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

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

For the sake of offering another solution:

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

I hope this is quick and easy :)

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.