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comp.lang.ruby

keep changes in string outside of scope?

Matthew Margolis

9/27/2004 9:39:00 PM

The code below is supposed to scan through a string and put a ','
before all zip codes. The code properly detects the zip codes and makes
the change in x but the change is never reflected in thestring. How can
I get these changes to exist outside of x's scope?

thestring.each do |x|
if x =~ /\d{5}/
puts "match"
x[-6] = ","
puts x
end
end


Thank you,
Matthew Margolis
9 Answers

James Gray

9/27/2004 9:46:00 PM

0

On Sep 27, 2004, at 4:39 PM, Matthew Margolis wrote:

> The code below is supposed to scan through a string and put a ','
> before all zip codes. The code properly detects the zip codes and
> makes the change in x but the change is never reflected in thestring.
> How can I get these changes to exist outside of x's scope?
>
> thestring.each do |x|
> if x =~ /\d{5}/
> puts "match"
> x[-6] = ","
> puts x
> end
> end

What about:

thestring.gsub!(/(\d{5})/, '\1')

That help?

James Edward Gray II

dblack

9/27/2004 9:52:00 PM

0

dblack

9/27/2004 9:54:00 PM

0

James Gray

9/27/2004 9:55:00 PM

0

> What about:
>
> thestring.gsub!(/(\d{5})/, '\1')

Oops, forgot the period. The replace string should be '.\1'. Sorry

James Edward Gray II

> That help?
>
> James Edward Gray II
>

Matthew Margolis

9/27/2004 10:03:00 PM

0

David A. Black wrote:

>Hi --
>
>On Tue, 28 Sep 2004, David A. Black wrote:
>
>
>
>> thestring.sub!(/\d{5}/) {|x| "#{x},"}
>>
>>
>
>Amendment: as James G. suggested, use gsub! if you have more than one
>in one string :-) Also make sure that you have no street numbers with
>five digits in a row.
>
>
>David
>
>
>
Excellent. I didn't know that each returned new objects. Thanks a
bunch guys.

-Matt Margolis

dblack

9/27/2004 10:15:00 PM

0

Matthew Margolis

9/27/2004 10:59:00 PM

0

David A. Black wrote:

>Hi --
>
>On Tue, 28 Sep 2004, Matthew Margolis wrote:
>
>
>
>>David A. Black wrote:
>>
>>
>>
>>>Hi --
>>>
>>>On Tue, 28 Sep 2004, David A. Black wrote:
>>>
>>>
>>>
>>>
>>>
>>>> thestring.sub!(/\d{5}/) {|x| "#{x},"}
>>>>
>>>>
>>>>
>>>>
>>>Amendment: as James G. suggested, use gsub! if you have more than one
>>>in one string :-) Also make sure that you have no street numbers with
>>>five digits in a row.
>>>
>>>
>>>David
>>>
>>>
>>>
>>>
>>>
>>Excellent. I didn't know that each returned new objects. Thanks a
>>bunch guys.
>>
>>
>
>Just to clarify: it (each) doesn't always return new objects; it
>depends on the particular case. For example, in the case of
>Array#each, you do get the actual array element. With strings, the
>effect is like splitting the string into lines, i.e., substrings.
>
>
>David
>
>
>
Understood.

-Matt Margolis

Robert Klemme

9/28/2004 7:22:00 AM

0


"James Edward Gray II" <james@grayproductions.net> schrieb im Newsbeitrag
news:CFC5333A-10CF-11D9-8CFB-000A95BA45F8@grayproductions.net...
> > What about:
> >
> > thestring.gsub!(/(\d{5})/, '\1')
>
> Oops, forgot the period. The replace string should be '.\1'. Sorry

He wanted a comma...
:-)

And without diving too far into escape handling I'd use

thestring.gsub!(/\d{5}/, ',\\&')

Note also, that you don't need the grouping.

robert

Robert Klemme

9/28/2004 7:29:00 AM

0


"David A. Black" <dblack@wobblini.net> schrieb im Newsbeitrag
news:Pine.LNX.4.44.0409271452500.8281-100000@wobblini...
> Hi --
>
> On Tue, 28 Sep 2004, David A. Black wrote:
>
> > thestring.sub!(/\d{5}/) {|x| "#{x},"}
>
> Amendment: as James G. suggested, use gsub! if you have more than one
> in one string :-) Also make sure that you have no street numbers with
> five digits in a row.

The block variant isn't necessary in this case. A remark about
efficiency:

require 'benchmark'

REP = 1000

thestring = ("foo 12345 bar baz" * 100).freeze

Benchmark.bm(15) do |b|
b.report "direct" do
REP.times { thestring.gsub(/\d{5}/, ',\\&') }
end

b.report "direct group" do
REP.times { thestring.gsub(/(\d{5})/, ',\\1') }
end

b.report "block" do
REP.times { thestring.gsub(/\d{5}/) {|m| ",#{m}"} }
end
end

user system total real
direct 0.516000 0.000000 0.516000 ( 0.527000)
direct group 0.546000 0.000000 0.546000 ( 0.537000)
block 1.829000 0.000000 1.829000 ( 1.844000)


Kind regards

robert