如何在Swift中获取2个数组的常用元素列表？ [英] How to get list of common elements of 2 array in Swift?

问题描述

我有两个数组：

fruitsArray = ["apple", "mango", "blueberry", "orange"]
vegArray = ["tomato", "potato", "mango", "blueberry"]

如何获取这两个数组中的常用项目列表

How can I get the list of common items in those two array which gives

ouptput = ["mango", "blueberry"]

如果包含（数组，字符串）我不能使用 因为我想比较2个数组。

I can't use if contains(array, string) as I want to compare 2 arrays.

推荐答案

您还可以使用过滤器和包含：

let fruitsArray = ["apple", "mango", "blueberry", "orange"]
let vegArray = ["tomato", "potato", "mango", "blueberry"]

// only Swift 1
let output = fruitsArray.filter{ contains(vegArray, $0) }

// in Swift 2 and above
let output = fruitsArray.filter{ vegArray.contains($0) }
// or
let output = fruitsArray.filter(vegArray.contains)

---

设置 vs 数组用于公共元素的单次计算

我们考虑以下代码片段：

---

Set vs Array for a single computation of common elements

We consider the following code snippet:

let array1: Array = ...
let array2: Array = ...

// `Array`
let commonElements = array1.filter(array2.contains)

// vs `Set`
let commonElements = Array(Set(array1).intersection(Set(array2)))
// or (performance wise equivalent)
let commonElements: Array = Set(array1).filter(Set(array2).contains)

我用 Int 和短/长字符串 s（10）做了一些（人为的）基准测试到100 字符 s）（所有随机生成的）。我总是使用 array1.count == array2.count

I have made some (artificial) benchmarks with Int and short/long Strings (10 to 100 Characters) (all randomly generated). I always use array1.count == array2.count

我得到以下结果：

如果你有超过关键＃（元数）元素转换为设置更可取

If you have more than critical #(number of) elements converting to a Set is preferable

data         |  critical #elements
-------------|--------------------
         Int |        ~50
short String |       ~100
 long String |       ~200

结果说明

使用数组方法使用暴力搜索，其中时间复杂度 O（N ^ 2）其中 N = array1.count = array2.count 其中与 Set 方法 O（N）相反。但是，对于大数据，从数组到 Set 的转换非常昂贵，这解释了<$ c $的增加c>对于更大的数据类型，关键#elements 。

Explanation of the results

Using the Array approach uses "Brute force"-search which has time complexity O(N^2) where N = array1.count = array2.count which is in contrast to the Set approach O(N). However the conversion from Array to Set and back is very expensive for large data which explains the increase of critical #elements for bigger data types.

对于包含大约100个元素的小数组，数组方法很好但是对于较大的那些你应该使用 Set 方法。

For small Arrays with about 100 elements the Array approach is fine but for larger ones you should use the Set approach.

如果你想使用这个共同元素 - 操作多个如果可能的话，建议使用设置 s （元素的类型必须是 Hashable ）。

If you want to use this "common elements"-operation multiple times it is advisable to use Sets only if possible (the type of the elements has to be Hashable).

从数组转换到设置从 Set 转换为数组时价格昂贵相比之下非常便宜。

A conversion from Array to Set is kind of expensive while the conversion from Set to Array is in contrast very inexpensive.

使用过滤器与 .filter（array1.contains）的性能明显快于 .filter {array1.contains（$ 0）} 自：

Using filter with .filter(array1.contains) is performance wise faster than .filter{ array1.contains($0) } since:

• 最后一个创建一个新的闭包（只有一次）而第一个只传递一个函数指针


• ，对于最后一个，闭包的调用会创建一个额外的堆栈帧，这会花费空间和时间（多次： O（N））

• the last one creates a new closure (only once) whereas the first one passes only a function pointer
• for the last one the call of the closure creates an additional stack frame which costs space and time (multiple times: O(N))

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