Question

我正在尝试优化返回NSMutableDictionary的函数，如下所示：

 -(NSMutableDictionary *)getValuses{

    NSNumber *n1 = [NSNumber numberWithInt:-1];
    NSNumber *n2 = [NSNumber numberWithInt:-1];
    NSNumber *n3 = [NSNumber numberWithInt:-1];
    NSNumber *n4 = [NSNumber numberWithInt:-1];
    NSNumber *n5 = [NSNumber numberWithInt:-1];

    if (self.k1)
        n1 = self.k1;
    if (self.k2)
        n2 = self.k2;
    if (self.k3)
        n3 = self.k3;
    if (self.k4)
        n4 = self.k4;
    if (self.k5)
        n5 = self.k5;

    NSMutableDictionary * dictionary = [[NSMutableDictionary alloc]initWithObjectsAndKeys:n1,[NSNumber numberWithInt:2],n2,[NSNumber numberWithInt:3],n3,[NSNumber numberWithInt:4],n4,[NSNumber numberWithInt:5],n5,[NSNumber numberWithInt:6], nil];

    return dictionary;
}

我在循环中运行此函数超过1 000 000次，因此任何优化都是好的。它有效，但我希望它能更快地工作。

Answer 1

-(NSMutableDictionary *)getValuses{

    NSNumber *n1 = [NSNumber numberWithInt:-1];
   NSMutableDictionary * dictionary = [[NSMutableDictionary alloc]initWithObjectsAndKeys:(self.k1)? self.k1:n1,[NSNumber numberWithInt:2],(self.k2)? self.k2:n1,[NSNumber numberWithInt:3],(self.k3)? self.k3:n1,[NSNumber numberWithInt:4],(self.k4)? self.k4:n1,[NSNumber numberWithInt:5],(self.k5)? self.k5:n1,[NSNumber numberWithInt:6], nil];

    return dictionary;
}

尝试以上代码......

Answer 2

你真的需要带有-1值的字典吗？你可以避免所有“if / then”的东西（我听说你的cpu速度很慢）如果你这样做的话

    NSMutableDictionary * dictionary = [[NSMutableDictionary alloc]initWithObjectsAndKeys:k1,[NSNumber numberWithInt:2],k2,[NSNumber numberWithInt:3],k3,[NSNumber numberWithInt:4],k4,[NSNumber numberWithInt:5],k5,[NSNumber numberWithInt:6], nil];
    // then you can do things like this
    id obj = [dictionary objectForKey:@2];
    if (obj)
        NSLog(@"dict with good values");
    else
        NSLog(@"old dict with -1");

Answer 3

您可以尝试这样的事情（未经测试）：

-(NSMutableDictionary *)getValuses {
    NSNumber *n = [NSNumber numberWithInt:-1];
    NSMutableDictionary * dictionary = 
        [[NSMutableDictionary alloc] initWithObjectsAndKeys: 
            self.k1 ? self.k1 : n,[NSNumber numberWithInt:2],
            self.k2 ? self.k2 : n,[NSNumber numberWithInt:3]...
    return dictionary;
}

Answer 4

您可以减少创建的NSNumber个对象的数量，并且可以使用新的文字语法至少使代码更短和更短。更容易阅读。您还可以将属性访问次数减半。无论这些是否会对性能产生重大影响，您都必须找到答案。

-(NSMutableDictionary *)getValuses
{
   NSNumber *n = @(-1);

   return @{ @2: (self.k1 ?: n), @3: (self.k2 ?: n), @4: (self.k3 ?: n),
             @5: (self.k4 ?: n), @6: (self.k5 ?: n)
           };
}

（表达式a ?: b是a ? a : b 的简写，但 a只会被评估一次，因此属性访问次数减半。）< / p>

Answer 5

试试这个！ dispatch_appy方法用于创建循环并在并发队列中执行代码。这是执行大循环的更好方法。

dispatch_apply blocks - Apple Documentation

我没有编译它，但它应该工作。希望有助于给你一个线索。祝你好运！

-(NSMutableDictionary *)getValuses{

    //add the values here
    NSMutableArray *array = [@[self.k1,self.k2,self.k3,self.k4,self.k5]mutableCopy];
    NSMutableDictionary * dictionary = [@{}mutableCopy];

    //this kind of block is better for big loops...
    size_t count = array.count;
    dispatch_queue_t queue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0);

    dispatch_apply(count, queue, ^(size_t i) {
        id value = [array objectAtIndex:i];
        [dictionary setObject:value?value:(@-1) forKey:@(i)];
    });

    return dictionary;
}

通过创建NSDictionarry来优化功能

5 个答案: