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Replace number with repeated characters


Replace number with repeated characters

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Replace number with repeated characters
Tag : regex , By : lhoBas
Date : November 28 2020, 08:01 AM


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preg_replace php to replace repeated characters


Tag : php , By : barefootChild
Date : March 29 2020, 07:55 AM
Hope that helps So I have a list of values like that goes like this: , No need for regex for this:
join(",", array_unique(split(",", $values)))

How to replace repeated pattern of characters


Tag : python , By : AJacques
Date : March 29 2020, 07:55 AM
wish helps you You need backreferences here in order to repeat the match that was actually made, as opposed to trying to make a new match with the same pattern:
([a-z0-9]{2})\1\1
>>> import re
>>> re.sub(r'([a-z0-9]{2})\1\1', r'', "abababwhatevercdcdcd")
'whatever'
>>> re.sub(r'([a-z0-9]{2})\1\1', r'', "wabababhatevercdcdcd")
'whatever'

Regex to replace repeated characters


Tag : java , By : Umang
Date : March 29 2020, 07:55 AM
Hope this helps Can someone give me a Java regex to replace the following. , Change your regex like below.
string.replaceAll("((.)\\2{2})\\2+","$1");

Reducing repeated characters in a string while having a maximum number of repeated characters allowed


Tag : c , By : Bas
Date : March 29 2020, 07:55 AM
may help you . I have no idea how I would even begin to tackle this problem, , Simple solution:
void x(char *s, int n)
{
    char *cp= s;
    int i, j;

    while (*cp) {
        i= 1;
        while (*(cp+i) && ((*(cp+i))&~32)==((*cp)&~32)) i++;    // count character repeat
        for (j=0; j<n && j<i; j++)              // repeat at most max repeat (n)
            *s++ = *(cp+j);
        cp += i;
    }
    *s= '\0';
}

How to group repeated characters by number of the times those got repeated?


Tag : javascript , By : mylonov
Date : October 05 2020, 08:00 AM
I hope this helps you . Objective: , Try (we use characters as keys in h={} and count them)
const randomChars = 'ABaaBCDdeGFAAR';

let h= {};
randomChars.toUpperCase().split('').forEach(c => h[c]=++h[c]||1);
let r = Object.keys(h).sort().map(k => h[k]+k).join(' ');

console.log(r);
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