获取按值排序的 NSDictionary 键的 NSArray，然后按键

Question

假设我有一个包含键（单词）和值（分数）的字典，如下所示：

GOD    8
DONG   16
DOG    8
XI     21

我想创建一个字典键（单词）的 NSArray，它首先按分数排序，然后按字母顺序排序。从上面的示例中，这将是：

XI
DONG
DOG
GOD

实现这一目标的最佳方法是什么？

Answer 1

我会使用 NSDictionaries 和 NSArray，然后使用 NSSortDescriptors 实现它：

NSSortDescriptor *sdScore = [NSSortDescriptor alloc] initWithKey:@"SCORE" ascending:NO];
NSSortDescriptor *sdName = [NSSortDescriptor alloc] initWithKey:@"NAME" ascending:YES];
NSArray *sortedArrayOfDic = [unsortedArrayOfDic sortedArrayUsingDescriptors:[NSArray arrayWithObjects: sdScore, sdName, nil]];

Answer 2

Carlos 的回答是正确的，我只是发布了我最终得到的完整代码，以防万一有人感兴趣：

NSDictionary *dataSourceDict = [NSDictionary dictionaryWithObjectsAndKeys:
                      [NSNumber numberWithInt:8], @"GOD",
                      [NSNumber numberWithInt:16], @"DONG",
                      [NSNumber numberWithInt:8], @"DOG",
                      [NSNumber numberWithInt:21], @"XI", nil];

NSSortDescriptor *scoreSort = [NSSortDescriptor sortDescriptorWithKey:@"SCORE" ascending:NO];
NSSortDescriptor *wordSort = [NSSortDescriptor sortDescriptorWithKey:@"WORD" ascending:YES];
NSArray *sorts = [NSArray arrayWithObjects:scoreSort, wordSort, nil];


NSMutableArray *unsortedArrayOfDict = [NSMutableArray array];

for (NSString *word in dataSourceDict)
{
    NSString *score = [dataSourceDict objectForKey:word];
    [unsortedArrayOfDict addObject: [NSDictionary dictionaryWithObjectsAndKeys:word, @"WORD", score, @"SCORE",  nil]];
}
NSArray *sortedArrayOfDict = [unsortedArrayOfDict sortedArrayUsingDescriptors:sorts];

NSDictionary *sortedDict = [sortedArrayOfDict valueForKeyPath:@"WORD"];

NSLog(@"%@", sortedDict);

相关：NSDictionary split into two arrays (objects and keys) and then sorted both by the objects array (or a similar solution)

Answer 3

我无法对此进行测试，因为我不在 Mac 上（对不起，如果我拼错了什么），但是：

NSDictionary *dic1 = [NSDictionary dictionaryWithObjectsAndKeys:@"GOD", @"WORD", [NSNumber numberWithInt:8], @"SCORE", nil];
NSDictionary *dic2 = [NSDictionary dictionaryWithObjectsAndKeys:@"DONG", @"WORD", [NSNumber numberWithInt:16], @"SCORE", nil];
NSDictionary *dic3 = [NSDictionary dictionaryWithObjectsAndKeys:@"DOG", @"WORD", [NSNumber numberWithInt:8], @"SCORE", nil];
NSDictionary *dic4 = [NSDictionary dictionaryWithObjectsAndKeys:@"XI", @"WORD", [NSNumber numberWithInt:21], @"SCORE", nil];

NSSortDescriptor *scoreSort = [NSSortDescriptor sortDescriptorWithKey:@"SCORE" ascending:NO];
NSSortDescriptor *wordSort = [NSSortDescriptor sortDescriptorWithKey:@"WORD" ascending:YES];

NSArray *sortedArrayOfDic = [[NSArray arrayWithObjects:dic1, dic2, dic3, dic4, nil] sortedArrayUsingDescriptors:[NSArray arrayWithObjects:scoreSort, wordSort, nil]];

NSLog(@"%@", [sortedArrayOfDict valueForKeyPath:@"WORD"]);

这会做同样的事情，但会有所减少并避免迭代。