1

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?

4

3 回答 3

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]];
于 2012-09-24T18:28:41.150 回答
0

卡洛斯的回答是正确的,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 分成两个数组(对象和键),然后按对象数组(或类似的解决方案)排序

于 2012-09-24T19:46:36.867 回答
0

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。

于 2012-09-24T20:47:13.730 回答