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Remove all occurences of several characters from a string


Remove all occurences of several characters from a string

By : Jose Burgos
Date : November 22 2020, 02:42 PM
it fixes the issue Usually i opt for a framework solution over a regex one, but in your case regex makes for the most 'elegant' solution:
code :
var str = "AD 123 (+45) AS 6.7(+8%)";

var doubles = new List<double>();
foreach (var item in Regex.Matches(str, @"(\d|\.)+"))
{
    doubles.Add(double.Parse(item.ToString()));
}

//doubles = [123, 45, 6.7, 8] 


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Counting occurences of characters in a String using tail recursion

Counting occurences of characters in a String using tail recursion


By : Billy Rogers
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Possible Solutions:
A String is a list of characters. Group them by identity which is (x => x), then count them. Normally groupBy returns a Map which can by transformed to a list of tuples by toList.
code :
def times(s: String) = s.groupBy(identity).mapValues(_.size).toList 
times: (s: String)List[(Char, Int)]
times("explanation")
res1: List[(Char, Int)] = List((e,1), (x,1), (n,2), (t,1), (a,2), (i,1), (l,1), (p,1), (o,1))
import scala.annotation.tailrec

def myTimes(s: String) : List[(Char,Int)] = {

  @tailrec // comiler cheks if really tailrecursive
  def timesTail(chars: List[Char], res: List[(Char,Int)])  : List[(Char,Int)] = 
    chars match {
      case Nil => res      // we are done when there are no characters left
      case char :: rest => {
          // otherwise 
          val newCharCount = 
            res.
              find (_._1 == char). //check if we already have seen the character
              map{ case (c,count) => (c,count + 1)  }. // if yes, we raise take the old count and raise it by one
              getOrElse( (char,1) ) // otherwise we count one occurrence
          // here it gets recursive 
          timesTail(
                rest, // remaining Characters
                newCharCount :: res.filterNot(_._1 == char) // add the new count to list, remove old if present
          )
      }
    }
  // initial call with empty lsit to start the helper function  
  timesTail(s.toList,List())

}
Reduce multiple occurences of any non-alphanumeric characters in string down to one, **NOT REMOVE**

Reduce multiple occurences of any non-alphanumeric characters in string down to one, **NOT REMOVE**


By : user3673050
Date : March 29 2020, 07:55 AM
around this issue Use a backreference. \1 in the regex refers to the first match group in the expression.
code :
var str = 'HMQ 2.. Elizabetttth      .';

var regex = /([^A-Za-z0-9])\1+/g;

var trimmed = str.replace(regex, "$1");
console.log( trimmed );
C# regex to extract string between two characters but first character has multiple occurences

C# regex to extract string between two characters but first character has multiple occurences


By : thiago r sa
Date : March 29 2020, 07:55 AM
Does that help I'm trying to extract string between two characters. First charachter has multiple occurences and I need the last occurence of first character. I've already checked similar questions but all suggested regexes don't work in case of multiple occurences of first character. , Let's try a bit different pattern:
code :
(?<=-)([^-_]+?)(?=_)
Counting the number of occurences of characters in a string

Counting the number of occurences of characters in a string


By : Ismail Bangulzai
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I am trying to write a Java program which takes in as input a string and counts the number of occurrences of characters in a string and then prints a new string having the character followed by the no of occurrences. , This is the problem:
Remove all characters existing between first n occurences of a specific character in each line in shell

Remove all characters existing between first n occurences of a specific character in each line in shell


By : Mahmoud
Date : March 29 2020, 07:55 AM
This might help you Say I have txt file with characters as follows: , This would be a typical task for cut
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