14

我有一个 NSArray，数组中的每个对象都有一个 groupId 和一个名称。每个对象都是唯一的，但有许多具有相同的 groupId。有没有办法可以将数组拆开并重建它，以便将名称分组到具有相应 groubId 的单个对象中？这是数组当前的样子：

2013-03-12 20:50:05.572 appName[4102:702] the  array:  (
        {
        groupId = 1;
        name = "Dan";
    },
        {
        groupId = 1;
        name = "Matt";
    },
        {
        groupId = 2;
        name = "Steve";
    },
        {
        groupId = 2;
        name = "Mike";
    },
        {
        groupId = 3;
        name = "John";

    },
        {
        groupId = 4;
        name = "Kevin";
    }
)

这就是我希望它看起来的样子：

2013-03-12 20:50:05.572 appName[4102:702] the  array:  (
        {
        groupId = 1;
        name1 = "Dan";
        name2 = "Matt";
    },
        {
        groupId = 2;
        name1 = "Steve";
        name2 = "Mike";
    },
        {
        groupId = 3;
        name = "John";

    },
        {
        groupId = 4;
        name = "Kevin";
    }
)

编辑：我已经尝试过很多次但都失败了，大部分都是这样的（草率的娱乐，但给出一个想法）：

int idNum = 0;
for (NSDictionary *arrObj in tempArr){
    NSString *check1 = [NSString stringWithFormat:@"%@",[arrObj valueForKey:@"groupId"]];
    NSString *check2 = [NSString stringWithFormat:@"%@",[[newDict valueForKey:@"groupId"]];
    if (check1 == check2){
        NSString *nameStr = [NSString stringWithFormat:@"name_%d",idNum];
        [newDict setValue:[arrObj valueForKey:@"name"] forKey:nameStr];
    }
    else {
        [newDict setValue:arrObj forKey:@"object"];
    }
    idNum++;
}
4

5 回答 5

34 
NSArray *array = @[@{@"groupId" : @"1", @"name" : @"matt"},
                   @{@"groupId" : @"2", @"name" : @"john"},
                   @{@"groupId" : @"3", @"name" : @"steve"},
                   @{@"groupId" : @"4", @"name" : @"alice"},
                   @{@"groupId" : @"1", @"name" : @"bill"},
                   @{@"groupId" : @"2", @"name" : @"bob"},
                   @{@"groupId" : @"3", @"name" : @"jack"},
                   @{@"groupId" : @"4", @"name" : @"dan"},
                   @{@"groupId" : @"1", @"name" : @"kevin"},
                   @{@"groupId" : @"2", @"name" : @"mike"},
                   @{@"groupId" : @"3", @"name" : @"daniel"},
                   ];

NSMutableArray *resultArray = [NSMutableArray new];
NSArray *groups = [array valueForKeyPath:@"@distinctUnionOfObjects.groupId"];
for (NSString *groupId in groups)
{
    NSMutableDictionary *entry = [NSMutableDictionary new];
    [entry setObject:groupId forKey:@"groupId"];

    NSArray *groupNames = [array filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"groupId = %@", groupId]];
    for (int i = 0; i < groupNames.count; i++)
    {
        NSString *name = [[groupNames objectAtIndex:i] objectForKey:@"name"];
        [entry setObject:name forKey:[NSString stringWithFormat:@"name%d", i + 1]];
    }
    [resultArray addObject:entry];
}

NSLog(@"%@", resultArray);

输出：

    (
        {
        groupId = 3;
        name1 = steve;
        name2 = jack;
        name3 = daniel;
    },
        {
        groupId = 4;
        name1 = alice;
        name2 = dan;
    },
        {
        groupId = 1;
        name1 = matt;
        name2 = bill;
        name3 = kevin;
    },
        {
        groupId = 2;
        name1 = john;
        name2 = bob;
        name3 = mike;
    }
 )
于 2013-03-13T02:02:00.333 回答
2

这需要 NSArrays 的 NSDictionary。没有快速而优雅的方式 - 您必须滚动浏览源代码。

NSMutableDictionary *d = [NSMutableDictionary dictionaryWithCapacity:10]; //Or use alloc/init
for(SomeObject o in appname) //What's the type of objects? you tell me
{
    NSObject *ID = [o objectForKey: @"groupId"];
    NSMutableArray *a = [d objectForKey: ID];
    if(a == nil)
    {
        a = [NSMutableArray arrayWithCapacity: 10];
        [d setObject:a forKey: ID];
    }
    [a addObject: [o objectForKey: @"name"]];
}

编辑：编辑为不假定密钥的数据类型。

于 2013-03-13T01:47:35.690 回答
1

这类似于 Seva 的答案，但可以将其添加为 NSArray 上的类别方法：

/// @return A dictionary of NSMutableArrays
- (NSDictionary *)abc_groupIntoDictionary:(id<NSCopying>(^)(id object))keyFromObjectCallback {
    NSParameterAssert(keyFromObjectCallback);
    NSMutableDictionary *result = [NSMutableDictionary dictionary];
    for (id object in self) {
        id<NSCopying> key = keyFromObjectCallback(object);
        NSMutableArray *array = [result objectForKey:key];
        if (array == nil) {
            array = [NSMutableArray new];
            [result setObject:array forKey:key];
        }
        [array addObject:object];
    }
    return [result copy];
}

你可以像这样使用它：

NSDictionary *groups = [people abc_groupIntoDictionary:^id<NSCopying>(NSDictionary *person) {
    return person[@"groupId"];
}];

这与原始答案不完全相同，因为它将人员字典保留为数组中的值，但是您可以从中读取 name 属性。

于 2014-10-06T23:48:10.667 回答
1

为我的菜鸟们迅速实施 Sergery 的回答。

class People: NSObject {
    var groupId: String
    var name : String
    init(groupId: String, name: String){
        self.groupId = groupId
        self.name = name
    }
}


let matt = People(groupId: "1", name: "matt")
let john = People(groupId: "2", name: "john")
let steve = People(groupId: "3", name: "steve")
let alice = People(groupId: "4", name: "alice")
let bill = People(groupId: "1", name: "bill")
let bob = People(groupId: "2", name: "bob")
let jack = People(groupId: "3", name: "jack")
let dan = People(groupId: "4", name: "dan")
let kevin = People(groupId: "1", name: "kevin")
let mike = People(groupId: "2", name: "mike")
let daniel = People(groupId: "3", name: "daniel")

let arrayOfPeople = NSArray(objects: matt, john, steve, alice, bill, bob, jack, dan, kevin, mike, daniel)

var resultArray = NSMutableArray()
let groups = arrayOfPeople.valueForKeyPath("@distinctUnionOfObjects.groupId") as [String]


for groupId in groups {
    var entry = NSMutableDictionary()
    entry.setObject(groupId, forKey: "groupId")
    let predicate = NSPredicate(format: "groupId = %@", argumentArray: [groupId])
    var groupNames = arrayOfPeople.filteredArrayUsingPredicate(predicate)
    for i in 0..<groupNames.count {
        let people = groupNames[i] as People
        let name = people.name
        entry.setObject(name, forKey: ("name\(i)"))
    }
    resultArray.addObject(entry)
}

println(resultArray)

注意 valueForKeyPath 中的 @ 符号。这让我有点绊倒:)

于 2015-01-27T08:24:40.230 回答
0

下面的代码将通过将该数组中每个字典中的任何匹配键对对象进行分组来重建 NSArray

//only to a take unique keys. (key order should be maintained)
NSMutableArray *aMutableArray = [[NSMutableArray alloc]init];

NSMutableDictionary *dictFromArray = [NSMutableDictionary dictionary];

for (NSDictionary *eachDict in arrOriginal) {
  //Collecting all unique key in order of initial array
  NSString *eachKey = [eachDict objectForKey:@"roomType"];
  if (![aMutableArray containsObject:eachKey]) {
   [aMutableArray addObject:eachKey];
  }

  NSMutableArray *tmp = [grouped objectForKey:key];
  tmp  = [dictFromArray objectForKey:eachKey];

 if (!tmp) {
    tmp = [NSMutableArray array];
    [dictFromArray setObject:tmp forKey:eachKey];
 }
[tmp addObject:eachDict];

}

//NSLog(@"dictFromArray %@",dictFromArray);
//NSLog(@"Unique Keys :: %@",aMutableArray);

//再次从字典转换为数组...

self.finalArray = [[NSMutableArray alloc]init];
for (NSString *uniqueKey in aMutableArray) {
   NSDictionary *aUniqueKeyDict = @{@"groupKey":uniqueKey,@"featureValues":[dictFromArray objectForKey:uniqueKey]};
   [self.finalArray addObject:aUniqueKeyDict];
}

希望它可以帮助..

于 2016-07-20T05:01:18.543 回答