Parse a number using regex

Question

I would like to take a number and format it as price (as a string). For example I want to take 250000 and display $250,000. How can I accomplish this using regex?

AFAIK, regex can't loop. Are you guaranteed an upper bound on the number? — beatgammit, Mar 11 '13 at 15:09
To put the commas you need to know the number of digits, this can't be achieved with regex only — ichigolas, Mar 11 '13 at 15:18

rvalvik · Accepted Answer · 2013-03-11T15:46:37.107

5

For adding the commas you could give something like this a try:

/(\d)(?=(?:\d{3})+$)/

Then replace every match with \1,.

As such:

"12345512312".gsub(/(\d)(?=(?:\d{3})+$)/,'\1,')  => "12,345,512,312"

This would match any digit followed by an arbitrary number of 3 digit groups.

E.g. the first 2 in the above example is followed by 3 groups: 345, 512 and 312. The first 5 is followed by 2 groups: 512 and 312, etc.

Not sure if you'll be able to add the $ in the same regex though.

edited Mar 11 '13 at 15:46

answered Mar 11 '13 at 15:28

rvalvik

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Would you mind explaining how the `?:` works? It doesn't seem like it is necessary. – Addie Mar 13 '13 at 01:14
1

Strictly speaking the `?:` is not necessary, all it does is make the grouping `\d{3}` a passive group (meaning no back-reference is created). We need to make this a group because we want to match 1 or more occurrences of `\d{3}`. We could've done `(\d{3})+`, but then `\2` would've back-referenced the matching numbers. In this case it would've made no difference, but it keeps it cleaner and makes the intent clearer when we use passive groups. – rvalvik Mar 13 '13 at 08:53

score 2 · Answer 2 · answered Mar 11 '13 at 15:29

2

Try this one out (disclaimer- not a regex):

def prettify(n)
  int, dec = n.to_s.split('.')
  int = int.reverse.scan(/.{1,3}/).join(',').reverse
  [int, dec].reject(&:empty?).join('.')
end

Probably a gem out there for this kind of thing though

answered Mar 11 '13 at 15:29

AJcodez

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score 2 · Answer 3 · edited May 23 '17 at 12:04

2

Matching with regex it's a pain in the #@! in this case since regex engines begin matching from the beginning of the string, not the end (finding 3 digit patterns beginning at the end of the string and going backwards). I'd suggest either doing something like this:

format_int = ->(s) do
  str = s.reverse.scan(/\d{1,3}/).join(',').reverse
  "$#{str}"
end

format_int['2500600'] => "$2,500,600"

... using Kernel#sprintf (this can be a little tricky) or, as you wanted: I was wrong, it can be achieved with regex as shown here and here.

edited May 23 '17 at 12:04

Community

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answered Mar 11 '13 at 15:31

ichigolas

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This is a cleaner solution – bjhaid Mar 11 '13 at 17:37

score 0 · Answer 4 · answered Dec 20 '13 at 20:49

you should be using the number_to_currency helper

number_to_currency(1234567890.50)                    # => $1,234,567,890.50
number_to_currency(1234567890.506)                   # => $1,234,567,890.51
number_to_currency(1234567890.506, precision: 3)     # => $1,234,567,890.506
number_to_currency(1234567890.506, locale: :fr)      # => 1 234 567 890,51 €
number_to_currency("123a456")                        # => $123a456

number_to_currency("123a456", raise: true)           # => InvalidNumberError

number_to_currency(-1234567890.50, negative_format: "(%u%n)")
# => ($1,234,567,890.50)
number_to_currency(1234567890.50, unit: "&pound;", separator: ",", delimiter: "")
# => &pound;1234567890,50
number_to_currency(1234567890.50, unit: "&pound;", separator: ",", delimiter: "", format: "%n %u")
# => 1234567890,50 &pound;

for more info look at http://api.rubyonrails.org/classes/ActionView/Helpers/NumberHelper.html

Parse a number using regex

4 Answers4