Question

我想查找字符串是否具有指数格式。目前我正在检查如下。

var s = "1.23E+04";
var hasExponential = s.Contains("E");

但我知道这不是正确的方法。那么有人可以指导正确和快速的方式来实现我的要求吗？

Answer 1

如果您还想确保它确实是一个数字，而不仅仅是一个带有＆＃39; E＆＃39;在其中，也许这样的功能可能会有所帮助。逻辑仍然很简单。

private bool IsExponentialFormat(string str)
{
    double dummy;
    return (str.Contains("E") || str.Contains("e")) && double.TryParse(str, out dummy);
}

Answer 2

尝试使用正则表达式？

string s = "1.23E+04";
string pattern = @"^\d{1}.\d+(E\+)\d+$";
Regex rgx = new Regex(pattern, RegexOptions.IgnoreCase);
bool hasExponential = rgx.IsMatch(s);

Answer 3

您可以随时在e上分开。或者使用double.TryParse。两者都很好。（我敢打赌ParseExponential2对于有效的ParseExponential1更快，而对于无效的public static void _Main(string[] args) { string[] exponentials = new string[] { "1.23E+4", "1.23E+04", "1.23e+4", "1.23", "1.23e+4e+4", "abce+def", "1.23E-04" }; for (int i = 0; i < exponentials.Length; i++) Console.WriteLine("Input: {0}; Result 1: {1}; Result 2: {2}; Result 3: {3}", exponentials[i], (ParseExponential1(exponentials[i]) ?? 0), (ParseExponential2(exponentials[i]) ?? 0), (ParseExponential3(exponentials[i]) ?? 0)); } public static double? ParseExponential1(string input) { if (input.Contains("e") || input.Contains("E")) { string[] inputSplit = input.Split(new char[] { 'e', 'E' }); if (inputSplit.Length == 2) // If there were not two elements split out, it's an invalid exponential. { double left = 0; int right = 0; if (double.TryParse(inputSplit[0], out left) && int.TryParse(inputSplit[1], out right) // Parse the values && (left >= -5.0d && left <= 5.0d && right >= -324) // Check that the values are within the range of a double, this is the minimum. && (left >= -1.7d && left <= 1.7d && right <= 308)) // Check that the values are within the range of a double, this is the maximum. { double result = 0; if (double.TryParse(input, out result)) return result; } } } return null; } public static double? ParseExponential2(string input) { if (input.Contains("e") || input.Contains("E")) { double result = 0; if (double.TryParse(input, out result)) return result; } return null; } public static double? ParseExponential3(string input) { double result = 0; if (double.TryParse(input, out result)) return result; return null; } 可以更快。）

ParseExponential1

如果ParseExponential2，ParseExponential3或null返回ParseExponential3，则无效。这也允许您同时解析它并获得指示的值。

1.23的唯一问题是，如果它们不是指数的，它也会返回有效数字。（即，对于1.23案例，它将返回double。）

您还可以删除Regex范围的支票。他们只是为了完整。

另外，为了证明这一点，我只是使用下面的代码运行基准测试，而nevermoi在另一个答案中的500ms选项使exponentials在ParseExponential1上运行了100,000次}数组，547ms选项使用ParseExponential2，317ms选项346ms。 sstan的方法花了ParseExponential3。最后，最快的是我在134ms的{{1}}方法。

string[] exponentials = new string[] { "1.23E+4", "1.23E+04", "1.23e+4", "1.23", "1.23e+4e+4", "abce+def", "1.23E-04" };

Stopwatch sw = new Stopwatch();
sw.Start();
for (int round = 0; round < 100000; round++)
    for (int i = 0; i < exponentials.Length; i++)
        ParseExponential1(exponentials[i]);
sw.Stop();
Console.WriteLine("Benchmark 1 (ParseExponential1) complete: {0}ms", sw.ElapsedMilliseconds);

sw.Reset();

sw.Start();
for (int round = 0; round < 100000; round++)
    for (int i = 0; i < exponentials.Length; i++)
        ParseExponential2(exponentials[i]);
sw.Stop();
Console.WriteLine("Benchmark 2 (ParseExponential2) complete: {0}ms", sw.ElapsedMilliseconds);
sw.Reset();
string pattern = @"^\d{1}.\d+(E\+)\d+$";
Regex rgx = new Regex(pattern, RegexOptions.IgnoreCase);

sw.Start();
for (int round = 0; round < 100000; round++)
    for (int i = 0; i < exponentials.Length; i++)
        rgx.IsMatch(exponentials[i]);
sw.Stop();
Console.WriteLine("Benchmark 3 (Regex Parse) complete: {0}ms", sw.ElapsedMilliseconds);
sw.Reset();

sw.Start();
for (int round = 0; round < 100000; round++)
    for (int i = 0; i < exponentials.Length; i++)
        IsExponentialFormat(exponentials[i]);
sw.Stop();
Console.WriteLine("Benchmark 4 (IsExponentialFormat) complete: {0}ms", sw.ElapsedMilliseconds);

sw.Start();
for (int round = 0; round < 100000; round++)
    for (int i = 0; i < exponentials.Length; i++)
        ParseExponential3(exponentials[i]);
sw.Stop();
Console.WriteLine("Benchmark 5 (ParseExponential3) complete: {0}ms", sw.ElapsedMilliseconds);

并添加了以下方法：

private static bool IsExponentialFormat(string str)
{
    double dummy;
    return (str.Contains("E") || str.Contains("e")) && double.TryParse(str, out dummy);
}

如何在C＃中以指数格式查找字符串值？

3 个答案: