Regex replacement problem---replace every n-th


#1

Hi all

I have the string '1 20 3 400 5 60 7 800 9 0 ’ and need to replace
every n-th space in the string with another character. So if n=2 and
the replacement character is ‘\n’, then the above would become

‘1 20\n3 400\n5 60\n7 800\n9 0\n’.

I’m sure it must be possible (easy, even) to do this with a regex
substitution, but unfortunately I’m no regex ninja.

So far I have:

'1 20 3 400 5 60 7 800 9 0 '.sub(/(\d* \d* )/) {|s| s + “\n”}
=> "1 20 \n3 400 5 60 7 800 9 0 "

But how do I get the substitution to be ‘repeated’? Changing the regexp
to /(\d* \d*)*/ does not do what I want.

Also, I’m planning to run this on a large string, so I’d like it to be
efficient if possible.

Thanks in advance

C


#2

On Wed, 2006-03-01 at 21:03 +0900, removed_email_address@domain.invalid wrote:

Hi all

I have the string '1 20 3 400 5 60 7 800 9 0 ’ and need to replace
every n-th space in the string with another character. So if n=2 and
the replacement character is ‘\n’, then the above would become

‘1 20\n3 400\n5 60\n7 800\n9 0\n’.

I’m sure it must be possible (easy, even) to do this with a regex
substitution, but unfortunately I’m no regex ninja.

There’s probably a better way to do this, but here’s a ‘metaregex’ idea:

str =  "1 20 3 400 5 60 7 800 9 0 "
# => "1 20 3 400 5 60 7 800 9 0 "

n = 2
# => 2

r = Regexp.new('(' + ('\d+\s' * (n - 1)) + '\d+)(\s)')
# => /(\d+\s\d+)(\s)/

str.gsub(r) { $1 + rep }
# => "1 20\n3 400\n5 60\n7 800\n9 0\n"

n = 3
# => 3

r = Regexp.new('(' + ('\d+\s' * (n - 1)) + '\d+)(\s)')
# => /(\d+\s\d+\s\d+)(\s)/

str.gsub(r) { $1 + rep }
# => "1 20 3\n400 5 60\n7 800 9\n0 "

#3

On Wed, 2006-03-01 at 21:36 +0900, I forgot:

There’s probably a better way to do this, but here’s a ‘metaregex’ idea:

  • rep = “\n”

#4

Hi,

every n-th space in the string with another character. So if n=2 and

=> "1 20 3 400 5 60 7 800 9 0 "

n = 3

=> 3

r = Regexp.new(’(’ + (’\d+\s’ * (n - 1)) + ‘\d+)(\s)’)

=> /(\d+\s\d+\s\d+)(\s)/

str.gsub® { $1 + rep }

=> "1 20 3\n400 5 60\n7 800 9\n0 "

Here’s another idea:

str = "1 20 3 400 5 60 7 800 9 0 "
rep = “\n”
n = 2
str.gsub(/((\d+\s){#{n}})/){$1.chop+rep}

Regards,

Park H.


#5

removed_email_address@domain.invalid wrote:

Thanks in advance

C

How about:

text = '1 20 3 400 5 60 7 800 9 0 ’
n = 2
text.gsub!(/(\d+\s+){#{n-1}}(\d+)(\s+)/) { $1 + $2 + “\n” }

or completely different:

text = '1 20 3 400 5 60 7 800 9 0 ’
n = 2
i = 0
text.gsub!(/\s+/) { |s| (i = (i+1) % n).zero? ? “\n” : s }

Robin


#6

Thanks all.

C


#7

removed_email_address@domain.invalid wrote:

'1 20 3 400 5 60 7 800 9 0 '.sub(/(\d* \d* )/) {|s| s + “\n”}
=> "1 20 \n3 400 5 60 7 800 9 0 "

But how do I get the substitution to be ‘repeated’? Changing the regexp
to /(\d* \d*)*/ does not do what I want.

The answer you’re looking for is String#gsub:
'1 20 3 400 5 60 7 800 9 0 '.gsub(/(\d* \d* )/) {|s| s + “\n”}
=> "1 20 \n3 400\n 5 60\n 7 800\n 9 0\n "

Cheers,
Dave


#8

On Wed, 2006-03-01 at 22:58 +0900, Dave B. wrote:

removed_email_address@domain.invalid wrote:

'1 20 3 400 5 60 7 800 9 0 '.sub(/(\d* \d* )/) {|s| s + “\n”}
=> "1 20 \n3 400 5 60 7 800 9 0 "

But how do I get the substitution to be ‘repeated’? Changing the regexp
to /(\d* \d*)*/ does not do what I want.

The answer you’re looking for is String#gsub:
'1 20 3 400 5 60 7 800 9 0 '.gsub(/(\d* \d* )/) {|s| s + “\n”}
=> "1 20 \n3 400\n 5 60\n 7 800\n 9 0\n "

Well, that’ll teach me not to try and do stuff before my all-important
Caffotine breakfast. I still had the extra * on the end and couldn’t
make it work… Duh :frowning: