CODEWITHSUNDEEP
pythonregex
J33nn
J33nn

Reputation: 3234

How to find line with regex and remove any preceding lines

I'm looking for way to remove every lines before line which contains specific string in multiline string like this:

string1 
string2
string3
==== bump
string4
string5
string6
==== bump

But only first matching one...

At the end I would like to have this as an output:

==== bump
string4
string5
string6
==== bump

Upvotes: 1

Views: 101

Answers (4)

Todd A. Jacobs
Todd A. Jacobs

Reputation: 84343

Alternative Language: Use Perl's Flip-Flop Operator

Assuming that you've stored your text in /tmp/corpus, you could use the following Perl one-liner:

perl -ne 'print if /\A==== bump/ ... /\A==== bump/' /tmp/corpus

This leverages the power of Perl's range operator. If you want to capture the output from Perl within your Python program, you can use the Python subprocess module. For example:

import subprocess
result = subprocess.check_output(
    "perl -ne 'print if /\A==== bump/ ... /\A==== bump/' /tmp/corpus",
    shell=True)
print result

Upvotes: 1

falsetru
falsetru

Reputation: 369064

lines = '''
string1 
string2
string3
==== bump
string4
string5
string6
==== bump
'''

import re
sep = '==== bump'
matched = re.search('{0}.*?{0}'.format(re.escape(sep)), lines, flags=re.S)
print(matched.group(0))

Upvotes: 0

falsetru
falsetru

Reputation: 369064

import io
import itertools
import sys

lines = io.StringIO(u'''\
string1 
string2
string3
==== bump
string4
string5
string6
==== bump
''')

sep = '==== bump'
it = itertools.dropwhile(lambda line: not line.startswith(sep), lines)
sys.stdout.writelines(it)

Output

==== bump
string4
string5
string6
==== bump

Upvotes: 1

unutbu
unutbu

Reputation: 879451

import re
text = '''\
string1 
string2
string3
==== bump
string4
string5
string6
==== bump'''

print(re.split(r'(=== bump)', text, maxsplit=1)[-1])

yields

string4
string5
string6
==== bump

Upvotes: 1

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