递归解析字符串

时间:2012-02-01 13:47:30

标签: c# regex

我正在尝试从字符串中提取信息 - 特定的fortran格式字符串。字符串的格式如下:

F8.3, I5, 3(5X, 2(A20,F10.3)), 'XXX'

格式化字段由“,”和括号内的格式化组分隔,括号前面的数字表示格式化模式重复连续的次数。因此,上面的字符串扩展为:

F8.3, I5, 5X, A20,F10.3, A20,F10.3, 5X, A20,F10.3, A20,F10.3, 5X, A20,F10.3, A20,F10.3, 'XXX'

我正在尝试在C#中创建一些能够扩展符合该模式的字符串。我已经开始使用大量的开关和if语句,但我想知道我是不是以错误的方式去做?

我基本上想知道一些Regex wizzard是否认为正则表达式可以在一个整齐的一举中做到这一点?我对正则表达式一无所知,但如果这可以解决我的问题,我正在考虑花一些时间来学习如何使用它们......另一方面,如果正则表达式无法解决这个问题,那么我宁愿花我的时间看另一种方法。

4 个答案:

答案 0 :(得分:1)

这必须适用于Regex :) 我已经扩展了我之前的示例,并且可以很好地测试您的示例。

// regex to match the inner most patterns of n(X) and capture the values of n and X.
private static readonly Regex matcher = new Regex(@"(\d+)\(([^(]*?)\)", RegexOptions.None);

// create new string by repeating X n times, separated with ','
private static string Join(Match m)
{
    var n = Convert.ToInt32(m.Groups[1].Value); // get value of n
    var x = m.Groups[2].Value; // get value of X
    return String.Join(",", Enumerable.Repeat(x, n));
}

// expand the string by recursively replacing the innermost values of n(X).
private static string Expand(string text)
{
    var s = matcher.Replace(text, Join);
    return (matcher.IsMatch(s)) ? Expand(s) : s;
}

// parse a string for occurenses of n(X) pattern and expand then.
// return the string as a tokenized array.
public static string[] Parse(string text)
{
    // Check that the number of parantheses is even.
    if (text.Sum(c => (c == '(' || c == ')') ? 1 : 0) % 2 == 1)
        throw new ArgumentException("The string contains an odd number of parantheses.");

    return Expand(text).Split(new[] { ',', ' ' }, StringSplitOptions.RemoveEmptyEntries);
}

答案 1 :(得分:0)

就个人而言,我建议使用递归函数。每次敲击左括号时,再次调用该函数来解析该部分。我不确定你是否可以使用正则表达式来匹配递归数据结构。

编辑:删除了错误的正则表达式)

答案 2 :(得分:0)

我建议使用类似下面示例的重复方法(未经测试):

ResultData Parse(String value, ref Int32 index)
{
    ResultData result = new ResultData();
    Index startIndex = index; // Used to get substrings

    while (index < value.Length) 
    {
        Char current = value[index];

        if (current == '(')
        {
            index++;
            result.Add(Parse(value, ref index));
            startIndex = index;
            continue;
        }
        if (current == ')')
        {
            // Push last result
           index++;
           return result;
        }

        // Process all other chars here
    }

    // We can't find the closing bracket
    throw new Exception("String is not valid");
}

您可能需要修改代码的某些部分,但我在编写简单编译器时使用了此方法。虽然还没有完成,只是一个例子。

答案 3 :(得分:0)

今天结束了改写。事实证明,这可以通过一种方法完成:

private static string ExpandBrackets(string Format)
    {
        int maxLevel = CountNesting(Format);

        for (int currentLevel = maxLevel; currentLevel > 0; currentLevel--)
        {
            int level = 0;
            int start = 0;
            int end = 0;

            for (int i = 0; i < Format.Length; i++)
            {
                char thisChar = Format[i];
                switch (Format[i])
                {
                    case '(':
                        level++;
                        if (level == currentLevel)
                        {
                            string group = string.Empty;
                            int repeat = 0;

                            /// Isolate the number of repeats if any
                            /// If there are 0 repeats the set to 1 so group will be replaced by itself with the brackets removed
                            for (int j = i - 1; j >= 0; j--)
                            {
                                char c = Format[j];
                                if (c == ',')
                                {
                                    start = j + 1;
                                    break;
                                }
                                if (char.IsDigit(c))
                                    repeat = int.Parse(c + (repeat != 0 ? repeat.ToString() : string.Empty));
                                else
                                    throw new Exception("Non-numeric character " + c + " found in front of the brackets");
                            }
                            if (repeat == 0)
                                repeat = 1;

                            /// Isolate the format group
                            /// Parse until the first closing bracket. Level is decremented as this effectively takes us down one level
                            for (int j = i + 1; j < Format.Length; j++)
                            {
                                char c = Format[j];
                                if (c == ')')
                                {
                                    level--;
                                    end = j;
                                    break;
                                }
                                group += c;
                            }

                            /// Substitute the expanded group for the original group in the format string
                            /// If the group is empty then just remove it from the string
                            if (string.IsNullOrEmpty(group))
                            {
                                Format = Format.Remove(start - 1, end - start + 2);
                                i = start;
                            }
                            else
                            {
                                string repeatedGroup = RepeatString(group, repeat);
                                Format = Format.Remove(start, end - start + 1).Insert(start, repeatedGroup);
                                i = start + repeatedGroup.Length - 1;
                            }
                        }
                        break;

                    case ')':
                        level--;
                        break;
                }
            }
        }

        return Format;
    }

CountNesting()返回格式语句中括号嵌套的最高级别,但可以作为参数传递给方法。 RepeatString()只重复指定次数的字符串,并将其替换为格式字符串中括号内的组。