Problems with sys.stdout.write() with time.sleep() in a function

问题

What I wanted is printing out 5 dots that a dot printed per a second using time.sleep(), but the result was 5 dots were printed at once after 5 seconds delay.
Tried both print and sys.stdout.write, same result.

Thanks for any advices.

import time
import sys

def wait_for(n):
    """Wait for {n} seconds. {n} should be an integer greater than 0."""
    if not isinstance(n, int):
        print 'n in wait_for(n) should be an integer.'
        return
    elif n < 1:
        print 'n in wait_for(n) should be greater than 0.'
        return
    for i in range(0, n):
        sys.stdout.write('.')
        time.sleep(1)
    sys.stdout.write('\n')

def main():
    wait_for(5)    # FIXME: doesn't work as expected

if __name__ == '__main__':
    try:
        main()
    except KeyboardInterrupt:
        print '\nAborted.'

回答1:

You need to flush after writing.

sys.stdout.write('foo')
sys.stdout.flush()
wastetime()
sys.stdout.write('bar')
sys.stdout.flush()

回答2:

You should use sys.stderr.write for progress bars; stderr has the (not at all coincidental) advantage of not being buffered, so no sys.stderr.flush calls are needed.

See also this answer.

来源：https://stackoverflow.com/questions/2808832/problems-with-sys-stdout-write-with-time-sleep-in-a-function