Swift查找子字符串的所有匹配项
这里有一个Swift中String类的扩展,它返回给定子字符串的第一个字母的索引 有人能帮我做一下吗?这样它就可以返回所有事件的数组,而不仅仅是第一个事件 多谢各位Swift查找子字符串的所有匹配项,swift,string,swift3,Swift,String,Swift3,这里有一个Swift中String类的扩展,它返回给定子字符串的第一个字母的索引 有人能帮我做一下吗?这样它就可以返回所有事件的数组,而不仅仅是第一个事件 多谢各位 extension String { func indexOf(string : String) -> Int { var index = -1 if let range = self.range(of : string) { if !range.isEmpty
extension String {
func indexOf(string : String) -> Int {
var index = -1
if let range = self.range(of : string) {
if !range.isEmpty {
index = distance(from : self.startIndex, to : range.lowerBound)
}
}
return index
}
}
例如,不是返回值
50
而是像[50、74、91、103]
这样的函数,这里有两个函数。一个返回[Range]
,另一个返回[Range]
。如果你不需要前者,你可以把它保密。我将其设计为模仿范围(of:options:range:locale:)
方法,因此它支持所有相同的功能
import Foundation
extension String {
public func allRanges(
of aString: String,
options: String.CompareOptions = [],
range: Range<String.Index>? = nil,
locale: Locale? = nil
) -> [Range<String.Index>] {
// the slice within which to search
let slice = (range == nil) ? self[...] : self[range!]
var previousEnd = s.startIndex
var ranges = [Range<String.Index>]()
while let r = slice.range(
of: aString, options: options,
range: previousEnd ..< s.endIndex,
locale: locale
) {
if previousEnd != self.endIndex { // don't increment past the end
previousEnd = self.index(after: r.lowerBound)
}
ranges.append(r)
}
return ranges
}
public func allRanges(
of aString: String,
options: String.CompareOptions = [],
range: Range<String.Index>? = nil,
locale: Locale? = nil
) -> [Range<Int>] {
return allRanges(of: aString, options: options, range: range, locale: locale)
.map(indexRangeToIntRange)
}
private func indexRangeToIntRange(_ range: Range<String.Index>) -> Range<Int> {
return indexToInt(range.lowerBound) ..< indexToInt(range.upperBound)
}
private func indexToInt(_ index: String.Index) -> Int {
return self.distance(from: self.startIndex, to: index)
}
}
let s = "abc abc abc abc abc"
print(s.allRanges(of: "abc") as [Range<String.Index>])
print()
print(s.allRanges(of: "abc") as [Range<Int>])
<代码>导入基础
扩展字符串{
公共职能所有范围(
阿斯汀:弦,
选项:String.CompareOptions=[],
范围:范围?=零,
语言环境:语言环境?=nil
)->[范围]{
//要在其中搜索的切片
让切片=(范围==nil)?self[…]:self[range!]
var previousEnd=s.startIndex
变量范围=[范围]()
当r=slice.range时(
of:aString,options:options,
范围:previousEnd..您只需不断推进搜索范围,直到找不到子字符串的更多实例:
extension String {
func indicesOf(string: String) -> [Int] {
var indices = [Int]()
var searchStartIndex = self.startIndex
while searchStartIndex < self.endIndex,
let range = self.range(of: string, range: searchStartIndex..<self.endIndex),
!range.isEmpty
{
let index = distance(from: self.startIndex, to: range.lowerBound)
indices.append(index)
searchStartIndex = range.upperBound
}
return indices
}
}
let keyword = "a"
let html = "aaaa"
let indicies = html.indicesOf(string: keyword)
print(indicies) // [0, 1, 2, 3]
扩展字符串{
func indicatesof(字符串:字符串)->[Int]{
var指数=[Int]()
var searchStartIndex=self.startIndex
当searchStartIndex 让range=self.range(of:string,range:searchStartIndex..这可以通过递归方法完成。我使用了一个数字字符串来测试它。它返回一个可选数组Int
,这意味着如果找不到子字符串,它将为零
extension String {
func indexes(of string: String, offset: Int = 0) -> [Int]? {
if let range = self.range(of : string) {
if !range.isEmpty {
let index = distance(from : self.startIndex, to : range.lowerBound) + offset
var result = [index]
let substr = self.substring(from: range.upperBound)
if let substrIndexes = substr.indexes(of: string, offset: index + distance(from: range.lowerBound, to: range.upperBound)) {
result.append(contentsOf: substrIndexes)
}
return result
}
}
return nil
}
}
let numericString = "01234567890123456789012345678901234567890123456789012345678901234567890123456789"
numericString.indexes(of: "3456")
实际上没有一个内置函数来实现这一点,但是我们可以实现一个modified函数来获得我们想要匹配的字符串的所有索引。它也应该非常有效,因为我们不需要重复调用字符串上的range
extension String {
func indicesOf(string: String) -> [Int] {
// Converting to an array of utf8 characters makes indicing and comparing a lot easier
let search = self.utf8.map { $0 }
let word = string.utf8.map { $0 }
var indices = [Int]()
// m - the beginning of the current match in the search string
// i - the position of the current character in the string we're trying to match
var m = 0, i = 0
while m + i < search.count {
if word[i] == search[m+i] {
if i == word.count - 1 {
indices.append(m)
m += i + 1
i = 0
} else {
i += 1
}
} else {
m += 1
i = 0
}
}
return indices
}
}
扩展字符串{
func indicatesof(字符串:字符串)->[Int]{
//转换为utf8字符数组使标记和比较更加容易
让search=self.utf8.map{$0}
让word=string.utf8.map{$0}
var指数=[Int]()
//m-搜索字符串中当前匹配项的开头
//i-当前字符在我们试图匹配的字符串中的位置
var m=0,i=0
而m+i
我知道我们不是在玩代码高尔夫,但对于那些对不使用var
s或循环的函数式单行实现感兴趣的人来说,这是另一种可能的解决方案:
extension String {
func indices(of string: String) -> [Int] {
return indices.reduce([]) { $1.encodedOffset > ($0.last ?? -1) && self[$1...].hasPrefix(string) ? $0 + [$1.encodedOffset] : $0 }
}
}
我已经调整了接受的答案,以便可以配置区分大小写
extension String {
func allIndexes(of subString: String, caseSensitive: Bool = true) -> [Int] {
let subString = caseSensitive ? subString : subString.lowercased()
let mainString = caseSensitive ? self : self.lowercased()
var indices = [Int]()
var searchStartIndex = mainString.startIndex
while searchStartIndex < mainString.endIndex,
let range = mainString.range(of: subString, range: searchStartIndex..<mainString.endIndex),
!range.isEmpty
{
let index = distance(from: mainString.startIndex, to: range.lowerBound)
indices.append(index)
searchStartIndex = range.upperBound
}
return indices
}
}
扩展字符串{
func-allIndexes(子字符串的类型:String,区分大小写:Bool=true)->[Int]{
let subString=区分大小写?subString:subString.lowercased()
让mainString=区分大小写?self:self.lowercased()
var指数=[Int]()
var searchStartIndex=mainString.startIndex
当searchStartIndex 让range=mainString.range(of:subString,range:searchStartIndex..请检查以下答案,以便在多个位置查找多个项目
func indicesOf(string: String) -> [Int] {
var indices = [Int]()
var searchStartIndex = self.startIndex
while searchStartIndex < self.endIndex,
let range = self.range(of: string, range: searchStartIndex..<self.endIndex),
!range.isEmpty
{
let index = distance(from: self.startIndex, to: range.lowerBound)
indices.append(index)
searchStartIndex = range.upperBound
}
return indices
}
func attributedStringWithColor(_ strings: [String], color: UIColor, characterSpacing: UInt? = nil) -> NSAttributedString {
let attributedString = NSMutableAttributedString(string: self)
for string in strings {
let indexes = self.indicesOf(string: string)
for index in indexes {
let range = NSRange(location: index, length: string.count)
attributedString.addAttribute(NSAttributedString.Key.foregroundColor, value: color, range: range)
}
}
guard let characterSpacing = characterSpacing else {return attributedString}
attributedString.addAttribute(NSAttributedString.Key.kern, value: characterSpacing, range: NSRange(location: 0, length: attributedString.length))
return attributedString
}
并获得结果哨兵值,如-1
在Swift中没有位置。这就是选项的作用。而let
.brilliant.Hmm。这似乎失败了。let keyword=“hello”let html=“My websitehellohello”let indicies=html.indiciesof(string:keyword)print(indicies)
致命错误:无法增加超过endIndex的值
更简单的示例:let keyword=“A”let html=“A”let indicies=html。indiciesof(字符串:关键字)打印(标记)
致命错误:不能在endIndex之外递增
如果将两个相同的关键字放在一行中,它将只找到一个。但是如果将三个关键字放在一行中,它会找到两个。尽管如此,看起来要好得多。@Eugenio如果允许子字符串重叠,则每次将搜索开始索引提前一个字符:searchStartIndex=string.index(之后:searchStartIndex)
这不需要递归。这只是一个不必要的内存命中。Oooo非常好,我感谢您自己实现了这一点。我们不需要重复调用字符串上的range
这可能很好,很可能range
也只进行KMP搜索。这很好!我注意到一个小缺点:如果哟
let message = "Item 1 + Item 2 + Item 3"
message.attributedStringWithColor(["Item", "+"], color: UIColor.red)