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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;
}

我在循环中运行这个函数超过 1 000 000 次，所以任何优化都是好的。它可以工作，但我希望它工作得更快。

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5 回答 5

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-(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;
}

试试上面的代码......

于 2013-10-25T07:20:51.807 回答
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你真的需要 -1 值的字典吗？如果你只是这样做，你可以避免所有“如果/那么”的东西（我听说它对于 cpu 来说可能很慢）

    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");
于 2013-10-25T07:28:55.493 回答
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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;
}
于 2013-10-25T07:24:05.677 回答
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您可以减少NSNumber创建的对象数量，并且可以使用新的文字语法至少使代码更短且更易于阅读。您还可以将属性访问次数减半。这是否会对性能产生重大影响，您必须找出答案。

-(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)
           };
}

（该表达式a ?: b是简写，a ? a : b 但 a只会计算一次，因此属性访问次数减半。）

于 2013-10-25T07:57:04.817 回答
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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 块 - Apple 文档

我没有编译它，但它应该可以工作。希望对您有所帮助。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;
}
于 2013-10-25T07:59:47.790 回答