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 n
th 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()
orfor 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.