从另一个Array 过滤Array 的值

I have two arrays:

我有两个数组：

let arr1 = ["one.json", "two.json", "three.json"]
let arr2 = ["one.json", "three.json"]

Now I want to remove all values of arr2 in arr1, so my expected result in the example above would be let arrFiltered = ["two.json"]. I know how to handle this using a for-loop, however, I thought there may is an easier and more performance-oriented solution?

现在我想删除arr1中arr2的所有值，所以我在上面的例子中的预期结果将是arrFiltered = [“two.json”]。我知道如何使用for循环来处理这个问题，但是，我认为可能有一个更简单，更面向性能的解决方案？

2 个解决方案

#1

You have to use Set instead of Array in this case.

在这种情况下，您必须使用Set而不是Array。

let arr1 = Set(["one.json", "two.json", "three.json"])
let arr2 = Set(["one.json", "three.json"])

arr1.subtract(arr2)

Fundamental Set Operations

基本集合运算

The illustration below depicts two sets–a and b– with the results of various set operations represented by the shaded regions.

下图描绘了两组-a和b-，其中各种设定操作的结果由阴影区域表示。

Use the intersect(_:) method to create a new set with only the values common to both sets.
使用intersect（_ :)方法创建一个新集合，其中只包含两个集合共有的值。
Use the exclusiveOr(_:) method to create a new set with values in either set, but not both.
使用exclusiveOr（_ :)方法创建一个新集合，其中包含任一集合中的值，但不能同时创建两者。
Use the union(_:) method to create a new set with all of the values in both sets.
使用union（_ :)方法创建一个包含两个集合中所有值的新集合。
Use the subtract(_:) method to create a new set with values not in the specified set.
使用subtract（_ :)方法创建一个值不在指定集中的新集。

#2

Solution using the filter function

解决方案使用过滤功能

let arr1 = ["one.json", "two.json", "three.json"]
let arr2 = ["one.json", "three.json"]

let arrFiltered = arr1.filter{ !arr2.contains($0) }

#1