从文本文件读取双精度值

Question

尝试使用C＃应用程序从文本文件读取数据。有多行数据，每行以整数开头，然后是一堆双精度值。文本文件的一部分看起来像这样，

   33 0.573140941467E-01 0.112914262390E-03 0.255553577735E-02 0.497192659486E-04 0.141869181079E-01-0.147813598922E-03
   34 0.570076593453E-01 0.100112550891E-03 0.256427138318E-02-0.868691490164E-05 0.142821920093E-01-0.346011975369E-03
   35 0.715507714946E-01 0.316132133031E-03-0.106581466521E-01-0.920513736900E-04 0.138018668842E-01-0.212219497066E-03

33、34、35是整数值，后跟6个double值。并且不能保证这些双精度值之间有空格或其他定界符。即，如果双精度数为负数，则其前面将带有“-”，这将占用空间。因此，基本上所有6个double值都可能在一起。

现在的挑战是，如何优雅地提取它？

我试过的

String.Split(' ');

这将无法正常工作，因为不能保证在初始整数值与其余双精度值之间留有空格。

这可以使用C ++在C ++中轻松解决sscanf。

String.Split(' ');

包含双精度值的文本文件是由第三方工具生成的，我无法控制其输出。

有没有一种方法可以逐行优雅地提取整数和双精度值？

Answer 1

如果我认为正确，则您具有“固定宽度数据”格式。比您可以简单地解析这个事实。

即假设值在文件中d:\temp\doubles.txt：

void Main()
{
    var filename = @"d:\temp\doubles.txt";
    Func<string, string[]> split = (s) =>
    {
        string[] res = new string[7];
        res[0] = s.Substring(0, 2);
        for (int i = 0; i < 6; i++)
        {
            res[i + 1] = s.Substring(2 + (i * 19), 19);
        }
        return res;
    };
    var result = from l in File.ReadAllLines(filename)
                 let la = split(l)
                 select new
                 {
                    i = int.Parse(la[0]),
                     d1 = double.Parse(la[1]),
                     d2 = double.Parse(la[2]),
                     d3 = double.Parse(la[3]),
                     d4 = double.Parse(la[4]),
                     d5 = double.Parse(la[5]),
                     d6 = double.Parse(la[6])

                 };
    foreach (var e in result)
    {
        Console.WriteLine($"{e.i}, {e.d1}, {e.d2}, {e.d3}, {e.d4}, {e.d5}, {e.d6}");
    }
}

输出：

33, 0.0573140941467, 0.00011291426239, 0.00255553577735, 4.97192659486E-05, 0.0141869181079, -0.000147813598922
34, 0.0570076593453, 0.000100112550891, 0.00256427138318, -8.68691490164E-06, 0.0142821920093, -0.000346011975369
35, 0.0715507714946, 0.000316132133031, -0.0106581466521, -9.205137369E-05, 0.0138018668842, -0.000212219497066

PS：使用您的准确数据，int应该分配更多空间。

Answer 2

用正则表达式解决。我的第一枪是：

"[\s-+]\d+\.\d+E[+-]\d\d"

我只是这样尝试的：

using System;
using System.Globalization;
using System.Text.RegularExpressions;

namespace ConsoleApp1 {

    class Program {
        static void Main(string[] args) {
            var fileContents =
                  "33 0.573140941467E-01 0.112914262390E-03 0.255553577735E-02 0.497192659486E-04 0.141869181079E-01-0.147813598922E-03"
                + "34 0.570076593453E-01 0.100112550891E-03 0.256427138318E-02-0.868691490164E-05 0.142821920093E-01-0.346011975369E-03"
                + "35 0.715507714946E-01 0.316132133031E-03-0.106581466521E-01-0.920513736900E-04 0.138018668842E-01-0.212219497066E-03";

            var rex = new Regex(@"[\s-+]\d+\.\d+E[+-]\d\d", RegexOptions.Multiline);
            foreach (Match match in rex.Matches(fileContents)) {
                double d = double.Parse(match.Value.TrimStart(), NumberFormatInfo.InvariantInfo);
                Console.WriteLine("found a match: " + match.Value.TrimStart() + " => " + d);
            }

            Console.ReadLine();
        }
    }
}

使用以下输出（德语本地化，逗号作为小数点分隔符）：

found a match: 0.573140941467E-01 => 0,0573140941467
found a match: 0.112914262390E-03 => 0,00011291426239
found a match: 0.255553577735E-02 => 0,00255553577735
found a match: 0.497192659486E-04 => 4,97192659486E-05
found a match: 0.141869181079E-01 => 0,0141869181079
found a match: -0.147813598922E-03 => -0,000147813598922
found a match: 0.570076593453E-01 => 0,0570076593453
found a match: 0.100112550891E-03 => 0,000100112550891
found a match: 0.256427138318E-02 => 0,00256427138318
found a match: -0.868691490164E-05 => -8,68691490164E-06
found a match: 0.142821920093E-01 => 0,0142821920093
found a match: -0.346011975369E-03 => -0,000346011975369
found a match: 0.715507714946E-01 => 0,0715507714946
found a match: 0.316132133031E-03 => 0,000316132133031
found a match: -0.106581466521E-01 => -0,0106581466521
found a match: -0.920513736900E-04 => -9,205137369E-05
found a match: 0.138018668842E-01 => 0,0138018668842
found a match: -0.212219497066E-03 => -0,000212219497066