray *array1=@[ @{@"key1":@"A"},@{@"key1":@"C"},@{@"key1":@"E"},@{@"key1":@"G"}];
NSArray *array2=@[ @{@"key1":@"A"},@{@"key1":@"Z"}];
NSMutableArray *resultArray = [NSMutableArray arrayWithArray:array1];
[resultArray removeObjectsInArray:array2];
NSLog(@"array %@",resultArray);