替换数组中的多次出现 - Swift 4.1
Replacing Multiple Occurrences in Array - Swift 4.1
替换数组中的多次出现
Swift4.1,Xcode9.3
我想做一个扩展,就像我为 Array
为 String
做的扩展一样。
字符串扩展:
public extension String {
///
/// Replaces multiple occurences of strings/characters/substrings with their associated values.
/// ````
/// var string = "Hello World"
/// let newString = string.replacingMultipleOccurrences(using: (of: "l", with: "1"), (of: "o", with: "0"), (of: "d", with: "d!"))
/// print(newString) //"He110 w0r1d!"
/// ````
///
/// - Returns:
/// String with specified parts replaced with their respective specified values.
///
/// - Parameters:
/// - array: Variadic values that specify what is being replaced with what value in the given string
///
public func replacingMultipleOccurrences<T: StringProtocol, U: StringProtocol>(using array: (of: T, with: U)...) -> String {
var str = self
for (a, b) in array {
str = str.replacingOccurrences(of: a, with: b)
}
return str
}
}
用法:
var string = "Hello World"
let newString = string.replacingMultipleOccurrences(using: (of: "l", with: "1"), (of: "o", with: "0"), (of: "d", with: "d!"))
print(newString) //"He110 w0r1d!"
到目前为止,我对数组的相同尝试
public extension Array {
public func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
var newArr : Array<Element> = self
var arr = array.filter { (arg) -> Bool in
let (a, b) = arg
return newArr.contains(a)
}
for (i,e) in self.enumerated() {
for (a,b) in arr {
if e == a {
newArr[i] = b
}
}
}
return newArr
}
}
注意:目前,此扩展正在产生大量错误。
理想用法:
let arr = [1,2,3,4,5,6,7,8,9]
let newArr = arr.replacingMultipleOccurrences(using: (of: 2, with: 20), (of: 3, with: 30), (of: 5, with: 50), (of: 8, with: 80), (of: 9, with: 90))
print(newArr) //[1,20,30,4,50,6,7,80,90]
我如何实现这个理想(最好以最有效的方式)?
我认为你应该使用 map
运算符。
extension String {
func replace(mapping : [String : String]) -> String {
return self.map { char -> String in
if let newValue = mapping[String(char)] {
return newValue
} else {
return String(char)
}
}
}
}
extension Array where Element: Equatable {
func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
var newArr: Array<Element> = self
for replacement in array {
for (index, item) in self.enumerated() {
if item == replacement.of {
newArr[index] = replacement.with
}
}
}
return newArr
}
}
如果数组元素是 Hashable
那么我会创建一个替换
词典 第一。然后可以有效地完成替换
一次遍历(使用 map
)和一次(快速)字典查找:
public extension Array where Element: Hashable {
public func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
let replacements = Dictionary<Element, Element>(array, uniquingKeysWith: { })
return map { replacements[[=10=]] ?? [=10=] }
}
}
如果替换值恰好是
替换后的替换键 table。
示例:
let arr = [1, 2, 3, 2, 1]
let newArr = arr.replacingMultipleOccurrences(using: (of: 2, with: 3), (of: 3, with: 4))
print(newArr) // [1, 3, 4, 3, 1]
对于只有 Equatable
个元素的数组,它可以实现
简洁地使用 map
和 first(where:)
:
public extension Array where Element: Equatable {
public func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
return map { elem in array.first(where: { [=12=].of == elem })?.with ?? elem }
}
}
替换数组中的多次出现
Swift4.1,Xcode9.3
我想做一个扩展,就像我为 Array
为 String
做的扩展一样。
字符串扩展:
public extension String {
///
/// Replaces multiple occurences of strings/characters/substrings with their associated values.
/// ````
/// var string = "Hello World"
/// let newString = string.replacingMultipleOccurrences(using: (of: "l", with: "1"), (of: "o", with: "0"), (of: "d", with: "d!"))
/// print(newString) //"He110 w0r1d!"
/// ````
///
/// - Returns:
/// String with specified parts replaced with their respective specified values.
///
/// - Parameters:
/// - array: Variadic values that specify what is being replaced with what value in the given string
///
public func replacingMultipleOccurrences<T: StringProtocol, U: StringProtocol>(using array: (of: T, with: U)...) -> String {
var str = self
for (a, b) in array {
str = str.replacingOccurrences(of: a, with: b)
}
return str
}
}
用法:
var string = "Hello World"
let newString = string.replacingMultipleOccurrences(using: (of: "l", with: "1"), (of: "o", with: "0"), (of: "d", with: "d!"))
print(newString) //"He110 w0r1d!"
到目前为止,我对数组的相同尝试
public extension Array {
public func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
var newArr : Array<Element> = self
var arr = array.filter { (arg) -> Bool in
let (a, b) = arg
return newArr.contains(a)
}
for (i,e) in self.enumerated() {
for (a,b) in arr {
if e == a {
newArr[i] = b
}
}
}
return newArr
}
}
注意:目前,此扩展正在产生大量错误。
理想用法:
let arr = [1,2,3,4,5,6,7,8,9]
let newArr = arr.replacingMultipleOccurrences(using: (of: 2, with: 20), (of: 3, with: 30), (of: 5, with: 50), (of: 8, with: 80), (of: 9, with: 90))
print(newArr) //[1,20,30,4,50,6,7,80,90]
我如何实现这个理想(最好以最有效的方式)?
我认为你应该使用 map
运算符。
extension String {
func replace(mapping : [String : String]) -> String {
return self.map { char -> String in
if let newValue = mapping[String(char)] {
return newValue
} else {
return String(char)
}
}
}
}
extension Array where Element: Equatable {
func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
var newArr: Array<Element> = self
for replacement in array {
for (index, item) in self.enumerated() {
if item == replacement.of {
newArr[index] = replacement.with
}
}
}
return newArr
}
}
如果数组元素是 Hashable
那么我会创建一个替换
词典 第一。然后可以有效地完成替换
一次遍历(使用 map
)和一次(快速)字典查找:
public extension Array where Element: Hashable {
public func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
let replacements = Dictionary<Element, Element>(array, uniquingKeysWith: { })
return map { replacements[[=10=]] ?? [=10=] }
}
}
如果替换值恰好是 替换后的替换键 table。 示例:
let arr = [1, 2, 3, 2, 1]
let newArr = arr.replacingMultipleOccurrences(using: (of: 2, with: 3), (of: 3, with: 4))
print(newArr) // [1, 3, 4, 3, 1]
对于只有 Equatable
个元素的数组,它可以实现
简洁地使用 map
和 first(where:)
:
public extension Array where Element: Equatable {
public func replacingMultipleOccurrences(using array: (of: Element, with: Element)...) -> Array {
return map { elem in array.first(where: { [=12=].of == elem })?.with ?? elem }
}
}