Question

(
    {
        data =         (
            "<Place: 0x7ff9ba4af4f0>"
        );
        type = Aquarium;
    },
    {
        data =         (
            "<Place: 0x7ff9ba467d70>",
            "<Place: 0x7ff9ba4bdc10>",
            "<Place: 0x7ff9ba4c1a00>",
            "<Place: 0x7ff9ba4d4f60>",
            "<Place: 0x7ff9ba4e5090>"
        );
        type = "Amusement Park";
    },
    {
        data =         (
            "<Place: 0x7ff9ba444f60>",
            "<Place: 0x7ff9ba4dced0>",
            "<Place: 0x7ff9ba4e9000>",
            "<Place: 0x7ff9ba4f4ab0>"
        );
        type = Spa;
    },
    {
        data =         (
            "<Place: 0x7ff9ba465380>",
            "<Place: 0x7ff9ba4b9d10>",
            "<Place: 0x7ff9ba4d1140>",
            "<Place: 0x7ff9ba4d8f60>"
        );
        type = "Art Gallery";
    },
    {
        data =         (
            "<Place: 0x7ff9ba763100>",
            "<Place: 0x7ff9ba76f520>"
        );
        type = "Art Gallery";
    },
    {
        data =         (
            "<Place: 0x7ff9ba763160>",
            "<Place: 0x7ff9ba68ffe0>",
            "<Place: 0x7ff9ba6a56c0>"
        );
        type = "Amusement Park";
    },
    {
        data =         (
            "<Place: 0x7ff9ba7630a0>",
            "<Place: 0x7ff9ba7676f0>",

            "<Place: 0x7ff9ba6ae020>",
            "<Place: 0x7ff9ba69ba30>"
        );
        type = Bakery;
    },
    {
        data =         (
            "<Place: 0x7ff9ba678520>",
            "<Place: 0x7ff9ba76e550>"
        );
        type = Spa;
    }
)

这是我的代码。我试图将两个字典与相同的密钥组合。下面的代码和我跟随stackoverflow网站NSArray of NSDictionaries - merge dictionaries with same key value pair

-(NSMutableArray*)combinedArray:(NSMutableArray*)array
{
    NSMutableArray*     combined = [[NSMutableArray alloc]init];
    for(id key in [NSSet setWithArray:[array valueForKeyPath:@"type"]])
    {
        if([key isKindOfClass:[NSNull class]])
        continue;
        // Sub array with only id = key
        NSArray*        filtered = [array filteredArrayUsingPredicate:[NSPredicate predicateWithBlock:^BOOL(NSDictionary* evaluatedObject, NSDictionary *bindings)
            {
            return [evaluatedObject valueForKey:@"data"] && [[evaluatedObject valueForKey:@"type"] isEqual:key];
            }]];

        // Grab the data
        NSArray *arrtyp = [filtered valueForKeyPath:@"data"];

        // add the new dictionary
        [combined addObject:@{ @"type":key, @"data": arrtyp }];
    }
    NSLog(@"%@",combined);

    return combined;
}

我得到的输出如下：

{
    data =  (
                (
                    "<Place: 0x7fa4b164a3f0>",
                    "<Place: 0x7fa4b1651940>",
                    "<Place: 0x7fa4b14c7890>",
                    "<Place: 0x7fa4b149e6c0>",
                    "<Place: 0x7fa4b1664e70>",
                    "<Place: 0x7fa4b1678530>",
                    "<Place: 0x7fa4b1668c90>",
                    "<Place: 0x7fa4b16708b0>"
                ),
                (
                    "<Place: 0x7fa4b16e2b80>",
                    "<Place: 0x7fa4b2808470>"
                ),
                (
                    "<Place: 0x7fa4b17515b0>",
                    "<Place: 0x7fa4b1772b70>"
                )
            );
    type = "Art Gallery";
},
{
    data = (
            (
                "<Place: 0x7fa4b164a390>",
                "<Place: 0x7fa4b1674630>",
                "<Place: 0x7fa4b1680670>",
                "<Place: 0x7fa4b168bd20>"
            ),
            (
                "<Place: 0x7fa4b16ea8e0>",
                "<Place: 0x7fa4b280cc60>",
                "<Place: 0x7fa4b28191d0>"
            ),
            (
                "<Place: 0x7fa4b174c8e0>",
                "<Place: 0x7fa4b1751550>",
                "<Place: 0x7fa4b1765e90>"
            )
            );
    type = Spa;
},

我想要的是：

{
    data = (
        "<Place: 0x7fa4b164a3f0>",
        "<Place: 0x7fa4b1651940>",
        "<Place: 0x7fa4b14c7890>",
        "<Place: 0x7fa4b149e6c0>",
        "<Place: 0x7fa4b1664e70>",
        "<Place: 0x7fa4b1678530>",
        "<Place: 0x7fa4b1668c90>",
        "<Place: 0x7fa4b16708b0>"

        "<Place: 0x7fa4b16e2b80>",
        "<Place: 0x7fa4b2808470>"

        "<Place: 0x7fa4b17515b0>",
        "<Place: 0x7fa4b1772b70>"
    )
    );
    type = "Art Gallery";
},
{
    data = (
        "<Place: 0x7fa4b164a390>",
        "<Place: 0x7fa4b1674630>",
        "<Place: 0x7fa4b1680670>",
        "<Place: 0x7fa4b168bd20>"

        "<Place: 0x7fa4b16ea8e0>",
        "<Place: 0x7fa4b280cc60>",
        "<Place: 0x7fa4b28191d0>"

        "<Place: 0x7fa4b174c8e0>",
        "<Place: 0x7fa4b1751550>",
        "<Place: 0x7fa4b1765e90>"
    )
    );
    type = Spa;
},

那么如何将两个字典与相同的键组合起来呢？我怎样才能将这两个数组合二为一。我将拥有那些＆＃34;类型＆＃34;作为tableview中的部分和＆＃34;数据＆＃34;在行中。我被困住了。需要帮助谢谢。

Answer 1

放在下面的功能：

- (NSMutableArray *)combinedArray:(NSMutableArray *)array
{

    NSMutableArray *combined = [[NSMutableArray alloc]init];

    for (id key in [NSSet setWithArray:[array valueForKeyPath:@"type"]])
    {
        if ([key isKindOfClass:[NSNull class]])
            continue;

        NSArray *filtered = [array filteredArrayUsingPredicate:[NSPredicate predicateWithBlock:^BOOL(NSDictionary* evaluatedObject, NSDictionary *bindings) {
            return [evaluatedObject valueForKey:@"data"] && [[evaluatedObject valueForKey:@"type"] isEqual:key];
        }]];

        NSMutableArray *marrtyp = [[NSMutableArray alloc] init];

        for (int i = 0; i < filtered.count; i++) {
            [marrtyp addObjectsFromArray:[[filtered objectAtIndex:i] valueForKey:@"data"]];
        }

        [combined addObject:@{@"type": key, @"data": marrtyp}];
    }

    NSLog(@"%@",combined);
    return combined;
}

这很好用。。。

Answer 2

当使用for循环时，我们也需要处理分配的内存。如果不是没有问题，它将照顾ARC。但是，如下面的代码清除已分配的数组是最佳做法。

- (NSMutableArray *)combinedArray:(NSMutableArray *)array
{
    NSMutableArray *combined = [[NSMutableArray alloc]init];

    for (id key in [NSSet setWithArray:[array valueForKeyPath:@"type"]])
    {
        if ([key isKindOfClass:[NSNull class]])
            continue;

        NSArray *filtered = [array filteredArrayUsingPredicate:[NSPredicate predicateWithBlock:^BOOL(NSDictionary* evaluatedObject, NSDictionary *bindings) {
            return [evaluatedObject valueForKey:@"data"] && [[evaluatedObject valueForKey:@"type"] isEqual:key];
        }]];

        NSMutableArray *arr = [[NSMutableArray alloc] init];
        for (int i = 0; i < filtered.count; i++) {
            [arr addObjectsFromArray:[[filtered objectAtIndex:i] valueForKey:@"data"]];
        }
        [combined addObject:@{@"type": key, @"data": arr}];
        arr = nil;
        filtered = nil;
    }

    NSLog(@"%@",combined);
    return combined;
}

在ios中将两个数组合并为一个

2 个答案: