If I have a list l:
l = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Is there a way to control the following for loop so that the next element in the list is only printed one second after the previous?
for i in l:
print i
In other words, is there a way to elegantly slow down a loop in Python?
You can use time.sleep
import time
for i in l:
print i
time.sleep(1)
If you use time.sleep(1), your loops will run a little over a second since the looping and printing also takes some time. A better way is to sleep for the remainder of the second. You can calculate that by using -time.time()%1
>>> import time
>>> L = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> for i in L:
... time.sleep(-time.time()%1)
... print i
...
It's easy to observe this by using print i, repr(time.time())
>>> for i in L:
... time.sleep(-time.time()%1)
... print i, repr(time.time())
...
0 1368580358.000628
1 1368580359.001082
2 1368580360.001083
3 1368580361.001095
4 1368580362.001149
5 1368580363.001085
6 1368580364.001089
7 1368580365.001086
8 1368580366.001086
9 1368580367.001085
vs
>>> for i in L:
... time.sleep(1)
... print i, repr(time.time())
...
0 1368580334.104903
1 1368580335.106048
2 1368580336.106716
3 1368580337.107863
4 1368580338.109007
5 1368580339.110152
6 1368580340.111301
7 1368580341.112447
8 1368580342.113591
9 1368580343.114737
time.time and time.sleep is negligible, but the time of print is significant, and that sleep is accurate to the relevant scale, both of which are unlikely to be true on most platforms…
time.time() and time.sleep() doesn't matter much as long as it's fairly consistent. sleep() doesn't offer a guarantee for precise timing anyway.sleep isn't accurate or precise no matter what you do. It could easily sleep past the next .000 mark, which means your code will lose a whole second, while the "dumb" code would just lose 50us. Is that really an improvement? You need more complicated logic (basically, keep track of the time to sleep until, so you can correct for oversleep) to actually do this right. There are very few situations where sleep(1) is not good enough, but sleep(-time.time()%1) is.print and takes a substantial part of (but less than) a second, my version will run 3600 times an hour.You can pause the code execution using time.sleep:
import time
for i in l:
print(i)
time.sleep(1)
Use the time.sleep function. Just do time.sleep(1) in your function.
Use time.sleep(number):
for i in range(i)
print(1)
time.sleep(0.05)
sleepcalls.