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我试图找到一个很好的解决方案，将一个字典数组拆分为一个较小的字典，这些字典由它们之间的共同值键入。
这是我的 JSON 示例，我从这里开始：

{
  "field": [
    {
      "id": 6,
      "name": "Andrea"
    },
    {
      "id": 67,
      "name": "Francesco"
    },
    {
      "id": 8,
      "name": "Maria"
    },
    {
      "id": 6,
      "name": "Paolo"
    },
    {
      "id": 67,
      "name": "Sara"
    }
  ]
}

我想得到如下结果：

{
  "field": [
    {
      "6": [
        {
          "name": "Andrea",
          "id": 6
        },
        {
          "name": "Paolo",
          "id": 6
        }
      ],
      "67": [
        {
          "name": "Sara",
          "id": 67
        },
        {
          "name": "Francesco",
          "id": 67
        }
      ],
      "8": [
        {
          "name": "Maria",
          "id": 8
        }
      ]
    }
  ]
}

我使用此代码进行了管理，它可以工作，但我想知道是否存在更正确和快速的东西：

    NSArray * array = ...;
    NSSortDescriptor *sorter1=[[NSSortDescriptor alloc]initWithKey:@"id" ascending:YES selector:@selector(compare:)];
    NSSortDescriptor *sorter2=[[NSSortDescriptor alloc]initWithKey:@"name" ascending:YES selector:@selector(caseInsensitiveCompare:)];
    NSArray *sortDescriptors=[NSArray arrayWithObjects:sorter1,sorter2,nil];
    array = [array sortedArrayUsingDescriptors:sortDescriptors];    
    //////////////////////////////SPLITTER
    NSMutableArray * subcategorySplittedArray = [[NSMutableArray alloc]initWithCapacity:30];
    NSNumber * lastID=[[array objectAtIndex:0]objectForKey:@"id"];
    NSMutableArray * shopArray = [[NSMutableArray alloc]initWithCapacity:100];
    NSMutableDictionary * catDict = nil;
    for (NSDictionary * dict in array) {
        NSNumber * catID = [dict objectForKey:@"id"];
        if ([lastID isEqualToNumber:catID]) {
            [shopArray addObject:dict];
        }
        else {

            catDict = [[NSMutableDictionary alloc]init ];
            [catDict setObject:[shopArray copy] forKey:lastID];
            [subcategorySplittedArray addObject:catDict];
            [shopArray removeAllObjects];
            [shopArray addObject:dict];
            lastID = catID;
        }
    }
    catDict = [[NSMutableDictionary alloc]init ];
    [catDict setObject:[shopArray copy] forKey:lastID];
    [subcategorySplittedArray addObject:catDict];
    ////////////////////////////////////
    return subcategorySplittedArray;

}
4

1 回答 1

9 
NSMutableDictionary* result = [NSMutableDictionary dictionary];
NSArray* ids = [array valueWithKey:@"id"];
NSSet* uniqueIDs = [NSSet setWithArray:ids];
for (NSNumber* anID in uniqueIDs)
{
    NSPredicate* pred = [NSPredicate predicateWithFormat:@"id == %@", anID];
    NSArray* dictsForID = [array filteredArrayUsingPredicate:pred];
    [result setObject:dictsForID forKey:anID];
}

如果有很多 ID，您可以通过在循环外部构建一个带有变量引用的谓词，然后只需将变量替换为 in 来为每次通过循环生成特定于 id 的谓词来加快速度。

顺便说一句，在您的问题中，由于某种原因，结果“字段”仍然是一个数组。我认为没有必要。

更新为只通过一次：

NSMutableDictionary* result = [NSMutableDictionary dictionary];
for (NSDictionary* dict in array)
{
    NSNumber* anID = [dict objectForKey:@"id"];
    NSMutableArray* resultsForID = [result objectForKey:anID];
    if (!resultsForID)
    {
        resultsForID = [NSMutableArray array];
        [result setObject:resultsForID forKey:anID];
    }

    [resultsForID addObject:dict];
}
于 2012-04-14T09:03:54.363 回答