Python: Print x lines after a pattern including the line with pattern

If a line starts with a number(xyz) in a file, I need to print(or write to a file) this line and the next xyz+1 lines.

What's the best way to do this?

xyz

xyz+1

So far, I've been able to print the line that starts with an int. How do I print the next lines?

int

import glob, os, sys
import subprocess

file = 'filename.txt'
with open(file,'r') as f:
data = f.readlines()
for line in data:
if line[0].isdigit():
print int(line)


If I made an iterator out of data, the print function skips a line every time.

data

with open(file,'r') as f:
data = f.readlines()
x = iter(data)
for line in x:
if line[0].isdigit():
print int(line)
for i in range(int(line)):
print x.next()


How could I make it stop skipping lines?

xyz

int(line)

1 Answer
1

Use a flag, when you find the line set it to true, then use it to write all future lines:

can_write = False
with open('source.txt') as f, open('destination.txt', 'w') as fw:
for line in f:
if line.startswith(xyz):
can_write = True
if can_write:
fw.write(line)


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