检查字典数组的值是否与数组元素列表匹配

Question

我在Swift中有一个数组（items）和一组字典（数据）：

let items = [2, 6, 4]

var data = [
    ["id": "1", "title": "Leslie", "color": "brown"],
    ["id": "8", "title": "Mary", "color": "red"],
    ["id": "6", "title": "Joe", "color": "blue"],
    ["id": "2", "title": "Paul", "color": "gray"],
    ["id": "5", "title": "Stephanie", "color": "pink"],
    ["id": "9", "title": "Steve", "color": "purple"],
    ["id": "3", "title": "Doug", "color": "violet"],
    ["id": "4", "title": "Ken", "color": "white"],
    ["id": "7", "title": "Annie", "color": "black"]
]

我想创建一个包含那些＆＃34; id＆＃34;的字典数组的数组。等于＆＃39;项目中提供的数字＆＃39;阵列。 Aka我想最终得到一个阵列：

var result = [
    ["id": "6", "title": "Joe", "color": "blue"],
    ["id": "2", "title": "Paul", "color": "gray"],
    ["id": "4", "title": "Ken", "color": "white"]
]

我试图使用谓词，但在经历了严重的头痛和心脏骤停后，我并没有使用它们。对于这项任务来说，他们看起来非常复杂。我现在处于一个我想在一个简单的for-in循环中执行此操作的地方。

使用谓词或其他方法有一种聪明的方法吗？

Answer 1

像这样使用size_t和std：

filter

Answer 2

试试这个，应该有效：

let items = [2, 6, 4]
var data = [
    ["id": "1", "title": "Leslie", "color": "brown"],
    ["id": "8", "title": "Mary", "color": "red"],
    ["id": "6", "title": "Joe", "color": "blue"],
    ["id": "2", "title": "Paul", "color": "gray"],
    ["id": "5", "title": "Stephanie", "color": "pink"],
    ["id": "9", "title": "Steve", "color": "purple"],
    ["id": "3", "title": "Doug", "color": "violet"],
    ["id": "4", "title": "Ken", "color": "white"],
    ["id": "7", "title": "Annie", "color": "black"]
]

var result = [Dictionary<String, String>]()

for index in 0..<data.count {
    let myData = data[index]
    let dataID = data[index]["id"]
    for i in 0..<items.count {
        if dataID == "\(items[i])" {
            result.append(myData)
        }
    }
}

for i in 0..<result.count {
    print(result[i])
}

结果是：

Answer 3

感谢大家的回复！我认为Eric D的解决方案最接近我所追求的目标。

我终于能够在raywenderlich.com上找到具有良好描述的解决方案：https://www.raywenderlich.com/82599/swift-functional-programming-tutorial

使用该方法（再次，非常接近Eric D的建议），首先我可以将其分解为：

func findItems(value: [String: String]) -> Bool {
    return items.contains(Int(value["id"]!)!)
}
var result = data.filter(findItems)

我可以进一步减少：

var result = data.filter {
    (value) in items.contains(Int(value["id"]!)!)
}

再次，可以减少到一行，如：

var result = data.filter { items.contains(Int($0["id"]!)!) }

再次感谢大家！