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
我做了一些研究，发现理查德·沃森的解决方案在大多数情况下是最快的。这是文章中每个解决方案的结果表(除了那些使用Regex的，因为它在解析“test{test”这样的字符串时抛出异常)
Name | Short/char | Long/char | Short/short| Long/short | Long/long |
Inspite | 134| 1853| 95| 1146| 671|
LukeH_1 | 346| 4490| N/A| N/A| N/A|
LukeH_2 | 152| 1569| 197| 2425| 2171|
Bobwienholt | 230| 3269| N/A| N/A| N/A|
Richard Watson| 33| 298| 146| 737| 543|
StefanosKargas| N/A| N/A| 681| 11884| 12486|
可以看到，在短字符串(10-50个字符)中查找短子字符串(1-5个字符)的出现次数时，首选原算法。
同样，对于多字符子字符串，您应该使用以下代码(基于Richard Watson的解决方案)
int count = 0, n = 0;
if(substring != "")
{
while ((n = source.IndexOf(substring, n, StringComparison.InvariantCulture)) != -1)
{
n += substring.Length;
++count;
}
}
其他回答
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。
我想我会把我的扩展方法扔到戒指(更多信息见评论)。我没有做过任何正式的基准测试，但我认为在大多数情况下必须非常快。
EDIT: OK - so this SO question got me to wondering how the performance of our current implementation would stack up against some of the solutions presented here. I decided to do a little bench marking and found that our solution was very much in line with the performance of the solution provided by Richard Watson up until you are doing aggressive searching with large strings (100 Kb +), large substrings (32 Kb +) and many embedded repetitions (10K +). At that point our solution was around 2X to 4X slower. Given this and the fact that we really like the solution presented by Richard Watson, we have refactored our solution accordingly. I just wanted to make this available for anyone that might benefit from it.
我们最初的解决方案:
/// <summary>
/// Counts the number of occurrences of the specified substring within
/// the current string.
/// </summary>
/// <param name="s">The current string.</param>
/// <param name="substring">The substring we are searching for.</param>
/// <param name="aggressiveSearch">Indicates whether or not the algorithm
/// should be aggressive in its search behavior (see Remarks). Default
/// behavior is non-aggressive.</param>
/// <remarks>This algorithm has two search modes - aggressive and
/// non-aggressive. When in aggressive search mode (aggressiveSearch =
/// true), the algorithm will try to match at every possible starting
/// character index within the string. When false, all subsequent
/// character indexes within a substring match will not be evaluated.
/// For example, if the string was 'abbbc' and we were searching for
/// the substring 'bb', then aggressive search would find 2 matches
/// with starting indexes of 1 and 2. Non aggressive search would find
/// just 1 match with starting index at 1. After the match was made,
/// the non aggressive search would attempt to make it's next match
/// starting at index 3 instead of 2.</remarks>
/// <returns>The count of occurrences of the substring within the string.</returns>
public static int CountOccurrences(this string s, string substring,
bool aggressiveSearch = false)
{
// if s or substring is null or empty, substring cannot be found in s
if (string.IsNullOrEmpty(s) || string.IsNullOrEmpty(substring))
return 0;
// if the length of substring is greater than the length of s,
// substring cannot be found in s
if (substring.Length > s.Length)
return 0;
var sChars = s.ToCharArray();
var substringChars = substring.ToCharArray();
var count = 0;
var sCharsIndex = 0;
// substring cannot start in s beyond following index
var lastStartIndex = sChars.Length - substringChars.Length;
while (sCharsIndex <= lastStartIndex)
{
if (sChars[sCharsIndex] == substringChars[0])
{
// potential match checking
var match = true;
var offset = 1;
while (offset < substringChars.Length)
{
if (sChars[sCharsIndex + offset] != substringChars[offset])
{
match = false;
break;
}
offset++;
}
if (match)
{
count++;
// if aggressive, just advance to next char in s, otherwise,
// skip past the match just found in s
sCharsIndex += aggressiveSearch ? 1 : substringChars.Length;
}
else
{
// no match found, just move to next char in s
sCharsIndex++;
}
}
else
{
// no match at current index, move along
sCharsIndex++;
}
}
return count;
}
这是我们修改后的解决方案:
/// <summary>
/// Counts the number of occurrences of the specified substring within
/// the current string.
/// </summary>
/// <param name="s">The current string.</param>
/// <param name="substring">The substring we are searching for.</param>
/// <param name="aggressiveSearch">Indicates whether or not the algorithm
/// should be aggressive in its search behavior (see Remarks). Default
/// behavior is non-aggressive.</param>
/// <remarks>This algorithm has two search modes - aggressive and
/// non-aggressive. When in aggressive search mode (aggressiveSearch =
/// true), the algorithm will try to match at every possible starting
/// character index within the string. When false, all subsequent
/// character indexes within a substring match will not be evaluated.
/// For example, if the string was 'abbbc' and we were searching for
/// the substring 'bb', then aggressive search would find 2 matches
/// with starting indexes of 1 and 2. Non aggressive search would find
/// just 1 match with starting index at 1. After the match was made,
/// the non aggressive search would attempt to make it's next match
/// starting at index 3 instead of 2.</remarks>
/// <returns>The count of occurrences of the substring within the string.</returns>
public static int CountOccurrences(this string s, string substring,
bool aggressiveSearch = false)
{
// if s or substring is null or empty, substring cannot be found in s
if (string.IsNullOrEmpty(s) || string.IsNullOrEmpty(substring))
return 0;
// if the length of substring is greater than the length of s,
// substring cannot be found in s
if (substring.Length > s.Length)
return 0;
int count = 0, n = 0;
while ((n = s.IndexOf(substring, n, StringComparison.InvariantCulture)) != -1)
{
if (aggressiveSearch)
n++;
else
n += substring.Length;
count++;
}
return count;
}
string search = "/string";
var occurrences = (regex.Match(search, @"\/")).Count;
每次程序找到“/s”(区分大小写)时，都会进行计数 出现的次数将存储在变量“occurrences”中。
private int CountWords(string text, string word) {
int count = (text.Length - text.Replace(word, "").Length) / word.Length;
return count;
}
因为最初的解决方案，是最快的字符，我想它也将是字符串。这是我的贡献。
上下文:我在日志文件中寻找像“失败”和“成功”这样的词。
克 我
对于字符串分隔符的情况(而不是字符情况，如主题所述): 字符串源= "@@ once@@@upon@@ a@@@time@@@"; Int count = source。Split(new[] {"@@@"}， StringSplitOptions.RemoveEmptyEntries)。长度- 1; 海报的原始源值("/once/upon/a/time/")自然分隔符是一个字符'/'，并且响应确实解释了source. split (char[])选项…
