time module sequence in python

Question

import time

def threedot():
    time.sleep(0.5)
    print '.',
    time.sleep(0.5)
    print '.',
    time.sleep(0.5)
    print '.'

threedot()

When the above code is run, the interpreter waits 1.5 seconds and then prints '. . .'; rather than wait 0.5 seconds in-between printing '.'. Why is this? (on python 2.7.3)

You may also consider this http://stackoverflow.com/questions/881696/unbuffered-stdout-in-python-as-in-python-u-from-within-the-program — jamylak, Jun 15 '13 at 07:47

Ignacio Vazquez-Abrams · Accepted Answer · 2013-06-15T08:01:18.213

3

The output is being cached until one of several events which your code doesn't actually trigger until the final print. Flush the output (hint: sys.stdout.flush()) each time.

edited Jun 15 '13 at 08:01

answered Jun 15 '13 at 07:45

Ignacio Vazquez-Abrams

776,304
153
1,341
1,358

libc is buffering the output, not the OS. – Jonathon Reinhart Jun 15 '13 at 07:59

Jonathon Reinhart · Answer 2 · 2013-06-15T08:22:32.447

0

stdout is usually "line-buffered", meaning libc will buffer the output (and not write it to the console) until a newline is encountered, or the stream is closed.

Either write to stderr, or manually flush the output stream each time.

edited Jun 15 '13 at 08:22

answered Jun 15 '13 at 07:50

Jonathon Reinhart

132,704
33
254
328

This clarifies the problem I had. Thanks. – Pro542 Jun 15 '13 at 08:21

time module sequence in python

2 Answers2