# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility
import re
regex = (r"(?mx)\n"
r"( # the 1st capture group will contain this sequence\n"
r"1 # this sequence should begin with 1\n"
r"(?=(?:[01]{6,19}) # let's see that there are enough 0s and 1s in a line\n"
r" (.*$)) # the 2nd capture group will contain all characters to the end of a line\n"
r"(?:0*1){6}) # there must be six more 1s in the sequence\n"
r"(?=.{0,13} # complement the 1st capture group to 20 characters\n"
r"\2) # the rest of a line should be 2nd capture group")
test_str = ("0000000\n"
"101010101010111111100000000000001\n"
"00000001100000001110111100000000000000000000000000000000000000000000000000110000000111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001100\n"
"1111111\n"
"111111")
matches = re.finditer(regex, test_str, re.MULTILINE)
for matchNum, match in enumerate(matches, start=1):
print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))
for groupNum in range(0, len(match.groups())):
groupNum = groupNum + 1
print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))
# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.
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 Python, please visit: https://docs.python.org/3/library/re.html