我正在做一些事情，我意识到我想要在一个字符串中找到多少个/s，然后我突然想到，有几种方法可以做到这一点，但不能决定哪种是最好的(或最简单的)。
目前我想说的是:
string source = "/once/upon/a/time/";
int count = source.Length - source.Replace("/", "").Length;
但我一点都不喜欢，有人愿意吗?
我并不想为此挖掘出正则表达式，对吧?
我知道我的字符串将包含我要搜索的项，所以你可以假设…
当然对于长度为> 1的字符串，
string haystack = "/once/upon/a/time";
string needle = "/";
int needleCount = ( haystack.Length - haystack.Replace(needle,"").Length ) / needle.Length;
当前回答
我觉得我们缺少某些类型的子字符串计数，比如不安全的逐字节比较。我把原始海报的方法和我能想到的任何方法结合在一起。
这些是我做的字符串扩展。
namespace Example
{
using System;
using System.Text;
public static class StringExtensions
{
public static int CountSubstr(this string str, string substr)
{
return (str.Length - str.Replace(substr, "").Length) / substr.Length;
}
public static int CountSubstr(this string str, char substr)
{
return (str.Length - str.Replace(substr.ToString(), "").Length);
}
public static int CountSubstr2(this string str, string substr)
{
int substrlen = substr.Length;
int lastIndex = str.IndexOf(substr, 0, StringComparison.Ordinal);
int count = 0;
while (lastIndex != -1)
{
++count;
lastIndex = str.IndexOf(substr, lastIndex + substrlen, StringComparison.Ordinal);
}
return count;
}
public static int CountSubstr2(this string str, char substr)
{
int lastIndex = str.IndexOf(substr, 0);
int count = 0;
while (lastIndex != -1)
{
++count;
lastIndex = str.IndexOf(substr, lastIndex + 1);
}
return count;
}
public static int CountChar(this string str, char substr)
{
int length = str.Length;
int count = 0;
for (int i = 0; i < length; ++i)
if (str[i] == substr)
++count;
return count;
}
public static int CountChar2(this string str, char substr)
{
int count = 0;
foreach (var c in str)
if (c == substr)
++count;
return count;
}
public static unsafe int CountChar3(this string str, char substr)
{
int length = str.Length;
int count = 0;
fixed (char* chars = str)
{
for (int i = 0; i < length; ++i)
if (*(chars + i) == substr)
++count;
}
return count;
}
public static unsafe int CountChar4(this string str, char substr)
{
int length = str.Length;
int count = 0;
fixed (char* chars = str)
{
for (int i = length - 1; i >= 0; --i)
if (*(chars + i) == substr)
++count;
}
return count;
}
public static unsafe int CountSubstr3(this string str, string substr)
{
int length = str.Length;
int substrlen = substr.Length;
int count = 0;
fixed (char* strc = str)
{
fixed (char* substrc = substr)
{
int n = 0;
for (int i = 0; i < length; ++i)
{
if (*(strc + i) == *(substrc + n))
{
++n;
if (n == substrlen)
{
++count;
n = 0;
}
}
else
n = 0;
}
}
}
return count;
}
public static int CountSubstr3(this string str, char substr)
{
return CountSubstr3(str, substr.ToString());
}
public static unsafe int CountSubstr4(this string str, string substr)
{
int length = str.Length;
int substrLastIndex = substr.Length - 1;
int count = 0;
fixed (char* strc = str)
{
fixed (char* substrc = substr)
{
int n = substrLastIndex;
for (int i = length - 1; i >= 0; --i)
{
if (*(strc + i) == *(substrc + n))
{
if (--n == -1)
{
++count;
n = substrLastIndex;
}
}
else
n = substrLastIndex;
}
}
}
return count;
}
public static int CountSubstr4(this string str, char substr)
{
return CountSubstr4(str, substr.ToString());
}
}
}
接下来是测试代码…
static void Main()
{
const char matchA = '_';
const string matchB = "and";
const string matchC = "muchlongerword";
const string testStrA = "_and_d_e_banna_i_o___pfasd__and_d_e_banna_i_o___pfasd_";
const string testStrB = "and sdf and ans andeians andano ip and and sdf and ans andeians andano ip and";
const string testStrC =
"muchlongerword amuchlongerworsdfmuchlongerwordsdf jmuchlongerworijv muchlongerword sdmuchlongerword dsmuchlongerword";
const int testSize = 1000000;
Console.WriteLine(testStrA.CountSubstr('_'));
Console.WriteLine(testStrA.CountSubstr2('_'));
Console.WriteLine(testStrA.CountSubstr3('_'));
Console.WriteLine(testStrA.CountSubstr4('_'));
Console.WriteLine(testStrA.CountChar('_'));
Console.WriteLine(testStrA.CountChar2('_'));
Console.WriteLine(testStrA.CountChar3('_'));
Console.WriteLine(testStrA.CountChar4('_'));
Console.WriteLine(testStrB.CountSubstr("and"));
Console.WriteLine(testStrB.CountSubstr2("and"));
Console.WriteLine(testStrB.CountSubstr3("and"));
Console.WriteLine(testStrB.CountSubstr4("and"));
Console.WriteLine(testStrC.CountSubstr("muchlongerword"));
Console.WriteLine(testStrC.CountSubstr2("muchlongerword"));
Console.WriteLine(testStrC.CountSubstr3("muchlongerword"));
Console.WriteLine(testStrC.CountSubstr4("muchlongerword"));
var timer = new Stopwatch();
timer.Start();
for (int i = 0; i < testSize; ++i)
testStrA.CountSubstr(matchA);
timer.Stop();
Console.WriteLine("CS1 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrB.CountSubstr(matchB);
timer.Stop();
Console.WriteLine("CS1 and: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrC.CountSubstr(matchC);
timer.Stop();
Console.WriteLine("CS1 mlw: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrA.CountSubstr2(matchA);
timer.Stop();
Console.WriteLine("CS2 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrB.CountSubstr2(matchB);
timer.Stop();
Console.WriteLine("CS2 and: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrC.CountSubstr2(matchC);
timer.Stop();
Console.WriteLine("CS2 mlw: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrA.CountSubstr3(matchA);
timer.Stop();
Console.WriteLine("CS3 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrB.CountSubstr3(matchB);
timer.Stop();
Console.WriteLine("CS3 and: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrC.CountSubstr3(matchC);
timer.Stop();
Console.WriteLine("CS3 mlw: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrA.CountSubstr4(matchA);
timer.Stop();
Console.WriteLine("CS4 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrB.CountSubstr4(matchB);
timer.Stop();
Console.WriteLine("CS4 and: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrC.CountSubstr4(matchC);
timer.Stop();
Console.WriteLine("CS4 mlw: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrA.CountChar(matchA);
timer.Stop();
Console.WriteLine("CC1 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrA.CountChar2(matchA);
timer.Stop();
Console.WriteLine("CC2 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrA.CountChar3(matchA);
timer.Stop();
Console.WriteLine("CC3 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
timer.Restart();
for (int i = 0; i < testSize; ++i)
testStrA.CountChar4(matchA);
timer.Stop();
Console.WriteLine("CC4 chr: " + timer.Elapsed.TotalMilliseconds + "ms");
}
结果:CSX与CountSubstrX对应，CCX与CountCharX对应。“chr”搜索字符串中的“_”，“and”搜索字符串中的“and”，“mlw”搜索字符串中的“muchlongerword”
CS1 chr: 824.123ms
CS1 and: 586.1893ms
CS1 mlw: 486.5414ms
CS2 chr: 127.8941ms
CS2 and: 806.3918ms
CS2 mlw: 497.318ms
CS3 chr: 201.8896ms
CS3 and: 124.0675ms
CS3 mlw: 212.8341ms
CS4 chr: 81.5183ms
CS4 and: 92.0615ms
CS4 mlw: 116.2197ms
CC1 chr: 66.4078ms
CC2 chr: 64.0161ms
CC3 chr: 65.9013ms
CC4 chr: 65.8206ms
最后，我有了一个包含360万个字符的文件。“derp adfderdserp dfaerpderp deasderp”重复了10万次。我用上述方法在文件中搜索“derp”100次，得到这些结果。
CS1Derp: 1501.3444ms
CS2Derp: 1585.797ms
CS3Derp: 376.0937ms
CS4Derp: 271.1663ms
所以我的第四种方法肯定是赢家，但是，实际上，如果一个360万个字符的文件100次只需要1586ms作为最坏的情况，那么所有这些都是可以忽略不计的。
顺便说一下，我还用100次CountSubstr和CountChar方法在360万个字符的文件中扫描了'd'字符。结果……
CS1 d : 2606.9513ms
CS2 d : 339.7942ms
CS3 d : 960.281ms
CS4 d : 233.3442ms
CC1 d : 302.4122ms
CC2 d : 280.7719ms
CC3 d : 299.1125ms
CC4 d : 292.9365ms
原来海报的方法是非常糟糕的单个字符针在一个大草堆根据这一点。
注:所有值更新为发布版本输出。在我第一次发布这篇文章时，我不小心忘记了建立发布模式。我的一些声明已经被修改了。
其他回答
对于任何想要使用String扩展方法的人，
以下是我使用的基于张贴的最好的答案:
public static class StringExtension
{
/// <summary> Returns the number of occurences of a string within a string, optional comparison allows case and culture control. </summary>
public static int Occurrences(this System.String input, string value, StringComparison stringComparisonType = StringComparison.Ordinal)
{
if (String.IsNullOrEmpty(value)) return 0;
int count = 0;
int position = 0;
while ((position = input.IndexOf(value, position, stringComparisonType)) != -1)
{
position += value.Length;
count += 1;
}
return count;
}
/// <summary> Returns the number of occurences of a single character within a string. </summary>
public static int Occurrences(this System.String input, char value)
{
int count = 0;
foreach (char c in input) if (c == value) count += 1;
return count;
}
}
string source = "/once/upon/a/time/";
int count = 0, n = 0;
while ((n = source.IndexOf('/', n) + 1) != 0) count++;
这是Richard Watson的答案的一个变体，char在字符串中出现的次数越多，效率就会提高一点，代码也会更少!
虽然我必须说，在没有广泛测试每个场景的情况下，我确实看到了使用以下方法的显著速度提升:
int count = 0;
for (int n = 0; n < source.Length; n++) if (source[n] == '/') count++;
这两个都只适用于单字符搜索词…
countOccurences("the", "the answer is the answer");
int countOccurences(string needle, string haystack)
{
return (haystack.Length - haystack.Replace(needle,"").Length) / needle.Length;
}
也许更长的针头会更好…
但肯定有更优雅的方式。：）
我最初的想法是这样的:
public static int CountOccurrences(string original, string substring)
{
if (string.IsNullOrEmpty(substring))
return 0;
if (substring.Length == 1)
return CountOccurrences(original, substring[0]);
if (string.IsNullOrEmpty(original) ||
substring.Length > original.Length)
return 0;
int substringCount = 0;
for (int charIndex = 0; charIndex < original.Length; charIndex++)
{
for (int subCharIndex = 0, secondaryCharIndex = charIndex; subCharIndex < substring.Length && secondaryCharIndex < original.Length; subCharIndex++, secondaryCharIndex++)
{
if (substring[subCharIndex] != original[secondaryCharIndex])
goto continueOuter;
}
if (charIndex + substring.Length > original.Length)
break;
charIndex += substring.Length - 1;
substringCount++;
continueOuter:
;
}
return substringCount;
}
public static int CountOccurrences(string original, char @char)
{
if (string.IsNullOrEmpty(original))
return 0;
int substringCount = 0;
for (int charIndex = 0; charIndex < original.Length; charIndex++)
if (@char == original[charIndex])
substringCount++;
return substringCount;
}
使用替换和除法的大海捞针方法产生21秒以上，而这需要大约15.2秒。
在添加位后进行编辑，这将添加子字符串。长度- 1到charIndex(就像它应该的那样)，它在11.6秒。
编辑2:我使用了一个有26个双字符字符串的字符串，这里是更新到相同示例文本的时间:
大海捞针(OP版本):7.8秒
建议的机制:4.6秒。
编辑3:添加单个字符的大小写，它变成了1.2秒。
编辑4:作为上下文:使用了5000万次迭代。
string s = "65 fght 6565 4665 hjk";
int count = 0;
foreach (Match m in Regex.Matches(s, "65"))
count++;
