我有一系列可以选择的问题
[
{
"id": 1,
"title": "test",
"info": "Test123",
"is_selected": true
},
{
"id": 2,
"title": "test2",
"info": "test2",
"is_selected": false
},
{
"id": 3,
"title": "test23",
"info": "test23",
"is_selected": true
}
]
我们如何将具有多个键的字典数组简化为具有单个键的字典数组
[
{
"question_id": 1
},
{
"question_id": 2
}
]
答案 0 :(得分:2)
您可以使用以下内容映射数组:
let secondArray: [[String : Int]] = array.compactMap { dict in
guard let id = dict["id"] as? Int else { return nil }
return ["question_id" : id]
}
但是用一个键而不是仅仅一个值数组返回字典的意义是什么...(?)
let questionIds = array.compactMap { dict in
return dict["id"]
}
答案 1 :(得分:0)
您可以使用reduce
,然后在其闭包内部获取与键id
相对应的值,并将其存储在键question_id
下的新字典中。
let dictionariesWithSingleKey = dictionariesWithManyKeys.reduce(into: [[String:Int]](), { result, current in
guard let questionId = current["id"] as? Int else { return }
let singleKeyedDict = ["question_id": questionId]
result.append(singleKeyedDict)
})
print(dictionariesWithSingleKey)
答案 2 :(得分:0)
enum Const: String {
case id = "id"
case questionId = "question_id"
}
let reduced = dict.reduce([]) {
guard let element = $1[Const.id.rawValue] else {
return $0
}
return $0 + [[Const.questionId.rawValue: element]]
}
或者您可以将其简化为一个整数数组(仅适用于没有任何“键”的商店ID,因为每个字典中的ID都相同)