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Question

复杂的字典（关联数组）调用过滤器我有此数组：

class Filter { 

    var key = "" 
    var value = "" 

    init(key: String, value: String) { 
     self.key = key 
     self.value = value 
    } 

} 

let arry = [ 
    "a":[Filter(key:"city",value:"aachen"),Filter(key:"city",value:"augsburg")], 
    "b":[Filter(key:"city",value:"bremen"),Filter(key:"city",value:"berlin")] 
] 

，我想寻找奥格斯堡等输出看起来像这样从具有过滤功能的词典中删除：

let arry = [ 
    "a":[Filter(key:"city",value:"aachen")], 
    "b":[Filter(key:"city",value:"bremen"),Filter(key:"city",value:"berlin")] 
] 

我与许多过滤器和地图constelations尝试过，但我总是得到这样的结构，结果是：

let arry = [ 
    ["a":[Filter(key:"city",value:"aachen")]], 
    ["b":[Filter(key:"city",value:"bremen"),Filter(key:"city",value:"berlin")]] 
] 

例如使用此筛选器：

arry.map({ key,values in 

    return [key:values.filter{$0.value != "augsburg"}] 
}) 

这里有什么问题？我如何过滤和映射更复杂的对象？

Answer 1

也许你应该知道的一件事是map方法Dictionary返回Array而不是Dictionary。

public func map<T>(_ transform: (Key, Value) throws -> T) rethrows -> [T] 

所以，如果你想过滤的结果作为Dictionary，您可能需要使用reduce：

class Filter: CustomStringConvertible { 

    var key = "" 
    var value = "" 

    init(key: String, value: String) { 
     self.key = key 
     self.value = value 
    } 

    //For debugging 
    var description: String { 
     return "<Filter: key=\(key), value=\(value)>" 
    } 
} 

let dict = [ 
    "a":[Filter(key:"city",value:"aachen"),Filter(key:"city",value:"augsburg")], 
    "b":[Filter(key:"city",value:"bremen"),Filter(key:"city",value:"berlin")] 
] 

let filteredDict = dict.reduce([:]) {tempDict, nextPair in 
    var mutableDict = tempDict 
    mutableDict[nextPair.key] = nextPair.value.filter {$0.value != "augsburg"} 
    return mutableDict 
} 

（一般情况下，斯威夫特Dictionary是关联数组的哈希表基础实现，但你。应该更好地避免命名为arry变量Dictionary这样命名是如此混乱）

要不干脆用for-in循环：

var resultDict: [String: [Filter]] = [:] 
for (key, value) in dict { 
    resultDict[key] = value.filter {$0.value != "augsburg"} 
} 
print(resultDict) //->["b": [<Filter: key=city, value=bremen>, <Filter: key=city, value=berlin>], "a": [<Filter: key=city, value=aachen>]]