Python Regex - Capture match and previous two lines

Question

How can I capture a match and the previous two lines? I tried this but get:

unterminated subpattern at position 0 (line 1, column 1)

output = re.findall('(.*\r\n{2}random data.',f.read(), re.DOTALL)

What do you want to match? Repeated groups of two lines, like `line1\n line2\n line1\n line2\n`? Or lines with some spefific text? — mrzasa, Feb 22 '18 at 09:42

score 4 · Accepted Answer · answered Feb 22 '18 at 09:43

4

You may use

re.findall(r'(?:.*\r?\n){2}.*random data.*', s)

Note you can't use re.DOTALL or .* will match up to the end of the input and you will only get the last occurrence.

Pattern details

(?:.*\r?\n){2} - 2 occurrences of a sequence of
- .* - any 0+ chars other than line break chars, as many as possible (a line)
- \r?\n - a line ending (CRLF or LF)
.*random data.* - a line containing random data substring.

answered Feb 22 '18 at 09:43

Wiktor Stribiżew

What is the 's'? Gives me: NameError: name 's' is not defined – arealhobo Feb 22 '18 at 17:00
1

Worked like you said. – arealhobo Feb 22 '18 at 17:51

1 Answers1