ios - 根据键数组过滤NSDictionary

Question

我有一个“列入白名单”的键数组，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>";
}

有没有一种方法可以根据一组键过滤字典，而不会得到不存在的键。

Answer 1

更新

您可以通过过滤原始字典的所有键，然后从过滤后的键列表构建列入白名单的字典来获得功能解决方案:

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

或在两行中，通过为谓词声明一个单独的变量，以减小行的大小:

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

原始答案

半功能方法如下所示:

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

另外，您可以枚举字典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;
    }
}