在 Swift 中将 2 个字典分组到一个数组中

Question

当我的数字数组像 -> ["1","2","3","4","5","6","7"]

let numbers = ["1","2","3","4","5","6","7"]
let chunkSize = 2
let chunks = stride(from: 0, to: numbers.count, by: chunkSize).map {
    Array(numbers[$0..<min($0 + chunkSize, numbers.count)])
}

// prints as [["1", "2"], ["3", "4"], ["5", "6"], ["7"]]

但是当我的数字数组是这样的时候

   (
        {
        "facility_id" = 1;
        "options_id" = 3;
    },
        {
        "facility_id" = 3;
        "options_id" = 12;
    },
        {
        "facility_id" = 2;
        "options_id" = 7;
    },
        {
        "facility_id" = 3;
        "options_id" = 12;
    },
        {
        "facility_id" = 2;
        "options_id" = 6;
    },
        {
        "facility_id" = 1;
        "options_id" = 4;
    }
  ) 

它正在显示此错误：

无法将“CountableRange”类型的值转换为预期的参数类型“Int”

Answer 1

您的代码看起来正确。问题在于您的输入数组。来自您的一条评论

@adev -> var dataArray:NSMutableArray = NSMutableArray.init() -

您使用的是NSMutableArray，而不是快速的Array。 NSMutableArray 不能使用 CountableRange<Int> 下标

$0..<min($0 + chunkSize, numbers.count)  //returns CountableRange<Int>

您需要使用 swift Array 来完成这项工作

Answer 2

尝试以下解决方案：

let numbers = [["id":1],["id":2],["id":3],["id":4],["id":5],["id":6],["id":7]]//["1","2","3","4","5","6","7"]
let chunkSize = 2
let chunks = stride(from: 0, to: numbers.count, by: chunkSize).map  {
    Array(numbers[$0..<min($0 + chunkSize, numbers.count)])
}
print("chunks :: ", chunks)

// 打印为[[["id": 1], ["id": 2]], [["id": 3], ["id": 4]], [["id": 5], ["id": 6]], [["id": 7]]]

如果你只想要数字，试试下面的代码：

let numbersDictArray = [["id":1],["id":2],["id":3],["id":4],["id":5],["id":6],["id":7]]
let numbers = numbersDictArray.flatMap { $0.values }
let chunkSize = 2
let chunks = stride(from: 0, to: numbers.count, by: chunkSize).map  {
     Array(numbers[$0..<min($0 + chunkSize, numbers.count)])
}
print("chunks :: ", chunks)

// 打印为[[1, 2], [3, 4], [5, 6], [7]]

Answer 3

如果这段代码

[{'id':1},{'id':2},{'id':3},{'id':4},{'id':5},{'id':6},{'id':7}]

是Dictionary，那么你的策略应该是这样的：

1. 将所有值放入一个普通数组中；
2. 像你一样使用你的价值观。

例子：

let dict = [["id": 1], ["id": 2], ["id": 3], ["id": 4]]
let numbers = dict.flatMap { $0.values } // [1, 2, 3, 4]
let chunkSize = 2
let chunks = stride(from: 0, to: numbers.count, by: chunkSize).map {
    Array(numbers[$0..<min($0 + chunkSize, numbers.count)])
}
print(chunks) // [[1, 2], [3, 4]]

更新。

我看到numbers 是同一个变量，但它可以包含不同的值。在这种情况下，您可以检查numbers的类型，然后使用适当的方式进行分组。

var values: [Int] = []

if let numbers = numbers as? [String] {
    values = numbers.map { Int($0) }
} else if let numbers = numbers as? [[String, Int]] {
    values = numbers.compactMap { $0["id"] }
}

然后在values上使用跨步：

let chunkSize = 2
let chunks = stride(from: 0, to: numbers.count, by: chunkSize).map {
    Array(values[$0..<min($0 + chunkSize, numbers.count)])
}
print(chunks) // [[1, 2], [3, 4]]

注意，numbers 应该是 swift Array，而不是 NSMutableArray，因为 NSMutableArray 不能使用 CountableRange 来下标。