Regex strings in Ruby - ruby

Question

Input strings:
str1 = "$13.90 Price as Shown"
str2 = "$590.50 $490.00 Price as Selected"
str3 = "$9.90 or 5/$27.50 Price as Selected"
Output strings:
str1 = "13.90"
str2 = "490.00"
str3 = "9.90"
My code to get output:
str = str.strip.gsub(/\s\w{2}\s\d\/\W\d+.\d+/, "") # remove or 5/$27.50 from string
str = /\W\d+.\d+\s\w+/.match(str).to_s.gsub("$", "").gsub(" Price", "")
This code works fine for all 3 different types of strings. But how can I improve my code? Are there any better solutions?
Also guys can you give link to good regex guide/book?

Answer 1

A regex I suggested first is just a sum total of your regexps:
(?<=(?<!\/)\$)\d+.\d+(?=\s\w+)
See demo
Since it is next to impossible to compare numbers with regex, I suggest
Extracting all float numbers
Parse them as float values
Get the minimum one
Here is a working snippet:
def getLowestNumberFromString(input)
arr = input.scan(/(?<=(?<!\/)\$)\d+(?:\.\d+)?/)
arr.collect do |value|
value.to_f
end
return arr.min
end
puts getLowestNumberFromString("$13.90 Price as Shown")
puts getLowestNumberFromString("$590.50 $490.00 Price as Selected")
puts getLowestNumberFromString("$9.90 or 5/$27.50 Price as Selected")
The regex breakdown:
(?<=(?<!\/)\$) - assert that there is a $ symbol not preceded with / right before...
\d+ - 1 or more digits
(?:\.\d+)? - optionally followed with a . followed by 1 or more digits
Note that if you only need to match floats with decimal part, remove the ? and non-capturing group from the last subpattern (/(?<=(?<!\/)\$)\d+\.\d+/ or even /(?<=(?<!\/)\$)\d*\.?\d+/).

Answer 2

Supposing input can be relied upon to look like one of your three examples, how about this?
expr = /\$(\d+\.\d\d)\s+(?:or\s+\d+\/\$\d+\.\d\d\s+)?Price/
str = "$9.90 or 5/$27.50 Price as Selected"
str[expr, 1] # => "9.90"
Here it is on Rubular: http://rubular.com/r/CakoUt5Lo3
Explained:
expr = %r{
\$ # literal dollar sign
(\d+\.\d\d) # capture a price with two decimal places (assume no thousands separator)
\s+ # whitespace
(?: # non-capturing group
or\s+ # literal "or" followed by whitespace
\d+\/ # one or more digits followed by literal "/"
\$\d+\.\d\d # dollar sign and price
\s+ # whitespace
)? # preceding group is optional
Price # the literal word "Price"
}x
You might use it like this:
MATCH_PRICE_EXPR = /\$(\d+\.\d\d)\s+(?:or\s+\d+\/\$\d+\.\d\d\s+)?Price/
def match_price(input)
return unless input =~ MATCH_PRICE_EXPR
$1.to_f
end
puts match_price("$13.90 Price as Shown")
# => 13.9
puts match_price("$590.50 $490.00 Price as Selected")
# => 490.0
puts match_price("$9.90 or 5/$27.50 Price as Selected")
# => 9.9

Answer 3

My code works fine for all 3 types of strings. Just wondering how can
I improve that code
str = str.gsub(/ or \d\/[\$\d.]+/i, '')
str = /(\$[\d.]+) P/.match(str)
Ruby Live Demo
http://ideone.com/18XMjr

Answer 4

A better regex is probably: /\B\$(\d+\.\d{2})\b/
str = "$590.50 $490.00 Price as Selected"
str.scan(/\B\$(\d+\.\d{2})\b/).flatten.min_by(&:to_f)
#=> "490.00"

Answer 5

Assuming you simply want the smallest dollar value in each line:
r = /
\$ # match a dollar sign
\d+ # match one or more digits
\. # match a decimal point
\d{2} # match two digits
/x # extended mode
[str1, str2, str3].map { |s| s.scan(r).min_by { |s| s[1..-1].to_f } }
#=> ["$13.90", "$490.00", "$9.90"]
Actually, you don't have to use a regex. You could do it like this:
def smallest(str)
val = str.each_char.with_index(1).
select { |c,_| c == ?$ }.
map { |_,i| str[i..-1].to_f }.
min
"$%.2f" % val
end
smallest(str1) #=> "$13.90"
smallest(str2) #=> "$490.00"
smallest(str3) #=> "$9.90"