Forum: Ruby on Rails Regular Expression question

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northband (Guest)
on 2009-02-27 21:47
(Received via mailing list)
Hi -

I would like to use gsub() to strip decimals with trailing zeros from
a string.  My string looks like this:
--
19.0 " / 482.600 mm
--

I would like to end up with this:
--
19 " / 482.6 mm
--

Anyone have a regular expression that can do this?

Thanks!
Rob B. (Guest)
on 2009-02-27 23:15
(Received via mailing list)
On Feb 27, 2009, at 2:47 PM, northband wrote:

> --
> 19 " / 482.6 mm
> --
>
> Anyone have a regular expression that can do this?
>
> Thanks!
> >


It depends on how you make the string.

"19.0".sub(/\.?0+\z/,'')    #=> "19"
"482.600".sub(/\.?0+\z/,'') #=> "482.6"

If you replace \z with (\D|\z) and substitute '\1', it might work.
(You can try it out yourself.)

-Rob

Rob B.    http://agileconsultingllc.com
removed_email_address@domain.invalid
northband_101 (Guest)
on 2009-02-27 23:38
(Received via mailing list)
Awesome - this is a start - I'll take it from here.

Thanks!

On Feb 27, 4:14 pm, Rob B. <removed_email_address@domain.invalid>
pepe (Guest)
on 2009-02-28 17:13
(Received via mailing list)
string = '10.0'
string.sub!(/\.\d+/, '')

This will replace in place (sub!) any dot (\.) followed by at least
one number (\d+) with nothing ('').

Pepe
pepe (Guest)
on 2009-02-28 17:14
(Received via mailing list)
Sorry, I didn't read your first posting fully. My solution will not
work for the case of 482.600.

Pepe
pepe (Guest)
on 2009-02-28 18:49
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OK, got something working you might be able to use.

Just to make things more complicated:

s = '19.0 / 482.600 mm / 19.060 / 482.600 mm'
s.gsub!(/(\.0?[^0])?0+/, '\1').gsub!(/\.[\s\n]/, '')

Pepe
Matt J. (Guest)
on 2009-03-01 01:30
(Received via mailing list)
There appear to be some good solutions here, but I thought I'd jump
in
with a bit of non-Rails technical detail.

I'd double check with the source of this data - the zeros may be
significant.
(see http://en.wikipedia.org/wiki/Significant_figures)

The data given doesn't seem to match that (482.600 mm would be written
as
19.0000"), but it doesn't hurt to verify...

--Matt J.
pepe (Guest)
on 2009-03-01 18:51
(Received via mailing list)
Hi.

I just tested the regexp against 19.0000 and it works, but I got a
little problem with the [\s\n]I just solved:

s = '19.0000 / 482.600 mm / 19.060000 / 482.600 mm'
s.gsub!(/(\.0?[^0])?0+/, '\1').gsub!(/\.([\s\n])/, '\1')

This produces: '19 / 482.6 mm / 19.06 / 482.6 mm'

The "insignificant" trailing zeros after are always trimmed after a
decimal point.

Pepe
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