將NSArray轉換為NSDictionary

Question

如何將NSArray轉換為NSDictionary ，使用數組對象的int字段作為NSDictionary鍵？

Answer 1

試試這個魔法：

NSDictionary *dict = [NSDictionary dictionaryWithObjects:records 
                                   forKeys:[records valueForKey:@"intField"]];

僅供參考，因為這個內置功能：

@interface NSArray(NSKeyValueCoding)

/* Return an array containing the results of invoking -valueForKey: 
on each of the receiver's elements. The returned array will contain
NSNull elements for each instance of -valueForKey: returning nil.
*/
- (id)valueForKey:(NSString *)key;

Answer 2

- (NSDictionary *) indexKeyedDictionaryFromArray:(NSArray *)array 
{
  id objectInstance;
  NSUInteger indexKey = 0U;

  NSMutableDictionary *mutableDictionary = [[NSMutableDictionary alloc] init];
  for (objectInstance in array)
    [mutableDictionary setObject:objectInstance forKey:[NSNumber numberWithUnsignedInt:indexKey++]];

  return (NSDictionary *)[mutableDictionary autorelease];
}

Answer 3

這為NSArray添加了類別擴展。 需要C99模式（這是默認情況，但以防萬一）。

在一個.h文件的某個地方可以#import ed all ...

@interface NSArray (indexKeyedDictionaryExtension)
- (NSDictionary *)indexKeyedDictionary
@end

在.m文件中..

@implementation NSArray (indexKeyedDictionaryExtension)

- (NSDictionary *)indexKeyedDictionary
{
  NSUInteger arrayCount = [self count];
  id arrayObjects[arrayCount], objectKeys[arrayCount];

  [self getObjects:arrayObjects range:NSMakeRange(0UL, arrayCount)];
  for(NSUInteger index = 0UL; index < arrayCount; index++) { objectKeys[index] = [NSNumber numberWithUnsignedInteger:index]; }

  return([NSDictionary dictionaryWithObjects:arrayObjects forKeys:objectKeys count:arrayCount]);
}

@end

使用示例：

NSArray *array = [NSArray arrayWithObjects:@"zero", @"one", @"two", NULL];
NSDictionary *dictionary = [array indexKeyedDictionary];

NSLog(@"dictionary: %@", dictionary);

輸出：

2009-09-12 08:41:53.128 test[66757:903] dictionary: {
    0 = zero;
    1 = one;
    2 = two;
}

Answer 4

這是從員工列表NSMutableArray創建NSMutableDictionary的示例：

 NSMutableArray *emloyees = [[NSMutableArray alloc]initWithObjects:@"saman",@"Ruchira",@"Rukshan",@"ishan",@"Harsha",@"Ghihan",@"Lakmali",@"Dasuni", nil];

 NSMutableDictionary *dict = [NSMutableDictionary dictionary];
 for (NSString *word in emloyees) {
 NSString *firstLetter = [[word substringToIndex:1] uppercaseString];
 letterList  = [dict objectForKey:firstLetter];

if (!letterList) {
    letterList = [NSMutableArray array];
    [dict setObject:letterList forKey:firstLetter];
}
[letterList addObject:word];}  NSLog(@"dic %@",dict);

Answer 5

- (NSDictionary *) indexKeyedDictionaryFromArray:(NSArray *)array
{
    NSMutableDictionary *mutableDictionary = [[NSMutableDictionary alloc] init];
    [array enumerateObjectsUsingBlock:
     ^(id obj, NSUInteger idx, BOOL *stop){
         NSNumber *index = [NSNumber numberWithInteger:idx];
         [mutableDictionary setObject:obj forKey:index];
     }];
    NSDictionary *result = [NSDictionary.alloc initWithDictionary:mutableDictionary];
    return result;
}

Answer 6

我創建了一個簡單的庫，名為Linq to ObjectiveC ，它是一組方法，使這類問題更容易解決。 在您的情況下，您需要Linq-to-ObjectiveC toDictionary方法，其中您的'int'字段是通過鍵選擇器指定的：

NSDictionary* dictionary = [sourceArray toDictionaryWithKeySelector:^id(id item) {
    return [item intField];
}];

這將返回一個字典，其中鍵由源數組中的intField指定。