如何使用swift中的filter函数删除数组的字典[字符串：AnyObject]？

Question

var globalCountArray = [AnyObject]()
var assetDictionary = [String:AnyObject]()
globalCountArray.append(assetDictionary as AnyObject)

如何使用Anyobject数组过滤字典？我试过这个。

  globalCountArray = globalCountArray.filter {$0     as AnyObject != dict }

但我错了

二进制运算符“！=”无法应用于“AnyObject”和“[字符串：AnyObject]”类型的操作数

Answer 1

您可以为字典编写扩展名：

extension Dictionary where Key: ExpressibleByStringLiteral, Value: AnyObject {
    func isEqual(_ dictionary: [String: AnyObject]) -> Bool {
        return NSDictionary(dictionary: dictionary).isEqual(to: self)
    }

    func isNotEqual(_ dictionary: [String: AnyObject]) -> Bool {
        return !NSDictionary(dictionary: dictionary).isEqual(to: self)
    }
}

然后，您可以将字典与以下内容进行比较：

FirstDictionary.isEqual(SecondDictionary)
FirstDictionary.isNotEqual(SecondDictionary)

在您的情况下：

var globalCountArray = [AnyObject]()
var assetDictionary = [String:AnyObject]()
globalCountArray.append(assetDictionary as AnyObject)

globalCountArray = globalCountArray.filter { (($0 as! [String: AnyObject]).isNotEqual(assetDictionary)) }

Answer 2

尝试使用此

    var globalCountArray = [AnyObject]()
    var assetDictionary = [String:AnyObject]()
    globalCountArray.append(assetDictionary as AnyObject)
    let dict = [String:AnyObject]()




    globalCountArray = globalCountArray.filter({ (obj) -> Bool in

        if obj is[String:AnyObject] {

            return (obj as! [String:AnyObject]) != dict 


        }
        return false
    })

---------或者你也可以通过----------

globalCountArray = globalCountArray.filter({ (obj) -> Bool in

            if obj is[String:AnyObject] {

                return (obj as! [String:AnyObject]) == dict 


            }
            return true
        })

您需要将此方法添加到类定义之外。

public func !=(lhs: [String: AnyObject], rhs: [String: AnyObject] ) -> Bool {
    return !NSDictionary(dictionary: lhs).isEqual(to: rhs)
}

 public func ==(lhs: [String: AnyObject], rhs: [String: AnyObject] ) -> Bool {
    return NSDictionary(dictionary: lhs).isEqual(to: rhs)
}