How do you split multi-line string into lines?
I know this way
var result = input.Split("\n\r".ToCharArray(), StringSplitOptions.RemoveEmptyEntries);
looks a bit ugly and loses empty lines. Is there a better solution?
If it looks ugly, just remove the unnecessary ToCharArray
call.
If you want to split by either \n
or \r
, you've got two options:
Use an array literal – but this will give you empty lines for Windows-style line endings \r\n
:
var result = text.Split(new [] { '\r', '\n' });
Use a regular expression, as indicated by Bart:
var result = Regex.Split(text, "\r\n|\r|\n");
If you want to preserve empty lines, why do you explicitly tell C# to throw them away? ( StringSplitOptions
parameter) – use StringSplitOptions.None
instead.
using (StringReader sr = new StringReader(text)) {
string line;
while ((line = sr.ReadLine()) != null) {
// do something
}
}
This works great and is faster than Regex:
input.Split(new[] {"\r\n", "\r", "\n"}, StringSplitOptions.None)
It is important to have "\r\n"
first in the array so that it's taken as one line break. The above gives the same results as either of these Regex solutions:
Regex.Split(input, "\r\n|\r|\n")
Regex.Split(input, "\r?\n|\r")
Except that Regex turns out to be about 10 times slower. Here's my test:
Action<Action> measure = (Action func) => {
var start = DateTime.Now;
for (int i = 0; i < 100000; i++) {
func();
}
var duration = DateTime.Now - start;
Console.WriteLine(duration);
};
var input = "";
for (int i = 0; i < 100; i++)
{
input += "1 \r2\r\n3\n4\n\r5 \r\n\r\n 6\r7\r 8\r\n";
}
measure(() =>
input.Split(new[] {"\r\n", "\r", "\n"}, StringSplitOptions.None)
);
measure(() =>
Regex.Split(input, "\r\n|\r|\n")
);
measure(() =>
Regex.Split(input, "\r?\n|\r")
);
Output:
00:00:03.8527616
00:00:31.8017726
00:00:32.5557128
and here's the Extension Method:
public static class StringExtensionMethods
{
public static IEnumerable<string> GetLines(this string str, bool removeEmptyLines = false)
{
return str.Split(new[] { "\r\n", "\r", "\n" },
removeEmptyLines ? StringSplitOptions.RemoveEmptyEntries : StringSplitOptions.None);
}
}
Usage:
input.GetLines() // keeps empty lines
input.GetLines(true) // removes empty lines
You could use Regex.Split:
string[] tokens = Regex.Split(input, @"\r?\n|\r");
Edit: added |\r
to account for (older) Mac line terminators.
如果您想保留空行,只需删除 StringSplitOptions。
var result = input.Split(System.Environment.NewLine.ToCharArray());
string[] lines = input.Split(new[] { '\r', '\n' }, StringSplitOptions.RemoveEmptyEntries);
I had this other answer but this one, based on Jack's answer , is significantly faster might be preferred since it works asynchronously, although slightly slower.
public static class StringExtensionMethods
{
public static IEnumerable<string> GetLines(this string str, bool removeEmptyLines = false)
{
using (var sr = new StringReader(str))
{
string line;
while ((line = sr.ReadLine()) != null)
{
if (removeEmptyLines && String.IsNullOrWhiteSpace(line))
{
continue;
}
yield return line;
}
}
}
}
Usage:
input.GetLines() // keeps empty lines
input.GetLines(true) // removes empty lines
Test:
Action<Action> measure = (Action func) =>
{
var start = DateTime.Now;
for (int i = 0; i < 100000; i++)
{
func();
}
var duration = DateTime.Now - start;
Console.WriteLine(duration);
};
var input = "";
for (int i = 0; i < 100; i++)
{
input += "1 \r2\r\n3\n4\n\r5 \r\n\r\n 6\r7\r 8\r\n";
}
measure(() =>
input.Split(new[] { "\r\n", "\r", "\n" }, StringSplitOptions.None)
);
measure(() =>
input.GetLines()
);
measure(() =>
input.GetLines().ToList()
);
Output:
00:00:03.9603894
00:00:00.0029996
00:00:04.8221971
Slightly twisted, but an iterator block to do it:
public static IEnumerable<string> Lines(this string Text)
{
int cIndex = 0;
int nIndex;
while ((nIndex = Text.IndexOf(Environment.NewLine, cIndex + 1)) != -1)
{
int sIndex = (cIndex == 0 ? 0 : cIndex + 1);
yield return Text.Substring(sIndex, nIndex - sIndex);
cIndex = nIndex;
}
yield return Text.Substring(cIndex + 1);
}
You can then call:
var result = input.Lines().ToArray();
private string[] GetLines(string text)
{
List<string> lines = new List<string>();
using (MemoryStream ms = new MemoryStream())
{
StreamWriter sw = new StreamWriter(ms);
sw.Write(text);
sw.Flush();
ms.Position = 0;
string line;
using (StreamReader sr = new StreamReader(ms))
{
while ((line = sr.ReadLine()) != null)
{
lines.Add(line);
}
}
sw.Close();
}
return lines.ToArray();
}
Split a string into lines without any allocation.
public static LineEnumerator GetLines(this string text) {
return new LineEnumerator( text.AsSpan() );
}
internal ref struct LineEnumerator {
private ReadOnlySpan<char> Text { get; set; }
public ReadOnlySpan<char> Current { get; private set; }
public LineEnumerator(ReadOnlySpan<char> text) {
Text = text;
Current = default;
}
public LineEnumerator GetEnumerator() {
return this;
}
public bool MoveNext() {
if (Text.IsEmpty) return false;
var index = Text.IndexOf( '\n' ); // \r\n or \n
if (index != -1) {
Current = Text.Slice( 0, index + 1 );
Text = Text.Slice( index + 1 );
return true;
} else {
Current = Text;
Text = ReadOnlySpan<char>.Empty;
return true;
}
}
}
late to the party, but I've been using a simple collection of extension methods for just that, which leverages TextReader.ReadLine()
:
public static class StringReadLinesExtension
{
public static IEnumerable<string> GetLines(this string text) => GetLines(new StringReader(text));
public static IEnumerable<string> GetLines(this Stream stm) => GetLines(new StreamReader(stm));
public static IEnumerable<string> GetLines(this TextReader reader) {
string line;
while ((line = reader.ReadLine()) != null)
yield return line;
reader.Dispose();
yield break;
}
}
Using the code is really trivial:
// If you have the text as a string...
var text = "Line 1\r\nLine 2\r\nLine 3";
foreach (var line in text.GetLines())
Console.WriteLine(line);
// You can also use streams like
var fileStm = File.OpenRead("c:\tests\file.txt");
foreach(var line in fileStm.GetLines())
Console.WriteLine(line);
Hope this helps someone out there.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.