Question

我是正规表达的新手。我有格式的字符串，如
1）372万人（国家排名：第6位）（2004年估计）
2）10000人（2007年估计）

我想从这两种字符串中提取种群数和时间。我怎么能用C＃中的正则表达式来做呢。或者我需要编写多个正则表达式吗？

Answer 1

这是一个起点：

(?<population>\d\(.\d+)?)  #capturing group named "population"
                           #that is a number, optionally followed by a
                           #decimal point and at least one number
\s*                        #followed by one or more spaces
(?<magnitude>thousand|(m|b)illion)? #optional capturing group named "magnitude"
                                    # that matches "thousand", "million", or "billion"
\s*                        #one or more whitespace characters
people                     #the literal "people"
.*                         #match any number of characters
\(                         #Find literal opening parentheses...
   (?<year>\d{4})          #...followed by a four-digit year...
\s                         #...followed by a space...
estimate\)                 #...followed by the phrase "estimate)"
\s*$                       #followed by optional whitespace
                           #and the end of the string

显示用法的简单驱动程序：

class Program
{
/// Generate test strings
static IEnumerable<string> Generator()
{
    yield return "3.72 million people (country rank: 6th) (2004 estimate)";
    yield return "10000 people (2007 estimate)";
}

public static void Main()
{
    string expression = @"
(?<population>\d(.\d+)?)  #capturing group named 'population'
                           #that is a number, optionally followed by a
                           #decimal point and at least one number
\s*                        #followed by one or more spaces
(?<magnitude>thousand|(m|b)illion)? #optional capturing group named 'magnitude'
                                    # that matches 'thousand', 'million', or 'billion'
\s*                        #one or more whitespace characters
people                     #the literal 'people'
.*                         #match any number of characters
\(                         #Find literal opening parentheses...
   (?<year>\d{4})          #...followed by a four-digit year...
\s                         #...followed by a space...
estimate\)                 #...followed by the phrase 'estimate'
\s*$                       #followed by optional whitespace
                           #and the end of the string";

    RegexOptions options = 
        RegexOptions.IgnorePatternWhitespace // allow whitespace/comments
        | RegexOptions.IgnoreCase
        | RegexOptions.ExplicitCapture; // Only capture named groups

    Regex r = new Regex(expression, options);
    foreach (var test in Generator())
    {
        Match match = r.Match(test);
        if (!match.Success)
            Console.WriteLine("Could not match {0}", test);
        else
        {
            double population = double.Parse(match.Groups["population"].Value);
            if (match.Groups["magnitude"].Success) // magnitude is optional
                                                   // but if present, need to
                                                   // multiply population
            {
                switch (match.Groups["magnitude"].Value.ToLower())
                {
                    case "thousand": population *= 1000; break;
                    case "million": population *= 1E6; break;
                    case "billion": population *= 1E9; break;
                    default: throw new FormatException("Unexpected value in magnitude group");
                }
            }
            int year = int.Parse(match.Groups["year"].Value);
            Console.WriteLine("In {0}, population was {1} people.", year, population);
        }
    }
}

输出：

In 2004, population was 3720000 people.
In 2007, population was 10000 people.

Answer 2

尝试：

(?<number>\d+.\d*)(?: million)? people(?: \(country rank: 6th\))? \((?<year>\d+) estimate\)

在http://regexhero.net/tester/上，它会给出以下结果： enter image description here

在http://myregextester.com/index.php中你得到： enter image description here

Answer 3

如果你的目标是这种模式，请尝试下一个Regex：

[population/number and text] people [some text] ([date] estimate)

正则表达式：

var match = Regex.Match(inputString, 
                        @"(?<number>[\.\d]+(\s+\w+)?)\s+people .+\((?<date>\d+)\s+estimate\)");

var population = match.Groups["number"].Value;
var date = match.Groups["date"].Value;

Answer 4

您可能需要两个正则表达式，因为您希望以不同方式处理它们。我复制了你的两行，包括“1）”和“2）”。这是人口（开头有空间）：

 \d+(?!\w)\.?(?=\d*)\d*

如果空格后跟一个字母，后跟一个或零点，则后跟一个或多个数字的空格，仅在下一个字符是一个或多个数字后才有效，后跟数字。对于像万/千的单词，你必须用零替换它们。

然后是日期部分：

(?:\()\d{4}(?!\d)

匹配左括号而不记住它，如果第五个不是数字则匹配四位数。

希望有所帮助。老实说，我不太了解c＃，我用JavaScript测试过。

编辑：其他人有更完整的答案，他们实际上是在c＃，去看看。

正则表达式提取多个数字和浮点值

4 个答案: