有没有更好的方法来替换字符串?
我很惊讶Replace不接受字符数组或字符串数组。我想我可以写我自己的扩展,但我很好奇是否有更好的内置方式来做以下工作?注意最后一个Replace是一个字符串而不是字符。
myString.Replace(';', '\n').Replace(',', '\n').Replace('\r', '\n').Replace('\t', '\n').Replace(' ', '\n').Replace("\n\n", "\n");
有没有更好的方法来替换字符串?
我很惊讶Replace不接受字符数组或字符串数组。我想我可以写我自己的扩展,但我很好奇是否有更好的内置方式来做以下工作?注意最后一个Replace是一个字符串而不是字符。
myString.Replace(';', '\n').Replace(',', '\n').Replace('\r', '\n').Replace('\t', '\n').Replace(' ', '\n').Replace("\n\n", "\n");
当前回答
哦,表演太恐怖了! 答案有点过时了,但仍然……
public static class StringUtils
{
#region Private members
[ThreadStatic]
private static StringBuilder m_ReplaceSB;
private static StringBuilder GetReplaceSB(int capacity)
{
var result = m_ReplaceSB;
if (null == result)
{
result = new StringBuilder(capacity);
m_ReplaceSB = result;
}
else
{
result.Clear();
result.EnsureCapacity(capacity);
}
return result;
}
public static string ReplaceAny(this string s, char replaceWith, params char[] chars)
{
if (null == chars)
return s;
if (null == s)
return null;
StringBuilder sb = null;
for (int i = 0, count = s.Length; i < count; i++)
{
var temp = s[i];
var replace = false;
for (int j = 0, cc = chars.Length; j < cc; j++)
if (temp == chars[j])
{
if (null == sb)
{
sb = GetReplaceSB(count);
if (i > 0)
sb.Append(s, 0, i);
}
replace = true;
break;
}
if (replace)
sb.Append(replaceWith);
else
if (null != sb)
sb.Append(temp);
}
return null == sb ? s : sb.ToString();
}
}
其他回答
我知道这个问题非常古老,但我想提供两个更有效的选择:
首先,Paul Walls发布的扩展方法很好,但可以通过使用StringBuilder类来提高效率,StringBuilder类类似于字符串数据类型,但专门用于需要多次更改字符串值的情况。下面是我用StringBuilder做的扩展方法的一个版本:
public static string ReplaceChars(this string s, char[] separators, char newVal)
{
StringBuilder sb = new StringBuilder(s);
foreach (var c in separators) { sb.Replace(c, newVal); }
return sb.ToString();
}
我运行了这个操作10万次,使用StringBuilder花费了73毫秒,而使用string花费了81毫秒。所以区别通常是可以忽略不计的,除非你运行很多操作或使用一个巨大的字符串。
其次,这里有一个你可以使用的1线循环:
foreach (char c in separators) { s = s.Replace(c, '\n'); }
我个人认为这是最好的选择。它非常高效,并且不需要编写扩展方法。在我的测试中,这种方法在63毫秒内运行了10万次迭代,是最高效的。 下面是一个上下文中的例子:
string s = "this;is,\ra\t\n\n\ntest";
char[] separators = new char[] { ' ', ';', ',', '\r', '\t', '\n' };
foreach (char c in separators) { s = s.Replace(c, '\n'); }
本例的前两行要归功于Paul Walls。
如果你觉得自己特别聪明,不想使用Regex:
char[] separators = new char[]{' ',';',',','\r','\t','\n'};
string s = "this;is,\ra\t\n\n\ntest";
string[] temp = s.Split(separators, StringSplitOptions.RemoveEmptyEntries);
s = String.Join("\n", temp);
您也可以用一个扩展方法来包装它。
编辑:或者只要等2分钟,我还是会把它写完:)
public static class ExtensionMethods
{
public static string Replace(this string s, char[] separators, string newVal)
{
string[] temp;
temp = s.Split(separators, StringSplitOptions.RemoveEmptyEntries);
return String.Join( newVal, temp );
}
}
和瞧...
char[] separators = new char[]{' ',';',',','\r','\t','\n'};
string s = "this;is,\ra\t\n\n\ntest";
s = s.Replace(separators, "\n");
没有“替换”(仅限Linq):
string myString = ";,\r\t \n\n=1;;2,,3\r\r4\t\t5 6\n\n\n\n7=";
char NoRepeat = '\n';
string ByeBye = ";,\r\t ";
string myResult = myString.ToCharArray().Where(t => !"STOP-OUTSIDER".Contains(t))
.Select(t => "" + ( ByeBye.Contains(t) ? '\n' : t))
.Aggregate((all, next) => (
next == "" + NoRepeat && all.Substring(all.Length - 1) == "" + NoRepeat
? all : all + next ) );
我也摆弄了一下这个问题,发现这里的大多数解决方案都非常缓慢。最快的方法实际上是dodgy_coder发布的LINQ + Aggregate方法。
但我想,这也可能是相当沉重的内存分配取决于有多少旧字符。所以我得出了这个结论:
这里的想法是为当前线程缓存旧字符的替换映射,以实现安全分配。除此之外,只是处理输入的字符数组之后会再次以字符串的形式返回。而字符数组则被尽可能少地修改。
[ThreadStatic]
private static bool[] replaceMap;
public static string Replace(this string input, char[] oldChars, char newChar)
{
if (input == null) throw new ArgumentNullException(nameof(input));
if (oldChars == null) throw new ArgumentNullException(nameof(oldChars));
if (oldChars.Length == 1) return input.Replace(oldChars[0], newChar);
if (oldChars.Length == 0) return input;
replaceMap = replaceMap ?? new bool[char.MaxValue + 1];
foreach (var oldChar in oldChars)
{
replaceMap[oldChar] = true;
}
try
{
var count = input.Length;
var output = input.ToCharArray();
for (var i = 0; i < count; i++)
{
if (replaceMap[input[i]])
{
output[i] = newChar;
}
}
return new string(output);
}
finally
{
foreach (var oldChar in oldChars)
{
replaceMap[oldChar] = false;
}
}
}
对我来说,对于实际的输入字符串,这最多是两个分配。由于某些原因,StringBuilder对我来说要慢得多。它比LINQ变体快2倍。
在构建了自己的解决方案并查看这里使用的解决方案后,我利用了一个不使用复杂代码且通常对大多数参数有效的答案。
Cover base cases where other methods are more appropriate. If there are no chars to replacement, return the original string. If there is only one, just use the Replace method. Use a StringBuilder and initialize the capacity to the length of the original string. After all, the new string being built will have the same length of the original string if its just chars being replaced. This ensure only 1 memory allocation is used for the new string. Assuming that the 'char' length could be small or large will impact performance. Large collections are better with hashsets, while smaller collections are not. This is a near-perfect use case for Hybrid Dictionaries. They switch to using a Hash based lookup once the collection gets too large. However, we don't care about the value of the dictionary, so I just set it to "true". Have different methods for StringBuilder verse just a string will prevent unnecessary memory allocation. If its just a string, don't instantiate a StringBuilder unless the base cases were checked. If its already a StringBuilder, then perform the replacements and return the StringBuilder itself (as other StringBuilder methods like Append do). I put the replacement char first, and the chars to check at the end. This way, I can leverage the params keyword for easily passing additional strings. However, you don't have to do this if you prefer the other order.
namespace Test.Extensions
{
public static class StringExtensions
{
public static string ReplaceAll(this string str, char replacementCharacter, params char[] chars)
{
if (chars.Length == 0)
return str;
if (chars.Length == 1)
return str.Replace(chars[0], replacementCharacter);
StringBuilder sb = new StringBuilder(str.Length);
var searcher = new HybridDictionary(chars.Length);
for (int i = 0; i < chars.Length; i++)
searcher[chars[i]] = true;
foreach (var c in str)
{
if (searcher.Contains(c))
sb.Append(replacementCharacter);
else
sb.Append(c);
}
return sb.ToString();
}
public static StringBuilder ReplaceAll(this StringBuilder sb, char replacementCharacter, params char[] chars)
{
if (chars.Length == 0)
return sb;
if (chars.Length == 1)
return sb.Replace(chars[0], replacementCharacter);
var searcher = new HybridDictionary(chars.Length);
for (int i = 0; i < chars.Length; i++)
searcher[chars[i]] = true;
for (int i = 0; i < sb.Length; i++)
{
var val = sb[i];
if (searcher.Contains(val))
sb[i] = replacementCharacter;
}
return sb;
}
}
}