Need to exclude Numbers from String and returns cell arrays of strings in MATLAB
e.g str = 'abc76.5_pol0.00_Ev0.3'
output {'abc','pol','Ev'}
String is not specific to 'abc' etc, it could be an char long
Use regular expression.
str = 'abc76.5_pol0.00_Ev0.3';
C = regexp(str, '[a-zA-Z]*', 'match');
This is the solution that I found
output = regexp(str, '[^a-zA-Z]', 'split');
output(cellfun(#isempty,output)) = [];
You can also use strsplit with a RegularExpression option.
C = strsplit(str, '[^a-zA-Z]', 'DelimiterType', 'RegularExpression')
Related
Definitely I'm not good using regular expressions but are really cool!, Now I want to be able to get only the name "table" in this string:
[schema].[table]
I want to remove the schema name, the square brackets and the dot.
so I will get only the work table
I tried this:
string output = Regex.Replace(reader["Name"].ToString(), #"[\[\.\]]", "");
So you came up with a new idea?? Here is what you can try:
string input = "[schema].[table]";
// replacing the first thing into [] with the dot with empty
string one = Regex.Replace(input, #"^\[.*?\]\.", "");
// or replacing anything before the dot with empty
// string two = Regex.Replace(input, #".*[.]", "");
try this
string strRegex = #"^\[.*?\]\.";
Regex myRegex = new Regex(strRegex, RegexOptions.None);
string strTargetString = #"[schema].[table]";
string strReplace = #"";
var result=myRegex.Replace(strTargetString, strReplace);
Console.WriteLine(result);
Why do you want to do replace if you just want to extract part of string?
string table = Regex.Match("[schema].[table]", #"\w+(?=]$)").Value;
It works even in case if you don't have schema.
I have a string which is in the following format:
[0:2]={1.1,1,5.1.2}
My requirement here is to split the values inside curly braces after = operator, and store them in to a string array. I had tried to split part by using Substring() and IndexOf() methods, and it worked. But I needed a cleaner and elegant way to achieve this via regular expressions.
Does anybody having clues to achieve my requirement?
Here is your fully RegEx solution:
Dim input As String = "[0:2]={1.1,1,5.1.2}"
Dim match = Regex.Match(input, "\[\d:\d\]={(?:([^,]+),)*([^,]+)}")
Dim results = match.Groups(1).Captures.Cast(Of Capture).Select(Function(c) c.Value).Concat(match.Groups(2).Captures.Cast(Of Capture).Select(Function(c) c.Value)).ToArray()
Don't think it is more readable then standard split:
Dim startIndex = input.IndexOf("{"c) + 1
Dim length = input.Length - startIndex - 1
Dim results = input.Substring(startIndex, length).Split(",")
You could use a regular expression to extract the values inside the curly braces, and then use an ordinary Split:
Regex.Match("[0:2]={1.1,1,5.1.2}", "{(.*)}").Groups(1).Value.Split(","c)
Dim s As String = "[0:2]={1.1,1,5.1.2}";
Dim separatorChar as char = "="c;
Dim commaChar as char = ","c;
Dim openBraceChar as char = "{"c;
Dim closeBraceChar as char = "}"c;
Dim result() as String =
s.Split(separatorChar)(1)
.trim(openBraceChar)
.trim(closeBraceChar)
.split(commaChar);
(assuming it works! Typed on an iPad so can't verify syntax easily, but principal should be sound).
EDIT: updated to VB as downvoted for showing working .net methods in c# syntax.
if you want it using Regex
Dim s() As String=Regex.match(str,"(={)(.*)(})").Groups(1).Tostring.split(',');
Please let me how to remove double spaces and characters from below string.
String = Test----$$$$19****45#### Nothing
Clean String = Test-$19*45# Nothing
I have used regex "\s+" but it just removing the double spaces and I have tried other patterns of regex but it is too complex... please help me.
I am using vb.net
What you'll want to do is create a backreference to any character, and then remove the following characters that match that backreference. It's usually possible using the pattern (.)\1+, which should be replaced with just that backreference (once). It depends on the programming language how it's exactly done.
Dim text As String = "Test###_&aa&&&"
Dim result As String = New Regex("(.)\1+").Replace(text, "$1")
result will now contain Test#_&a&. Alternatively, you can use a lookaround to not remove that backreference in the first place:
Dim text As String = "Test###_&aa&&&"
Dim result As String = New Regex("(?<=(.))\1+").Replace(text, "")
Edit: included examples
For a faster alternative try:
Dim text As String = "Test###_&aa&&&"
Dim sb As New StringBuilder(text.Length)
Dim lastChar As Char
For Each c As Char In text
If c <> lastChar Then
sb.Append(c)
lastChar = c
End If
Next
Console.WriteLine(sb.ToString())
Here is a perl way to substitute all multiple non word chars by only one:
my $String = 'Test----$$$$19****45#### Nothing';
$String =~ s/(\W)\1+/$1/g;
print $String;
output:
Test-$19*45# Nothing
Here's how it would look in Java...
String raw = "Test----$$$$19****45#### Nothing";
String cleaned = raw.replaceAll("(.)\\1+", "$1");
System.out.println(raw);
System.out.println(cleaned);
prints
Test----$$$$19****45#### Nothing
Test-$19*45# Nothing
FOR EXAMPLE: Given a string... "1,2,3,4"
I need to be able to remove a given number and the comma after/before depending on if the match is at the end of the string or not.
remove(2) = "1,3,4"
remove(4) = "1,2,3"
Also, I'm using javascript.
As jtdubs shows, an easy way is is to use a split function to obtain an array of elements without the commas, remove the required element from the array, and then rebuild the string with a join function.
For javascript something like this might work:
function remove(array,to_remove)
{
var elements=array.split(",");
var remove_index=elements.indexOf(to_remove);
elements.splice(remove_index,1);
var result=elements.join(",");
return result;
}
var string="1,2,3,4,5";
var newstring = remove(string,"4"); // newstring will contain "1,2,3,5"
document.write(newstring+"<br>");
newstring = remove(string,"5");
document.write(newstring+"<br>"); // will contain "1,2,3,4"
You also need to consider the behavior you want if you have repeats, say the string is "1,2,2,4" and I say "remove(2)" should it remove both instances or just the first? this function will remove only the first instance.
Just use multiple substitutions.
s/^$removed,//;
s/,$removed$//;
s/,$removed,/,/;
This will be easier than trying to invent a single replacement that handles all those cases.
string input = "1,2,3,4";
List<string> parts = new List<string>(input.Split(new char[] { ',' }));
parts.RemoveAt(2);
string output = String.Join(",", parts);
Instead of using regex, I would do something like:
- split on comma
- delete the right element
- join with comma
Here is a perl script that does the job:
#!/usr/bin/perl
use 5.10.1;
use strict;
use warnings;
my $toremove = 5;
my $string = "1,2,3,4,5";
my #tmp = split/,/, $string;
#tmp = grep{ $_ != $toremove }#tmp;
$string =join',', #tmp;
say $string;
Output:
1,2,3,4
Javascript has improved since this question was posted.
I use the following regex to remove items from a csv string
let searchStr = "359";
let regex = new RegExp("^" + searchStr + ",?|," + searchStr);
csvStr = csvStr.replace(regex, "");
If the child_id is the start, middle or end, or only item it is replaced.
If the searchStr is at the start of the csvStr it and any trailing comma is replaced. Else if the searchStr is anywhere else in the csvStr it must be preceded with a comma so the searchStr and its preceding comma are replaced by an empty string.
I am trying to separate numbers from a string which includes %,/,etc for eg (%2459348?:, or :2434545/%). How can I separate it, in VB.net
you want only the numbers right?
then you could do it like this
Dim theString As String = "/79465*44498%464"
Dim ret = Regex.Replace(theString, "[^0-9]", String.Empty)
hth
edit:
or do you want to split by all non number chars?
then it would go like this
Dim ret = Regex.Split(theString, "[^0-9]")
You could loop through each character of the string and check the .IsNumber() on it.
This should do:
Dim test As String = "%2459348?:"
Dim match As Match = Regex.Match(test, "\d+")
If match.Success Then
Dim result As String = match.Value
' Do something with result
End If
Result = 2459348
Here's a function which will extract all of the numbers out of a string.
Public Function GetNumbers(ByVal str as String) As String
Dim builder As New StringBuilder()
For Each c in str
If Char.IsNumber(c) Then
builder.Append(c)
End If
Next
return builder.ToString()
End Function