How to read specific lines from a file (by line number)?

Question

If the file to read is big, and you don't want to read the whole file in memory at once:

fp = open("file")
for i, line in enumerate(fp):
    if i == 25:
        # 26th line
    elif i == 29:
        # 30th line
    elif i > 29:
        break
fp.close()

Note that i == n-1 for the nth line.

---

In Python 2.6 or later:

with open("file") as fp:
    for i, line in enumerate(fp):
        if i == 25:
            # 26th line
        elif i == 29:
            # 30th line
        elif i > 29:
            break

The quick answer:

f=open('filename')
lines=f.readlines()
print lines[25]
print lines[29]

or:

lines=[25, 29]
i=0
f=open('filename')
for line in f:
    if i in lines:
        print i
    i+=1

There is a more elegant solution for extracting many lines: linecache (courtesy of "python: how to jump to a particular line in a huge text file?", a previous stackoverflow.com question).

Quoting the python documentation linked above:

>>> import linecache
>>> linecache.getline('/etc/passwd', 4)
'sys:x:3:3:sys:/dev:/bin/sh\n'

Change the 4 to your desired line number, and you're on. Note that 4 would bring the fifth line as the count is zero-based.

If the file might be very large, and cause problems when read into memory, it might be a good idea to take @Alok's advice and use enumerate().

To Conclude:

• Use fileobject.readlines() or for line in fileobject as a quick solution for small files.
• Use linecache for a more elegant solution, which will be quite fast for reading many files, possible repeatedly.
• Take @Alok's advice and use enumerate() for files which could be very large, and won't fit into memory. Note that using this method might slow because the file is read sequentially.

For the sake of offering another solution:

import linecache
linecache.getline('Sample.txt', Number_of_Line)

I hope this is quick and easy :)

A fast and compact approach could be:

def picklines(thefile, whatlines):
  return [x for i, x in enumerate(thefile) if i in whatlines]

this accepts any open file-like object thefile (leaving up to the caller whether it should be opened from a disk file, or via e.g a socket, or other file-like stream) and a set of zero-based line indices whatlines, and returns a list, with low memory footprint and reasonable speed. If the number of lines to be returned is huge, you might prefer a generator:

def yieldlines(thefile, whatlines):
  return (x for i, x in enumerate(thefile) if i in whatlines)

which is basically only good for looping upon -- note that the only difference comes from using rounded rather than square parentheses in the return statement, making a list comprehension and a generator expression respectively.

Further note that despite the mention of "lines" and "file" these functions are much, much more general -- they'll work on any iterable, be it an open file or any other, returning a list (or generator) of items based on their progressive item-numbers. So, I'd suggest using more appropriately general names;-).