re = /.*[a-zA-Z.].*(*SKIP)(*FAIL)|\B(?=(\d{3})+(?!\d))/m
str = 'Good
1
23
123
1234
12345
752892
1234567
Bad
123123456.000000
12a3123456.0000000000
123456.12345a12345'
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