re = /(\d)(?=(\d{3})+\b)/
str = '1
10
100
1000
10000
100000
1000000
10000000
100000000
1000000000
10000000000
100000000000
'
subst = '$1,'
result = str.gsub(re, subst)
# Print the result of the substitution
puts result
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Ruby, please visit: http://ruby-doc.org/core-2.2.0/Regexp.html