How to parse substring between last set of parentheses in string in ruby.

Question

In my ruby on rails app, I am trying to build a parser to extract some metadata out of a string.

Let's say the sample string is:

The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20).

I want to extract the substring out of the last occurence of the ( ).

So, I want to get "ralph, 20" no matter how many ( ) are in the string.

Is there a best way to create this ruby string extraction ... regexp?

Thanks,

John

Answer 1

I would try this (here my regex assumes the first value is alphanumeric and the second value is a digit, adjust accordingly). Here the scan gets all occurrences as an array and the -1 tells us to grab just the last one, which seems to be just what you're asking for:

>> foo = "The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20)."
=> "The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20)."
>> foo.scan(/\(\w+, ?\d+\)/)[-1]
=> "(ralph, 20)"

Answer 2

It looks like you want a sexeger. They work by reversing the string, running a reversed regex against the string, and then reversing the results. Here is an example (pardon the code, I don't really know Ruby):

#!/usr/bin/ruby

s = "The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20).";

reversed_s = s.reverse;
reversed_s =~ /^.*?\)(.*?)\(/;
result = $1.reverse;
puts result;

The fact that this is getting no up votes tells me nobody clicked through to read why you want to use a sexeger, so here is are the results of a benchmark:

do they all return the same thing?
ralph, 20
ralph, 20
ralph, 20
ralph, 20
                        user     system      total        real
scan greedy         0.760000   0.000000   0.760000 (  0.772793)
scan non greedy     0.750000   0.010000   0.760000 (  0.760855)
right index         0.760000   0.000000   0.760000 (  0.770573)
sexeger non greedy  0.400000   0.000000   0.400000 (  0.408110)

And here is the benchmark:

#!/usr/bin/ruby

require 'benchmark'

def scan_greedy(s)
    result = s.scan(/\([^)]*\)/x)[-1]
    result[1 .. result.length - 2]
end

def scan_non_greedy(s)
    result = s.scan(/\(.*?\)/)[-1]
    result[1 .. result.length - 2]
end

def right_index(s)
    s[s.rindex('(') + 1 .. s.rindex(')') -1]
end

def sexeger_non_greedy(s)
    s.reverse =~ /^.*?\)(.*?)\(/
    $1.reverse
end

s = "The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20).";

puts "do they all return the same thing?", 
    scan_greedy(s), scan_non_greedy(s), right_index(s), sexeger_non_greedy(s)

n = 100_000
Benchmark.bm(18) do |x|
    x.report("scan greedy")        { n.times do; scan_greedy(s); end }
    x.report("scan non greedy")    { n.times do; scan_non_greedy(s); end }
    x.report("right index")        { n.times do; scan_greedy(s); end }
    x.report("sexeger non greedy") { n.times do; sexeger_non_greedy(s); end }
end

Answer 3

A simple non regular expression solution:

string = "The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20)."
string[string.rindex('(')..string.rindex(')')]

Example:

irb(main):001:0> string =  "The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20)."
=> "The quick red fox (frank,10) jumped over the lazy brown dog (ralph, 20)."
irb(main):002:0> string[string.rindex('(')..string.rindex(')')]
=> "(ralph, 20)"

And without the parentheses:

irb(main):007:0> string[string.rindex('(')+1..string.rindex(')')-1]
=> "ralph, 20"