Question

我有一组“白名单”键，我想根据这些键过滤NSDictionary;

NSArray *whiteList = @["one", "two"];
NSDictionary *dictionary = @{
    @"one" : @"yes",
    @"two" : @"yes",
    @"three" : @"no"
};
NSDictionary *whiteDictionary = [dictionary dictionaryWithValuesForKeys:whiteList];

产生：

{
    one = "yes";
    two = "yes;
}

但是，如果我的字典不包含一个或多个键：

NSArray *whiteList = @["one", "two"];
NSDictionary *dictionary = @{
    @"one" : @"yes"
};
NSDictionary *whiteDictionary = [dictionary dictionaryWithValuesForKeys:whiteList];

我回来了：

{
    one = "yes";
    two = "<null>";
}

有没有办法根据一组键过滤字典而不会获得不存在的键。

Answer 1

更新

您可以通过过滤原始字典的所有密钥，然后从过滤的密钥列表构建白名单字典来获得功能解决方案：

NSDictionary *whiteDictionary = [dictionary dictionaryWithValuesForKeys:[dictionary.allKeys filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"self IN %@", whiteList]]];,

或两行，通过为谓词声明一个单独的变量，来减小行的大小：

NSPredicate *whitelistPredicate = [NSPredicate predicateWithFormat:@"self IN %@", whiteList];
NSDictionary *whiteDictionary = [dictionary dictionaryWithValuesForKeys:[dictionary.allKeys filteredArrayUsingPredicate:whitelistPredicate]];,

原始回答

半功能方法看起来像这样：

NSDictionary *whiteDictionary = [dictionary dictionaryWithValuesForKeys:whiteList];
whiteDictionary = [whiteDictionary dictionaryWithValuesForKeys:[whiteDictionary keysOfEntriesPassingTest:^(NSString *key, id obj, BOOL *stop) { return obj != [NSNull null]; }]];

或者，您可以通过字典枚举并构建过滤后的字典：

NSMutableDictionary *whiteDictionary = [NSMutableDictionary dictionary];
[dictionary enumerateKeysAndObjectsUsingBlock:^(NSString *key, id obj, BOOL *stop) {
    if ([whiteList containsObject:key]) {
        whiteDictionary[key] = obj;
    }
}

Answer 2

如果您不关心在字典中对集合进行递归操作，那么这样的事情应该有效：

- (NSDictionary *)dictionaryWithNonNullValuesForKeys:(NSArray *)keys {
    NSMutableDictionary *mutableDictionary = [NSMutableDictionary dictionary];
    for (id key in keys) {
        id value = self[key];
        if ([value isKindOfClass:[NSNull class]]) {
            NSLog(@"Warning, value for key was null, skipping key %@", key);
        }
        else {
            mutableDictionary[key] = value;
        }
    }
    return [mutableDictionary copy];
}

根据键数组过滤NSDictionary

2 个答案: