Just had a quick look at the documentation for open: http://docs.python.org/lib/built-in-funcs.html I think that: sys.stdout = open("yourfile","w", 0) will give you no buffer.
On 06/06/07, André Pang <[EMAIL PROTECTED]> wrote:
On 06/06/2007, at 9:18 PM, Lindsay Holmwood wrote: > When I run the program it prints out each line to the shell with > the delay as expected, but when I redirect stdout to a file the > output is written in big chunks of 30+ lines. > [..] > The Python source: > > #!/usr/bin/env python > # > # logfeeder.py > # reads a specified file and outputs it in delayed chunks > # > > import sys > import random > import time > > if sys.argv < 1: > print 'Usage logfeeder.py <input>' > > lines = file(sys.argv[1]).readlines() > for line in lines: > print line.strip() > time.sleep(random.random() * 5) I'm no Python nor UNIX tty expert, but try adding a flush() after your print(). Terminal I/O (i.e. console stdout) is typically line- buffered; file I/O is typically block-buffered. Your "big chunks of 30+" lines might correspond very nicely to 1024/2048/4096 bytes. See also the setvbuf(3) manpage, setlinebuf(3) and friends... they control line buffering on FILE* handles in C. No idea what the equivalent in Python is, but I'm sure there is one. Hope that helps! -- % Andre Pang : trust.in.love.to.save <http://www.algorithm.com.au/> _______________________________________________ coders mailing list coders@slug.org.au http://lists.slug.org.au/listinfo/coders
-- Michael Connors
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