# Replacing verticle white spaces?

**URL:** <https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247>\
**Category:** Ruby\
**Created:** [July 17, 2009, 2:04pm UTC](https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247 "2009-07-17T14:04:31Z")\
**Posts on this page:** 6\
**Page:** 1

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**Author:** ![georgeuoa](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/g/7ab992/32.png) [@georgeuoa](https://www.ruby-forum.com/u/georgeuoa)\
**Post date:** [July 17, 2009, 2:04pm UTC](https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247/1 "2009-07-17T14:04:31Z")

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Any ideas on how to replace a verticle white space?

for example

tetetetetetetete  
tetetettetetetete

rererererererere  
rerererererererr

such that you end up with

tetetetetetetettete  
etetetetetetetetete

rererererererererer  
erererererererereer

Thank you

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**Author:** ![georgeuoa](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/g/7ab992/32.png) [@georgeuoa](https://www.ruby-forum.com/u/georgeuoa)\
**Post date:** [July 17, 2009, 2:12pm UTC](https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247/2 "2009-07-17T14:12:25Z")

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puts s.gsub(/\r\n(\r\n)+/,“\r\n\r\n”)

Just adjust the number of \r\n 😉

p.s. have a look mainly at (\r\n)+ : a serie of one or more \r\n

2009/7/17 George G. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid)

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<div class="post-metadata">

**Author:** ![georgeuoa](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/g/7ab992/32.png) [@georgeuoa](https://www.ruby-forum.com/u/georgeuoa)\
**Post date:** [July 17, 2009, 5:50pm UTC](https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247/3 "2009-07-17T17:50:56Z")

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At 2009-07-17 08:11AM, “Mirko Viviani” wrote:

> 2009/7/17 George G. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid)
> 
> > Any ideas on how to replace a verticle white space?  
> > […]  
> > puts s.gsub(/\r\n(\r\n)+/,“\r\n\r\n”)

I would write:  
s.gsub(/((?:\r?\n){2})(?:\r?\n)\*/, ‘\1’)

In Ruby, are newlines represented just as “\n” regardless of your  
platform?

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<div class="post-metadata">

**Author:** ![georgeuoa](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/g/7ab992/32.png) [@georgeuoa](https://www.ruby-forum.com/u/georgeuoa)\
**Post date:** [July 17, 2009, 5:55pm UTC](https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247/4 "2009-07-17T17:55:02Z")

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Well… I just tried out with \r\n… neverthless the solution is the  
same  
😉

2009/7/17 Glenn J. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid)

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<div class="post-metadata">

**Author:** ![georgeuoa](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/g/7ab992/32.png) [@georgeuoa](https://www.ruby-forum.com/u/georgeuoa)\
**Post date:** [July 23, 2009, 2:12pm UTC](https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247/5 "2009-07-23T14:12:49Z")

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Hi,  
Thanks you for the replies. Both approaches did not seem to achieve what  
i envisaged. Is because of reading the input as a string? rather than as  
a file?

Maybe to rephrase what i really wanted was to be able to delete the  
repeated lines and only remain with a single empty line.  
meaning if a file had the lines;

tetetetetetetete  
tetetettetetetete

rererererererere  
rerererererererr

remove the empty lines leaving only a single one

tetetetetetetete  
tetetettetetetete

rererererererere  
rerererererererr

Thank you  
George

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<div class="post-metadata">

**Author:** ![georgeuoa](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/g/7ab992/32.png) [@georgeuoa](https://www.ruby-forum.com/u/georgeuoa)\
**Post date:** [July 23, 2009, 3:36pm UTC](https://www.ruby-forum.com/t/replacing-verticle-white-spaces/172247/6 "2009-07-23T15:36:51Z")

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On Thu, Jul 23, 2009 at 8:12 AM, George  
George[removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote:

> rererererererere  
> rerererererererr

Do you have a more specific test case (with code) that is failing?  
Reading from a file and using Glenn J.'s gsub, seems to produce  
your desired output:

> cat z  
> tetetetetetetete  
> tetetettetetetete

rererererererere  
rerererererererr

> ruby -ve ‘puts File.new(“z”).read.gsub(/((?:\r?\n){2})(?:\r?\n)\*/, “\1”)’  
> ruby 1.8.6 (2007-09-24 patchlevel 111) [i386-linux]  
> tetetetetetetete  
> tetetettetetetete

rererererererere  
rerererererererr

> unix2dos z  
> unix2dos: converting file z to DOS format …  
> ruby -ve ‘puts File.new(“z”).read.gsub(/((?:\r?\n){2})(?:\r?\n)\*/, “\1”)’  
> ruby 1.8.6 (2007-09-24 patchlevel 111) [i386-linux]  
> tetetetetetetete  
> tetetettetetetete

rererererererere  
rerererererererr