Question

如何从句子中提取带有整数，小数和分数的字符串数组？请在iOS Swift中找到以下输入和输出。

输入：字符串数组[＆＃34;宽度32.3＆＃34;，＆＃34;长度61 1/4＆＃34;，＆＃34;高度23 4/5＆＃34;，＆＃34;措施5.23＆＃34;]

输出：[＆＃34; 32.3＆＃34;，＆＃34; 61 1/4＆＃34;，＆＃34; 23 4/5＆＃34;，＆＃34; 5.23＆＃34;]

Answer 1

使用正则表达式的另一种方法

let array = ["width 32.3", "Length 61 1/4", "height 23 4/5", "measure 5.23"]
let pattern = "\\d+[\\s.][\\d/]+"

let regex = try! NSRegularExpression(pattern: pattern)
var result = [String]()

for item in array {
    if let match = regex.firstMatch(in: item, range: NSRange(location:0, length: item.characters.count)) {
        result.append((item as NSString).substring(with: match.range))
    }
}

print(result)

模式搜索

一个或多个数字
后跟一个空格字符或点
后跟一个或多个数字或斜线

Answer 2

添加此项以支持字符串的正则表达式匹配：

extension String {
    mutating func stringByRemovingRegexMatches(pattern: String, replaceWith: String = "") {
        do {
            let regex = try NSRegularExpression(pattern: pattern, options: NSRegularExpression.Options.caseInsensitive)
            let range = NSMakeRange(0, self.characters.count)
            self = regex.stringByReplacingMatches(in: self, options: [], range: range, withTemplate: replaceWith)
        } catch {
            return
        }
    }
}

然后：

var str = ["width 32.3", "Length 61 1/4", "height 23 4/5", "measure 5.23"]
var vreturn:[String]=[]

for var s in str {
    s.stringByRemovingRegexMatches(pattern:"[a-zA-Z]", replaceWith: "")
    vreturn.append(s)
}

print( vreturn)

来源：Swift replace substring regex

从字符串的句子中提取整数，小数和小数

2 个答案: