Question

好的，这就是我遇到的问题：使用C＃我想通过打印一定数量的字符来格式化字符串，然后跳过一定数量的字符然后再打印，跳过。

例如：我想通过跳过下一个2来打印3个字符。

所以这个：
ABCDEFGHIJKL

会变成这样： ABCFGHKL

我只是说它可以跳过每2,3,4等角色，我想不出如何进一步解决这个问题。

到目前为止，我有什么

string text;
    int print = 3;
    int skip = 2;

    StreamReader file = new StreamReader(@"c:\test.txt");

    while ((text = file.Readtext()) != null)
    {
            string[] stringArray = new string[text.Length];
            char ch;

            for (int i = 0; i < text.Length; i++)
            {
                ch = text[i];
                stringArray[i] = ch.ToString();
            }

            for (int i = 0; i < stringArray.Length; i+=skip)
            {
                Console.Write(stringArray[i]);
            }
}

感谢。

感谢大家提供各种出色的解决方案，非常感谢您的帮助！谢谢！

Answer 1

int take = 3;
int skip = 2;
string s = "ABCDEFGHIJKL";

var newstr = String.Join("", s.Where((c,i) => i % (skip + take) < take));

修改

以下是我的测试结果......

int take = 3; int skip = 2; string s = String.Join("",Enumerable.Repeat("0123456789", 200)); var t1 = Measure(10000, () => { var newstr = String.Join("", s.Where((c, i) => i % (skip + take) < take)); }); var t2 = Measure(10000, () => { var newstr = new string(s.Where((c, i) => i % (skip + take) < take).ToArray()); }); var t3 = Measure(10000, () => { var newstr = GetString(s, take, skip); }); long Measure(int n,Action action) { action(); //JIT??? var sw = Stopwatch.StartNew(); for (int i = 0; i < n; i++) { action(); } return sw.ElapsedMilliseconds; }

<强>输出：

1665 ms. (String.Join) 1154 ms. (new string()) 7457 ms. (Sayse's GetString)

编辑2

修改Sayse的答案如下所示，我测试的代码中的结果最快。（ 311 ms）

public string GetString(string s, int substringLen, int skipCount) { StringBuilder newString = new StringBuilder(s.Length); for (int i = 0; i < s.Length; i += skipCount) { for (int j = 0; j < substringLen && i < s.Length; j++) { newString.Append(s[i++]); } } return newString.ToString(); }

Answer 2

我更喜欢I4V的答案，但这是实现这一目标的一种方法

    public string GetString(string s, int substringLen, int skipCount)
    {
        string newString = "";
        for (int i = 0; i < s.Length; i += skipCount)
        {
            for (int j = 0; j < substringLen && i < s.Length; j++)
            {
                newString += s[i++];
            }
        }
        return newString;
    }

编辑基准测试表明我的方式更快

        var newStr2 = new Program().GetString(a, take, skip);
        var newstr = String.Join("", a.Where((c, i) => i % (skip + take) < take));

        sw.Start();
        for (int i = 0; i < 1000000; i++)
        {
            newStr2 = new Program().GetString(a, take, skip);
        }
        sw.Stop();
        Console.WriteLine("my way: " + sw.Elapsed.ToString());

        sw.Reset();
        sw.Start();
        for (int i = 0; i < 1000000; i++)
        {
            newstr = String.Join("", a.Where((c, iv) => iv % (skip + take) < take));
        }
        sw.Stop();
        Console.WriteLine("I4V way: " + sw.Elapsed.ToString());

输出

我的方式：00：00：00.7634927

I4V方式：00：00：01.0183307

Answer 3

这是怎么做到的。如果您不熟悉Lambda表达式，我认为这可能会更清楚

        int print = 3;
        int skip = 2;
        string str = "ABCDEFGHILKL";

        StringBuilder stringBuilder = new StringBuilder();
        string res = String.Empty;
        int i = 0;
        while (i < str.Length)
        {
            stringBuilder.Append(str.Substring(i, Math.Min(print, str.Length - i)));
            i += print + skip;
        }

        Console.WriteLine(stringBuilder.ToString());

Answer 4

  int print = 3;
  int skip = 2;

  Boolean isPrintState = true;
  int statePosition = 0;

  StringBuilder Sb = new StringBuilder();

  using (TextReader reader = new StreamReader(@"c:\test.txt")) { 
    while (true) {
      int Ch = reader.Read();

      if (Ch < 0)
        break;

      statePosition += 1;

      if ((isPrintState && (statePosition > print)) || (!isPrintState && (statePosition > skip))) {
        statePosition = 1;
        isPrintState = !isPrintState;
      }

      if (isPrintState) 
        Sb.Append((Char) Ch);
    }
  }

  Console.Write(Sb.ToString());

Answer 5

在如此多的好看的答案中，我可能不会那么复杂，但我想我会发布它。简单的逻辑。

var take = 3;
var skip = 2;

StringBuilder source = new StringBuilder("ABCDEFGHIJKL");            
StringBuilder result = new StringBuilder();

while (source.Length > 0)
{
    if (source.Length >= take)
    {
        result.Append(source.ToString().Substring(0, take));
    }
    else
    {
        result.Append(source.ToString());
    }
    if (source.Length > skip + take)
    {
        source.Remove(0, skip + take);
    }
    else
        source.Clear();
}

C＃打印并跳过一定数量的字符

5 个答案: