Swift Dictionary获取值的关键

Question

我有一个字典定义为：

let drinks = [String:[String]]()

drinks = ["Soft Drinks": ["Cocoa-Cola", "Mountain Dew", "Sprite"], 
"Juice" :["Orange", "Apple", "Grape"]]

如何获取给定值的密钥。

let key = (drinks as NSDictionary).allKeysForObject("Orange") as! String
print(key)
//Returns an empty Array. Should return "Juice"

Answer 1

func findKeyForValue(value: String, dictionary: [String: [String]]) ->String?
{
    for (key, array) in dictionary
    {
        if (array.contains(value))
        {
            return key
        }
    }

    return nil
}

调用上面的函数，它将返回一个可选的String？

let drinks = ["Soft Drinks": ["Cocoa-Cola", "Mountain Dew", "Sprite"],
        "Juice" :["Orange", "Apple", "Grape"]]

print(self.findKeyForValue("Orange", dictionary: drinks))

此函数将仅返回已传递值的数组的第一个键。

Answer 2

在Swift 2.0中，您可以过滤字典，然后将结果映射到数组。

let keys = drinks.filter {
    return $0.1.contains("Orange")
}.map {
    return $0.0
}

结果将是一个String对象数组，表示匹配的键。

Answer 3

枚举所有字典条目并测试每个值列表中所需的值，并累积值所在的键。

例如，查找列表中包含所需值的所有饮品：

let drinks = [
    "Soft Drinks": ["Orange", "Cocoa-Cola", "Mountain Dew", "Sprite"],
    "Juice" :["Apple", "Grape"]
]

let value = "Orange"

var keys = [String]()
for (key, list) in drinks {
    if (list.contains(value)) {
        keys.append(key)
    }
}

print("keys: \(keys)")

键：[“软饮料”]

Answer 4

这是Swift 4中的一个简单解决方案：

let transport = ["bus": "red", "car": "white", "TARDIS": "blue"]

let key = (transport.filter { $0.value == "blue" }).first?.key
{
    print (key)
}

TARDIS

Answer 5

试试这个（快速）：

(dic as NSDictionary).allKeysForObject(<#T##anObject: AnyObject##AnyObject#>)

它对我有用