假设我们有一个名为imageFile的自定义类,这个类包含两个属性:

class imageFile  {
    var fileName = String()
    var fileID = Int()
}

很多都存储在一个数组中:

var images : Array = []

var aImage = imageFile()
aImage.fileName = "image1.png"
aImage.fileID = 101
images.append(aImage)

aImage = imageFile()
aImage.fileName = "image1.png"
aImage.fileID = 202
images.append(aImage)

我如何排序的图像数组由'fileID'升序或降序?


当前回答

使用KeyPath排序

你可以像这样通过KeyPath排序:

myArray.sorted(by: \.fileName, <) /* using `<` for ascending sorting */

通过实现这个有用的扩展。

extension Collection{
    func sorted<Value: Comparable>(
        by keyPath: KeyPath<Element, Value>,
        _ comparator: (_ lhs: Value, _ rhs: Value) -> Bool) -> [Element] {
        sorted { comparator($0[keyPath: keyPath], $1[keyPath: keyPath]) }
    }
}

希望Swift在不久的将来把这个添加到语言的核心中。

其他回答

如果你不使用自定义对象,而是使用值类型来实现可比协议(Int, String等),你可以简单地这样做:

myArray.sort(>) //sort descending order

一个例子:

struct MyStruct: Comparable {
    var name = "Untitled"
}

func <(lhs: MyStruct, rhs: MyStruct) -> Bool {
    return lhs.name < rhs.name
}
// Implementation of == required by Equatable
func ==(lhs: MyStruct, rhs: MyStruct) -> Bool {
    return lhs.name == rhs.name
}

let value1 = MyStruct()
var value2 = MyStruct()

value2.name = "A New Name"

var anArray:[MyStruct] = []
anArray.append(value1)
anArray.append(value2)

anArray.sort(>) // This will sort the array in descending order
var students = ["Kofi", "Abena", "Peter", "Kweku", "Akosua"]

students.sort(by: >)

print(students)

彼得Prints:“[””、“科菲Kweku”、“”、“Akosua Abena”、“]”

你也可以这样做

images = sorted(images) {$0.fileID > $1.fileID}

你的图像数组会以排序的方式存储

几乎每个人都给出了如何直接,让我来展示一下演变:

你可以使用Array的实例方法:

// general form of closure
images.sortInPlace({ (image1: imageFile, image2: imageFile) -> Bool in return image1.fileID > image2.fileID })

// types of closure's parameters and return value can be inferred by Swift, so they are omitted along with the return arrow (->)
images.sortInPlace({ image1, image2 in return image1.fileID > image2.fileID })

// Single-expression closures can implicitly return the result of their single expression by omitting the "return" keyword
images.sortInPlace({ image1, image2 in image1.fileID > image2.fileID })

// closure's argument list along with "in" keyword can be omitted, $0, $1, $2, and so on are used to refer the closure's first, second, third arguments and so on
images.sortInPlace({ $0.fileID > $1.fileID })

// the simplification of the closure is the same
images = images.sort({ (image1: imageFile, image2: imageFile) -> Bool in return image1.fileID > image2.fileID })
images = images.sort({ image1, image2 in return image1.fileID > image2.fileID })
images = images.sort({ image1, image2 in image1.fileID > image2.fileID })
images = images.sort({ $0.fileID > $1.fileID })

关于排序的工作原理的详细解释,请参见排序函数。

斯威夫特3

people = people.sorted(by: { $0.email > $1.email })