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Arrays 按特定映射筛选2个结构/数组

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Arrays 按特定映射筛选2个结构/数组,arrays,json,uitableview,swift3,Arrays,Json,Uitableview,Swift3,我是一个使用swift 3和xcode 8.3的新用户。当前面临筛选2阵列/结构的问题，其中控制台输出如下： A_List : Optional([117, 115, 18]) B_List : Optional([{ URL = "169.jpeg"; categories = "A"; description = "description XXX"; height = 128; id = 1; likes = "1.00"; name

我是一个使用swift 3和xcode 8.3的新用户。当前面临筛选2阵列/结构的问题，其中控制台输出如下：

A_List : Optional([117, 115, 18])

B_List : Optional([{
    URL = "169.jpeg";
    categories = "A";
    description = "description XXX";
    height = 128;
    id = 1;
    likes = "1.00";
    name = "Cake - Baked";
    price = "13.78";
    width = 128;
}, {
    URL = "1622.jpeg";
    categories = "A";
    description = "Baked till golden";
    height = 128;
    id = 15;
    likes = "1.00";
    name = "Croissant";
    price = "3.71";
    width = 128;
}, {
    URL = "11.jpeg";
    categories = "A";
    description = "description Crispy.";
    height = 128;
    id = 18;
    likes = "1.00";
    name = "Plain";
    price = "2.65";
    width = 128;
}, {
    URL = "1622.jpeg";
    categories = "A";
    description = "A ";
    height = 128;
    id = 103;
    likes = "1.00";
    name = "America Pie";
    price = "2.12";
    width = 128;
}, {
    URL = "11.jpeg";
    categories = "B";
    description = "Puff";
    height = 128;
    id = 115;
    likes = "1.00";
    name = "Puff";
    price = "2.12";
    width = 128;
}, {
    URL = "168.jpeg";
    categories = "C";
    description = "description YYY";
    height = 128;
    id = 117;
    likes = "1.00";
    name = "Normal";
    price = "3.18";
    width = 128;
}])

filtered_List : Optional([{
        URL = "11.jpeg";
        categories = "A";
        description = "description Crispy.";
        height = 128;
        id = 18;
        likes = "1.00";
        name = "Plain";
        price = "2.65";
        width = 128;
    }, {
        URL = "11.jpeg";
        categories = "B";
        description = "Puff";
        height = 128;
        id = 115;
        likes = "1.00";
        name = "Mini Puff";
        price = "2.12";
        width = 128;
    }, {
        URL = "168.jpeg";
        categories = "C";
        description = "description YYY";
        height = 128;
        id = 117;
        likes = "1.00";
        name = "Normal";
        price = "3.18";
        width = 128;
    }])

var filtered_List = [AnyObject]()
let fullrList = B_List?.map{$0["id"] as! String}.map{_ in A_List}
filtered_List.append(fullrList as AnyObject )
print("result :\(filtered_List)")
我想返回B_-List full info作为
var-filtered_-List=[AnyObject]（）
，其中仅包含一个列表id号
117、115和18，如下所示：


A_List : Optional([117, 115, 18])

B_List : Optional([{
    URL = "169.jpeg";
    categories = "A";
    description = "description XXX";
    height = 128;
    id = 1;
    likes = "1.00";
    name = "Cake - Baked";
    price = "13.78";
    width = 128;
}, {
    URL = "1622.jpeg";
    categories = "A";
    description = "Baked till golden";
    height = 128;
    id = 15;
    likes = "1.00";
    name = "Croissant";
    price = "3.71";
    width = 128;
}, {
    URL = "11.jpeg";
    categories = "A";
    description = "description Crispy.";
    height = 128;
    id = 18;
    likes = "1.00";
    name = "Plain";
    price = "2.65";
    width = 128;
}, {
    URL = "1622.jpeg";
    categories = "A";
    description = "A ";
    height = 128;
    id = 103;
    likes = "1.00";
    name = "America Pie";
    price = "2.12";
    width = 128;
}, {
    URL = "11.jpeg";
    categories = "B";
    description = "Puff";
    height = 128;
    id = 115;
    likes = "1.00";
    name = "Puff";
    price = "2.12";
    width = 128;
}, {
    URL = "168.jpeg";
    categories = "C";
    description = "description YYY";
    height = 128;
    id = 117;
    likes = "1.00";
    name = "Normal";
    price = "3.18";
    width = 128;
}])



filtered_List : Optional([{
        URL = "11.jpeg";
        categories = "A";
        description = "description Crispy.";
        height = 128;
        id = 18;
        likes = "1.00";
        name = "Plain";
        price = "2.65";
        width = 128;
    }, {
        URL = "11.jpeg";
        categories = "B";
        description = "Puff";
        height = 128;
        id = 115;
        likes = "1.00";
        name = "Mini Puff";
        price = "2.12";
        width = 128;
    }, {
        URL = "168.jpeg";
        categories = "C";
        description = "description YYY";
        height = 128;
        id = 117;
        likes = "1.00";
        name = "Normal";
        price = "3.18";
        width = 128;
    }])



var filtered_List = [AnyObject]()
let fullrList = B_List?.map{$0["id"] as! String}.map{_ in A_List}
filtered_List.append(fullrList as AnyObject )
print("result :\(filtered_List)")


我在youtube上尝试了一些代码和阅读教程，但不幸的是并没有找到任何解决方案，它仅限于swift2示例

目前，我的代码尝试如下：


A_List : Optional([117, 115, 18])

B_List : Optional([{
    URL = "169.jpeg";
    categories = "A";
    description = "description XXX";
    height = 128;
    id = 1;
    likes = "1.00";
    name = "Cake - Baked";
    price = "13.78";
    width = 128;
}, {
    URL = "1622.jpeg";
    categories = "A";
    description = "Baked till golden";
    height = 128;
    id = 15;
    likes = "1.00";
    name = "Croissant";
    price = "3.71";
    width = 128;
}, {
    URL = "11.jpeg";
    categories = "A";
    description = "description Crispy.";
    height = 128;
    id = 18;
    likes = "1.00";
    name = "Plain";
    price = "2.65";
    width = 128;
}, {
    URL = "1622.jpeg";
    categories = "A";
    description = "A ";
    height = 128;
    id = 103;
    likes = "1.00";
    name = "America Pie";
    price = "2.12";
    width = 128;
}, {
    URL = "11.jpeg";
    categories = "B";
    description = "Puff";
    height = 128;
    id = 115;
    likes = "1.00";
    name = "Puff";
    price = "2.12";
    width = 128;
}, {
    URL = "168.jpeg";
    categories = "C";
    description = "description YYY";
    height = 128;
    id = 117;
    likes = "1.00";
    name = "Normal";
    price = "3.18";
    width = 128;
}])



filtered_List : Optional([{
        URL = "11.jpeg";
        categories = "A";
        description = "description Crispy.";
        height = 128;
        id = 18;
        likes = "1.00";
        name = "Plain";
        price = "2.65";
        width = 128;
    }, {
        URL = "11.jpeg";
        categories = "B";
        description = "Puff";
        height = 128;
        id = 115;
        likes = "1.00";
        name = "Mini Puff";
        price = "2.12";
        width = 128;
    }, {
        URL = "168.jpeg";
        categories = "C";
        description = "description YYY";
        height = 128;
        id = 117;
        likes = "1.00";
        name = "Normal";
        price = "3.18";
        width = 128;
    }])



var filtered_List = [AnyObject]()
let fullrList = B_List?.map{$0["id"] as! String}.map{_ in A_List}
filtered_List.append(fullrList as AnyObject )
print("result :\(filtered_List)")



如果有专家能在这里指导或给出您的解决方案，我们将不胜感激
 您应该将所需ID存储在一个集合中，而不是一个数组中。您只需执行一个简单的
过滤器
操作：


let desiredIds: Set = [117, 115, 18]

B_List.filter{ $0["id"].map{ desiredIds.contains($0) } ?? false } as [AnyObject]



您应该将所需的ID存储在一个集合中，而不是一个数组中。您只需执行一个简单的
过滤器
操作：


let desiredIds: Set = [117, 115, 18]

B_List.filter{ $0["id"].map{ desiredIds.contains($0) } ?? false } as [AnyObject]



感谢所有专门回复@Alexander这个话题的人。在这里，我的解决方案与其他人分享，可能并不完美


var resultAnyObject = [AnyObject]()
var aListIds = [String]()

for i in A_List! {
    aListIds.append(i as! String)
}

//@Alexander : thanks for code below:
let desiredIds = aListIds

let fullList = B_List?.filter{ $0["id"].map{desiredIds.contains($0 as! String) } ?? false };

resultAnyObject.append(fullList as AnyObject)

print("Result of filtered_List :\(resultAnyObject)")



感谢所有专门回复@Alexander这个话题的人。在这里，我的解决方案与其他人分享，可能并不完美


var resultAnyObject = [AnyObject]()
var aListIds = [String]()

for i in A_List! {
    aListIds.append(i as! String)
}

//@Alexander : thanks for code below:
let desiredIds = aListIds

let fullList = B_List?.filter{ $0["id"].map{desiredIds.contains($0 as! String) } ?? false };

resultAnyObject.append(fullList as AnyObject)

print("Result of filtered_List :\(resultAnyObject)")



Swift的惯例是对变量名使用lowerCamelCase，而不是snake_大小写，当然也不是Upper_snake_大小写。此外，你的名字“A名单”、“B名单”绝对没有传达任何关于其内容的信息。我强烈建议你付出更多names@Peterhdd不要将代码格式用于与技术版本或名称不同的代码@Alexander感谢您的好意建议。是的，我接受您的评论，并将在我的代码中做得更好。@Peterhdd感谢您的更正。谢谢。@jmattheis，感谢您的commentSwift约定，变量名使用lowerCamelCase，而不是snake_大小写，绝对不是Upper_snake_大小写。此外，你的名字“A名单”、“B名单”绝对没有传达任何关于其内容的信息。我强烈建议你付出更多names@Peterhdd不要将代码格式用于与技术版本或名称不同的代码@Alexander感谢您的好意建议。是的，我接受您的评论，并将在我的代码中做得更好。@Peterhdd感谢您的更正。谢谢。@jmattheis，非常感谢您的评论谢谢您的回复。它起作用了！。无论如何，如何将_列表设置为DesiredId？因为一个列表会改变。错误显示“无法将“[AnyObject]”类型的值转换为指定类型的“Set”。@D_9268通过
Set
初始值设定项运行它。（例如
Set（someArray）
）。为什么
是
[AnyObject]
类型的
列表，而不是
[Int]
？谢谢。其他程序从JSON格式过滤了A\u列表。最后我发现A/B\u列表中的所有信息都是字符串。因此，我必须确保“id”“也使用字符串映射。@D_9268无论何时使用JSON，请确保您远离
Any
，
AnyObject
，
NSArray
，
NSDictionary
，
[string:Any]
，等等的世界。您想尽快到达类型安全区，请给出建议并指出我的弱点！我将学习更多关于如何在Swift中处理JSON的教程。谢谢你的回复。它起作用了！。无论如何，如何将_列表设置为DesiredId？因为一个列表会改变。错误显示“无法将“[AnyObject]”类型的值转换为指定类型的“Set”。@D_9268通过
Set
初始值设定项运行它。（例如
Set（someArray）
）。为什么
是
[AnyObject]
类型的
列表，而不是
[Int]
？谢谢。其他程序从JSON格式过滤了A\u列表。最后我发现A/B\u列表中的所有信息都是字符串。因此，我必须确保“id”“也使用字符串映射。@D_9268无论何时使用JSON，请确保您远离
Any
，
AnyObject
，
NSArray
，
NSDictionary
，
[string:Any]
，等等的世界。您想尽快到达类型安全区，请给出建议并指出我的弱点！我将学习更多关于在Swift中处理JSON内容的教程。您只需要
一个列表作为[String]
，或者
map
。避免使用重复的
append
调用循环您只需要
A_列表作为[String]
，或者
map
。避免重复的
append
调用循环