re = /(?<=\d)(?=(?:\d{3})+(?!\d|\.\d))/m
str = '123456789123456789123456789
123456789
12345678
12345678.11'
subst = ','
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