Regex to exclude a specific pattern python

You should avoid patterns starting with .*: they cause too many backtracking steps and slow down the code execution.

In Python, you may always get lines either by reading a file line by line, or by splitting a line with splitlines() and then get the necessary lines by testing them against a pattern without .*s.

1. Reading a file line by line:

final_output = []
with open(filepath, 'r', newline="\n", encoding="utf8") as f:
  for line in f:
    if "fiction" in line and "non-fiction" not in line:
      final_output.append(line.strip())

Or, getting the lines even with non-fiction if there is fiction with no non- in front using a bit modified @jlesuffleur's regex:

import re
final_output = []
rx = re.compile(r'\b(?<!non-)fiction\b')
with open(filepath, 'r', newline="\n", encoding="utf8") as f:
  for line in f:
    if rx.search(line):
      final_output.append(line.strip())

2. Getting lines from a multiline string (with both approaches mentioned above):

import re
text = "Your input string line 1\nLine 2 with fiction\nLine 3 with non-fiction\nLine 4 with fiction and non-fiction"
rx = re.compile(r'\b(?<!non-)fiction\b')
# Approach with regex returning any line containing fiction with no non- prefix:
final_output = [line.strip() for line in text.splitlines() if rx.search(line)]
# => ['Line 2 with fiction']
# Non-regex approach that does not return lines that may contain non-fiction (if they contain fiction with no non- prefix):
final_output = [line.strip() for line in text.splitlines() if "fiction" in line and "non-fiction" not in line]
# => ['Line 2 with fiction', 'Line 4 with fiction and non-fiction']

See a Python demo.

What about a negative lookbehind?

s = 'fiction non-fiction'
res = re.findall("(?<!non-)fiction", s)
res