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I have String called "RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa". I want to know number of "a" characters available in the give String. As per my knowledge I found two ways to find the count. That are: 1) By using String.Split() 2) Linq Lambda Expression

My Observations:

1) If i use String.Split() it is returning wrong result 2) If i use Linq Lambda Expression it is returning correct result.

Here my doubt is how can i get the count of the given split character from the given string by using String.Split()

And also please suggest me which is the best way to get count of the given split character from the given string either "String.Split()" or "Linq Lambda" expression?

Please find the complete example:

                                                                                                                   using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace SplitData
{
    class Program
    {
        static void Main(string[] args)
        {
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'a');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'r');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'R');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'm');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'd');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'g');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 's');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'o');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'c');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'u');
            SplitData("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'f');


            Console.ReadKey();
        }

        private static void SplitData(string data,char split)
        {

            // using lambda expresion
            int len = data.AsEnumerable().Where(x => x.ToString().ToLower().Contains(split)).Count();  
            Console.WriteLine("Total '" + split + "' available are:{0} using lambda", len.ToString());

            //using normal split function
            len = data.Split(split).Length;

            Console.WriteLine("Total '" + split + "' available are:{0} using normal split", len.ToString());


        }
    }
}
4

4 回答 4

7 
string str = "RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa";
int countA = str.Count(r => r == 'a');

如果你想要不区分大小写的计数，那么：

string str = "RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa";
char searchChar = 'a';
int countA = str.Count(r => char.ToUpperInvariant(r) == char.ToUpperInvariant(searchChar));

string.Split如果您要求和 Linq之间的最佳选择Count，那么 IMO, LINQ 更具可读性。我不确定性能，但我怀疑 LINQ 版本更快。

如果您想使用string.Split并使其不区分大小写，则构造一个包含两个元素（大写和小写）的字符数组，然后使用Split如下：

string str = "RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa";
char searchChar = 'a';
char[] delimeters = new char[2];
delimeters[0] = char.ToLowerInvariant(searchChar);
delimeters[1] = char.ToUpperInvariant(searchChar);
var count = str.Split(delimeters).Length - 1;
于 2013-10-04T15:24:56.367 回答
2

你的意思是你想计算一个字母的出现次数？像这样？

String data = "RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa";
Char letter = 'a';
Int32 totalOccurances = data.Count(character => character == letter);
于 2013-10-04T15:24:13.093 回答
2

对于不区分大小写的比较，您可以使用StringComparer实例或等效StringComparison枚举。至于您想如何编写它，请选择您的毒药。=)

// caller specifies comparison type
int Count1(string str, char searchChar, StringComparison comparison = StringComparison.CurrentCultureIgnoreCase)
{
    string searchStr = searchChar.ToString();
    int count = 0;
    for (int i = 0; i < str.Length; i++)
        if (string.Equals(searchStr, str[i].ToString(), comparison))
            count++;
    return count;
}
// ordinal comparison
int Count2(string str, char searchChar)
{
    string searchStr = searchChar.ToString();
    int count = 0;
    for (int i = 0; i < str.Length; i++)
        if (searchChar == str[i])
            count++;
    return count;
}
// ordinal comparison
int Count3(string str, char searchChar)
{
    return str.Split(searchChar).Length - 1;
}
// ordinal comparison
int Count4(string str, char searchChar)
{
    return str.Count(c => c == searchChar);
}
// caller specifies comparison type
int Count5(string str, char searchChar, StringComparison comparison = StringComparison.CurrentCultureIgnoreCase)
{
    string searchStr = searchChar.ToString();
    return str.Count(c => string.Equals(c.ToString(), searchStr, comparison));
}
于 2013-10-04T15:38:59.497 回答
1

不是一个花哨的 LINQ 解决方案，但尽管如此

int count = CountChar("RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa", 'a');
.....

int CountChar(string input, char toFind)
{
    int count = 0;
    int pos = -1;
    while((pos = input.IndexOf(toFind, pos+1)) != -1)
        count++;
    return count;
}

String.IndexOf从一个位置开始
，还有不区分大小写的选项

编辑：现在我很好奇，决定用这个和 lambda 解决方案测量时间。
区别很明显......

void Main()
{
    string str = "RamaSubbaReddyabcdaacacakkkoooahgafffgahgghsa";
    Stopwatch sw = new Stopwatch();
    sw.Start();
    for(int i = 0; i < 10000000; i++)
    {
        int count = CountChar(str, 'a');
     }   
    sw.Stop();
    Console.WriteLine("Using IndexOf:" + sw.ElapsedMilliseconds.ToString());  
    sw.Reset();
    sw.Start();
    for(int i = 0; i < 10000000; i++)
    {
        int countA = str.Count(r => r == 'a');
    }
    sw.Stop();
    Console.WriteLine("Using Count:" + sw.ElapsedMilliseconds.ToString());
}

第一个循环在 1160 毫秒内结束，第二个循环在 6200 毫秒内结束。
有人能发现这个测量是否有问题吗？

于 2013-10-04T15:28:21.350 回答