ios - NSDictionary中的值总和

我正在寻找一种简单有效的方法来汇总NSDictionary中的所有值。

例如，如果我有NSDictionarys像：

NSDictionary * first = @{@"A": [NSNumber numberWithInt:1],
                         @"B": [NSNumber numberWithInt:2],
                         @"C": [NSNumber numberWithInt:3],
                         @"D": [NSNumber numberWithInt:4]};

NSDictionary * second = @{@"A": [NSNumber numberWithInt:1],
                         @"B": [NSNumber numberWithInt:2],
                         @"C": [NSNumber numberWithInt:3],
                         @"D": [NSNumber numberWithInt:4]};

我想获得一个NSD纪念日

NSDictionary * sum = @{@"A": [NSNumber numberWithInt:2],
                         @"B": [NSNumber numberWithInt:4],
                         @"C": [NSNumber numberWithInt:6],
                         @"D": [NSNumber numberWithInt:8]};

最佳答案

NSDictionary * first = @{@"A": [NSNumber numberWithInt:1],
                         @"B": [NSNumber numberWithInt:2],
                         @"C": [NSNumber numberWithInt:3],
                         @"D": [NSNumber numberWithInt:4]};

NSDictionary * second = @{@"A": [NSNumber numberWithInt:1],
                          @"B": [NSNumber numberWithInt:2],
                          @"C": [NSNumber numberWithInt:3],
                          @"D": [NSNumber numberWithInt:4]};
NSMutableDictionary *Third=[[NSMutableDictionary alloc]init];
for (NSString* key in [first allKeys]) {
    NSLog(@"%@",key);

    int a=[[first objectForKey:key]integerValue];
    int b=[[second objectForKey:key]integerValue];

    int c= a+b;

    [Third setObject:[NSNumber numberWithInt:c]  forKey:key];

    }

NSLog(@"%@",Third);

更新了唯一密钥

NSArray *Key_of_First=[first allKeys];
NSArray *Key_of_Second=[second allKeys];

for (int j=0; j<Key_of_First.count; j++) {

    int a=[[first objectForKey:[Key_of_First objectAtIndex:j]]integerValue];
    int b=[[second objectForKey:[Key_of_Second objectAtIndex:j]]integerValue];

    int c= a+b;

    [Third setObject:[NSNumber numberWithInt:c]  forKey:@"Your Key Name which you give"];

}

  NSLog(@"%@",Third);