Python nested loop - get next N lines

Viewed 1185

I'm new to Python and trying to do a nested loop. I have a very large file (1.1 million rows), and I'd like to use it to create a file that has each line along with the next N lines, for example with the next 3 lines:

1    2
1    3
1    4
2    3
2    4
2    5

Right now I'm just trying to get the loops working with rownumbers instead of the strings since it's easier to visualize. I came up with this code, but it's not behaving how I want it to:

with open('C:/working_file.txt', mode='r', encoding = 'utf8') as f: 
for i, line in enumerate(f):
     line_a = i
     lower_bound = i + 1
     upper_bound = i + 4
     with open('C:/working_file.txt', mode='r', encoding = 'utf8') as g:
        for j, line in enumerate(g):
            while j >= lower_bound and j <= upper_bound:
                line_b = j
                j = j+1
                print(line_a, line_b)

Instead of the output I want like above, it's giving me this:

990     991
990     992
990     993
990     994
990     992
990     993
990     994
990     993
990     994
990     994

As you can see the inner loop is iterating multiple times for each line in the outer loop. It seems like there should only be one iteration per line in the outer loop. What am I missing?

EDIT: My question was answered below, here is the exact code I ended up using:

from collections import deque
from itertools import cycle
log = open('C:/example.txt', mode='w', encoding = 'utf8') 
try:
    xrange 
except NameError: # python3
    xrange = range

def pack(d):
    tup = tuple(d)
    return zip(cycle(tup[0:1]), tup[1:])

def window(seq, n=2):
    it = iter(seq)
    d = deque((next(it, None) for _ in range(n)), maxlen=n)
    yield pack(d)
    for e in it:
        d.append(e)
        yield pack(d)

for l in window(open('c:/working_file.txt', mode='r', encoding='utf8'),100):
    for a, b in l:
        print(a.strip() + '\t' + b.strip(), file=log)
5 Answers
Related