如何在 iOS Swift 中使用过滤器删除字典中的元素

Question

我有一个看起来像这样的字典，我想删除 *所有包含 stock = 0 的元素。

    variations =     {
    "64FB5468-96E9-41A4-A779-3FB57FF1296B" =         {
        price = 10;
        sku = 123456;
        stock = 1;
        volume = 1L;
    };
    "A5EDB714-3D39-42DA-9A5C-2392D39C3AEB" =         {
        price = 20;
        sku = 254648;
        stock = 0;
        volume = 2L;
    };
    "BCA98B51-B058-4ABE-9010-54B080B819FB" =         {
        price = 30;
        sku = 284964;
        stock = 0;
        volume = 3L;
    };
};

我使用了下面的代码，但是，它在展开错误时返回 found nil 。 如何正确删除变体字典中库存为 0 的字典？ kSTOCK 是字符串“stock”的常量

    let filterAllItems = self.allItems.filter({(($0.variations as AnyObject)[$0] as! [String:Any])[kSTOCK] as! Int > 0})

Answer 1

假设allItems是一个[String:Any]字典，制作一个（可变） variations的副本，过滤对象，删除找到的键的值并将字典重新分配给allItems

if var variations = allItems["variations"] as? [String:[String:Any]] {
    let allItemsWithStock0 = variations.filter({($0.value["stock"] as? Int) == 0})
    for (key, _) in allItemsWithStock0 {
        variations.removeValue(forKey: key)
    }
    allItems["variations"] = variations
}

Answer 2

public extension Dictionary {
    @inlinable public mutating func removeIf(condition: (Self.Element) -> Bool) {
        for (key, _) in filter(condition) { removeValue(forKey: key) }
    }
}