通过将具有匹配的id号的对象分组来重buildNSArray？
我有一个NSArray，数组中的每个对象都有一个groupId和一个名字。 每个对象都是唯一的，但是有许多具有相同的groupId。 有没有一种方法，我可以拆分数组，并重build它，以便名称被分组到一个单一的对象与相应的groubId？ 这是数组目前的样子：
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"; } ) 这是我想它看起来像：
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：我试了很多尝试失败，大多数沿着这样的东西（马虎娱乐，但给一个想法）：
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++; }
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; } )
这需要NSArrays的NSDictionary。 没有快速和优雅的方式 – 你必须滚动来源。
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：编辑不承担该键的数据types。
这与Seva的答案类似，但可以在NSArray上添加为类别方法：
/// @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，因为它会将人物字典保留为数组中的值，但是您可以只读取名称属性。
快速执行Sergery的答案，为我的同伴noobs。
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)
请注意valueForKeyPath中的@符号。 那绊了我一点:)
下面的代码将通过将对象与该数组中的每个字典中的任何匹配键分组来重buildNSArray
//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..