用条件合并数组 [英] Merge arrays with condition

问题描述

首先我的结构在数组中：

  struct Item {
 var id：Int 
 var name：String 
 var value：Int 
} 
  

这两个数组的第二个元素：

  let fisrt = Item（id：1，name：Oleg，value：3）
 let second = Item（id：2，name：Olexander，value： 5）
 
 let fisrtInSecond = Item（id：1，name：Bogdan，value：6）
 let secondInSecond = Item（id：2，name：Max，value： 9）
  

阵列：

  var fisrtArray = [fisrt，second] 
 let secondArray = [fisrtInSecond，secondInSecond] 
  

我很想使用集合的 zip 和 map 函数来存档结果。结果是由 id 更新 fisrtArray 元素名称。
$ b

示例： fisrtArray = [Item（id：1，name：Bogdan，value：3），Item（id：2，name ：Max，value：5）]

我知道如何通过简单循环来实现。但我在寻找更高级的函数式编程用法Swift。

我的实验：

  fisrtArray = zip（fisrtArray，secondArray）.map（）
  

问题我不知道如何在 map 函数中写入条件。条件应该是：

  if（$ 0.id == $ 1.id）{
 $ 0.name = $ 1。名称
} 
  

从讨论的注释中可以突出显示 zip 不适用于我的情况，因为我们应该遍历所有数组，以查找是否有类似的id，它们的顺序不同。

解决方案

下面的代码通过2个数组中的元素的顺序独立工作

  firstArray = firstArray.map {（item） - > Item 
 guard 
 let index = secondArray.index（其中：{$ 0.id == item.id}）
 else {返回项目} 
 var item = item 
 item.name = secondArray [index] .name 
 return item 
} 
  

< blockquote>

[Item（id：1，name：Bogdan，value：3），Item（id：2，name：Max，value：5）] \\\
/ p>

更新

以下版本使用 first（其中：方法由 Martin R建议

  firstArray = firstArray.map {item  - > Item in 
 guard let secondElm = secondArray.first（其中：{$ 0.id == item.id}）else { return item} 
 var item = item 
 item.name = secondElm.name 
 return item 
} 
  




I would like to merge two arrays with specific condition and update objects that they are containing.

First my struct that is in arrays:

struct Item {
    var id:Int
    var name:String
    var value:Int
}

Second elements for the two arrays:

let fisrt = Item(id: 1, name: "Oleg", value: 3)
let second = Item(id: 2, name: "Olexander", value:5)

let fisrtInSecond = Item(id: 1, name: "Bogdan", value: 6)
let secondInSecond = Item(id: 2, name: "Max", value: 9)

Arrays:

 var fisrtArray = [fisrt, second]
 let secondArray = [fisrtInSecond, secondInSecond]

I woudl like to use zip and map functions of the collection to achive result. Result is that fisrtArray elements names are updated by id.

Example: fisrtArray = [Item(id: 1, name: "Bogdan", value:3), Item(id: 2, name: "Max", value:5)]

I know how to do this via simple loops. But i am looking for more advanced usage of the functional programing is Swift.

My experiment:

fisrtArray = zip(fisrtArray, secondArray).map()

The main problem i do not know how to write condition in the map function. Condition should be:

if ($0.id == $1.id) {
   $0.name = $1.name
}

From the comment discussing it is possible to highlight that zip is not suitable in my case because we should iterate over all array to find if we have similar id's that are not in the same order.

解决方案

The following code does work independently by the order of the elements inside the 2 arrays

firstArray = firstArray.map { (item) -> Item in
    guard
        let index = secondArray.index(where: { $0.id == item.id })
        else { return item }
    var item = item
    item.name = secondArray[index].name
    return item
}

"[Item(id: 1, name: "Bogdan", value: 3), Item(id: 2, name: "Max", value: 5)]\n"

Update

The following version uses the first(where: method as suggested by Martin R.

firstArray = firstArray.map { item -> Item in
    guard let secondElm = secondArray.first(where: { $0.id == item.id }) else { return item }
    var item = item
    item.name = secondElm.name
    return item
}

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