time.sleep in python exceution order

Question

print(7)
sleep(10)
print(8)

when i typed this in Spyder's command line, I expected 7 to be printed, then after 10 seconds 8, would be printed, however, what happened was that both 7 and 8 are printed after a 10 second pause, why is this the case?

The reason I need this function is that I want Python to keep on searching for a file in a directory until it eventually appears.

done = os.path.exists ("done.txt")
while not done:
    sleep(10)
    done = os.path.exists ("done.txt")

will this work?

EDIT: corrected exist as exists.

Possible duplicate of [Python3 sleep() problem](http://stackoverflow.com/questions/4456581/python3-sleep-problem) — Blorgbeard, Dec 29 '16 at 19:24
For your problem wouldn't `done` be initialized once in the beginning? `sleep(10)` will never occur then. Maybe you should check for `os.path.exist()` as the `while` condition. Like `while os.path.exists('/your/path/file.txt')`? — , Dec 29 '16 at 19:28
It should be `os.path.exists` not `os.path.exist`. Other than that, this should work fine. — Joshua Grigonis, Dec 29 '16 at 19:29

score 2 · Accepted Answer · edited May 23 '17 at 12:08

First: It should be os.path.exists not os.path.exist.

Other than that, this should work fine.

Second: I think kiran.konduru misunderstood the question, but you could certainly simplify this and remove the need for the done variable:

while not os.path.exists("done.txt"):
    sleep(10)

Finally: The time.sleep() function is getting buffered in the Spyder command line, see here. If you use sys.stdout.write() and sys.stdout.flush() rather than a simple print, things will work fine in that command line as well.

time.sleep in python exceution order

1 Answers1