如果两个数组包含相同的元素，无论这些元素出现的顺序如何，我如何检入Swift？

Let's say there are two arrays...

假设有两个阵列......

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a"]

I'd like the result of the comparison of these two arrays to be true, and the following...

我希望这两个数组的比较结果是真实的，以下是......

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a", "d"]

...to be false. How can I achieve that in Swift? I tried to convert both arrays to sets but for some reason Set() keeps removing some (usually duplicated) objects that the array contains.

......是假的。我怎样才能在Swift中实现这一目标？我试图将两个数组转换为集合，但由于某种原因，Set（）不断删除数组包含的一些（通常是重复的）对象。

Any help would be appreciated.

任何帮助，将不胜感激。

4 个解决方案

#1

Swift 3, 4

斯威夫特3,4

extension Array where Element: Comparable {
    func containsSameElements(as other: [Element]) -> Bool {
        return self.count == other.count && self.sorted() == other.sorted()
    }
}

// usage
let a: [Int] = [1, 2, 3, 3, 3]
let b: [Int] = [1, 3, 3, 3, 2]
let c: [Int] = [1, 2, 2, 3, 3, 3]

print(a.containsSameElements(as: b)) // true
print(a.containsSameElements(as: c)) // false

#2

you can do something like this:

你可以这样做：

  array1.sortInPlace()
  array2.sortInPlace()

  print(array1,array2)

  if array1 == array2 {
    print("equal")
  } else {
  print("not equal") 
  }

and if don't want change origional array we can do

如果不想改变原始阵列，我们可以做到

 let sorted1 = array1.sort()
 let sorted2 = array2.sort()

  if sorted1 == sorted2 {
    print("equal")
  }else {
    print("not equal")
  }

#3

Create function to compare them:

创建比较它们的函数：

func containSameElements(var firstArray firstArray: [String], var secondArray: [String]) -> Bool {
    if firstArray.count != secondArray.count {
        return false
    } else {
        firstArray.sortInPlace()
        secondArray.sortInPlace()
        return firstArray == secondArray
    }
}

Then:

然后：

var array1 = ["a", "a", "b"]
var array2 = ["a", "b", "a"]

var array3 = ["a", "b", "c"]
var array4 = ["b", "c", "a", "d"]

print(containSameElements(firstArray: array1, secondArray: array2)) //true
print(containSameElements(firstArray: array3, secondArray: array4)) //false
print(array1) //["a", "a", "b"]
print(array2) //["a", "b", "a"]
print(array3) //["a", "b", "c"]
print(array4) //["b", "c", "a", "d"]

#4

Here is a solution that does not require the element to be Comparable, but only Equatable. It is much less efficient than the sorting answers, so if your type can be made Comparable, use one of those.

这是一个不需要元素可比较的解决方案，但只有Equatable。它的效率远低于排序答案，因此如果您的类型可以制作可比较，请使用其中一种。

extension Array where Element: Equatable {
    func equalContents(to other: [Element]) -> Bool {
        guard self.count == other.count else {return false}
        for e in self{
          guard self.filter{$0==e}.count == other.filter{$0==e}.count else {
            return false
          }
        }
        return true
    }
}

#1