Replacing verticle white spaces?

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

puts s.gsub(/\r\n(\r\n)+/,"\r\n\r\n")

Just adjust the number of \r\n :wink:

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

2009/7/17 George G. [email protected]

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

2009/7/17 George G. [email protected]

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?

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

2009/7/17 Glenn J. [email protected]

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

On Thu, Jul 23, 2009 at 8:12 AM, George
George[email protected] 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

This forum is not affiliated to the Ruby language, Ruby on Rails framework, nor any Ruby applications discussed here.

| Privacy Policy | Terms of Service | Remote Ruby Jobs