Grep for a word, and if found print 10 lines before and 10 lines after the pattern match

Question

I am processing a huge file. I want to search for a word in the line and when found I should print 10 lines before and 10 lines after the pattern match. How can I do it in Python?

Answer 1

import collections
import itertools
import sys

with open('huge-file') as f:
    before = collections.deque(maxlen=10)
    for line in f:
        if 'word' in line:
            sys.stdout.writelines(before)
            sys.stdout.write(line)
            sys.stdout.writelines(itertools.islice(f, 10))
            break
        before.append(line)

used collections.deque to save up to 10 lines before match, and itertools.islice to get next 10 lines after the match.

---

UPDATE To exclude lines with ip/mac address:

import collections
import itertools
import re  # <---
import sys

addr_pattern = re.compile(
    r'\b\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\b|'
    r'\b[\da-f]{2}:[\da-f]{2}:[\da-f]{2}:[\da-f]{2}:[\da-f]{2}:[\da-f]{2}\b',
    flags=re.IGNORECASE
)  # <--

with open('huge-file') as f:
    before = collections.deque(maxlen=10)
    for line in f:
        if addr_pattern.search(line):  # <---
            continue                   # <---
        if 'word' in line:
            sys.stdout.writelines(before)
            sys.stdout.write(line)
            sys.stdout.writelines(itertools.islice(f, 10))
            break
        before.append(line)

Answer 2

Use grep with -C option, easiest solution:

grep -C 10 'what_to_search' file.txt