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: breakfp.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
The quick answer:
f=open('filename')lines=f.readlines()print lines[25]print lines[29]
or:
lines=[25, 29]i=0f=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.
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;-).