import time
start = time.clock()
while True:
elapsed = (time.clock() - start)
if elapsed > 10:
print("MOTION")
elapsed = 0
I start a timer, calculate elapsed time, and if 10 seconds have passed, I display "MOTION" and then reset elapsed to 0 so "MOTION" only displays every 10 seconds. For some reason, it doesn't work: MOTION does initially get displayed after 10 seconds, but after that, it keeps getting displayed on every iteration.
What did I do wrong?
You have two options; your code doesn't work because you are trying to reset the clock but instead you reset elapsed, which does nothing.
Using modulo division.
start = time.clock()
while True:
elapsed = (time.clock() - start)
if int(elapsed) % 10:
print("MOTION")
Resetting the clock.
start = time.clock()
while True:
elapsed = (time.clock() - start)
if elapsed >= 10:
print("MOTION")
start = time.clock()
You neglected to reset your reference time: change the basis, not the interval. On each iteration, you reset elapsed to 0, but then immediately go back to the original start time. Change the last line of you loop:
start = time.clock()
while True:
elapsed = (time.clock() - start)
if elapsed > 10:
print("MOTION")
start = time.clock()
Related
I want to work with exactly 20ms sleep time. When i was using time.sleep(0.02), i am facing many problems. It is not working what i want. If I had to give an example;
import time
i = 0
end = time.time() + 10
while time.time() < end:
i += 1
time.sleep(0.02)
print(i)
We wait to see "500" in console. But it is like "320". It is a huge difference. Because sleep time is not working true and small deviations occur every sleep time. It is increasing cumulatively and we are seeing wrong result.
And then, i want to create my new project for clock pulse. Is it that possible with time.time()?
import time
first_time = time.time() * 100 #convert seconds to 10 * miliseconds
first_time = int(first_time) #convert integer
first_if = first_time
second_if = first_time + 2 #for sleep 20ms
third_if = first_time + 4 #for sleep 40ms
fourth_if = first_time + 6 #for sleep 60ms
fifth_if = first_time + 8 #for sleep 80ms
end = time.time() + 8
i = 0
while time.time() < end:
now = time.time() * 100 #convert seconds to 10 * miliseconds
now = int(now) #convert integer
if i == 0 and (now == first_if or now > first_if):
print('1_' + str(now))
i = 1
if i == 1 and (now == second_if or now > second_if):
print('2_' + str(now))
i = 2
if i == 2 and (now == third_if or now > third_if):
print('3_' + str(now))
i = 3
if i == 3 and (now == fourth_if or now > fourth_if):
print('4_' + str(now))
i = 4
if i == 4 and (now == fifth_if or now > fifth_if):
print('5_' + str(now))
break
Out >> 1_163255259009
2_163255259011
3_163255259013
4_163255259015
5_163255259017
Is this project true logic? And If it is true logic, how can finish this projects with true loops?
Because i want these sleeps to happen all the time. Thank you in advice.
Let's say you want to count in increments of 20ms. You need to sleep for the portion of the loop that's not the comparison, increment, and print. Those operations take time, probably about 10ms based on your findings.
If you want to do it in a loop, you can't hard code all the possible end times. You need to do something more general, like taking a remainder.
Start with the time before the loop:
t0 = time.time()
while time.time() < end:
i += 1
Now you need to figure out how long to sleep so that the time between t0 and the end of the sleep is a multiple of 20ms.
(time.time() - t0) % 0.02 tells you how far past a 20ms increment you are because python conveniently supports floating point modulo. The amount of time to wait is then
time.sleep(0.02 - (time.time() - t0) % 0.02)
print(i)
Using sign rules of %, you can reduce the calculation to
time.sleep(-(time.time() - t0) % 0.02)
I am trying to create a countdown in python and I want very simple way of creating that. I watched a couple of videos but couldn't find a right solution for it.
This is the code which I am using right now.
import time
def countdown(t):
while t:
mins, secs = divmod(t, 60)
timer = '{:02d}:{:02d}'.format(mins, secs)
print(timer, end="\r")
time.sleep(1)
t -= 1
print('Time Over!!!!')
t = input("Enter the time in seconds: ")
countdown(int(t))
The problem is that when you sleep for 1 second, it will not be for exactly 1 second and theoretically over long enough time the errors could propagate enough such that you could conceivably be printing out an incorrect time. To correct this, your code needs to actually check in its loop how much time has actually elapsed since the start of the program running and use that to compute what the new value of t is, and it should do this frequently so that the countdown is smooth. For example:
import time
def countdown(t):
start_time = time.time()
start_t = t
# compute accurate new t value aprroximately every .05 seconds:
while t > 0:
mins, secs = divmod(t, 60)
timer = '{:02d}:{:02d}'.format(mins, secs)
print(timer, end="\r")
time.sleep(.05) # finer timing
now = time.time()
elapsed_time = int(now - start_time) # truncated to seconds
t = start_t - elapsed_time
print('Time Over!!!!')
t = input("Enter the time in seconds: ")
countdown(int(t))
This question already has answers here:
Python's time.clock() vs. time.time() accuracy?
(16 answers)
Closed 6 years ago.
I am new to Python programming. I started working on Project Euler this morning and I wanted to find out how long it takes to execute my solution. I have searched online for a solution to my
import time
class Solution(object):
def fibonacci(self,limit):
sum = 0
current = 1
next = 2
while(current <= limit):
if current % 2==0:
sum += current
current, next = next, current + next
return str(sum)
if __name__ == "__main__":
start = time.clock()
solution = Solution().fibonacci(4000000)
elapsed = time.clock()-start
print("Solution: %s"%(solution))
print("Time: %s seconds"%(elapsed))
Output:
Solution: 4613732
Time: 2.006085436846098e-05 seconds
import time
class Solution(object):
def fibonacci(self,limit):
sum = 0
current = 1
next = 2
while(current <= limit):
if current % 2==0:
sum += current
current, next = next, current + next
return str(sum)
if __name__ == "__main__":
start = time.time()
solution = Solution().fibonacci(4000000)
elapsed = time.time()-start
print("Solution: %s"%(solution))
print("Time: %s seconds"%(elapsed))
Output:
Solution: 4613732
Time: 0.0 seconds
My question is
Is the time calculated above correct?
What is the difference between time.time() vs time.clock(). If I use time.time() I get 0.0 as time.
In the Python time module, time.clock() measures the time since the first call of the function in seconds, and time.time() measures the time since January 1st, 1970, in seconds.
time.clock() is generally more precise, so using this is what I recommend. This is the reason why you have the tiny result in the first example, rounded down to zero in the second example.
How can I format the time elapsed from seconds to hours, mins, seconds?
My code:
start = time.time()
... do something
elapsed = (time.time() - start)
Actual Output:
0.232999801636
Desired/Expected output:
00:00:00.23
You could exploit timedelta:
>>> from datetime import timedelta
>>> str(timedelta(seconds=elapsed))
'0:00:00.233000'
If you want to include times like 0.232999801636 as in your input:
import time
start = time.time()
end = time.time()
hours, rem = divmod(end-start, 3600)
minutes, seconds = divmod(rem, 60)
print("{:0>2}:{:0>2}:{:05.2f}".format(int(hours),int(minutes),seconds))
Example:
In [12]: def timer(start,end):
....: hours, rem = divmod(end-start, 3600)
....: minutes, seconds = divmod(rem, 60)
....: print("{:0>2}:{:0>2}:{:05.2f}".format(int(hours),int(minutes),seconds))
....:
In [13]: timer(12345.242,12356.434)
00:00:11.19
In [14]: timer(12300.242,12600.5452)
00:05:00.30
In [19]: timer(0.343,86500.8743)
24:01:40.53
In [16]: timer(0.343,865000.8743)
240:16:40.53
In [17]: timer(0,0.232999801636)
00:00:00.23
The strftime function of time itself can be (ab)used with limitations (no millisec and <24 hr)
elapsed = 4*3600 + 13*60 + 6 # 15186 s
time.strftime("%Hh%Mm%Ss", time.gmtime(elapsed)) # '04h13m06s'
import time
start = time.time()
#do something
end = time.time()
temp = end-start
print(temp)
hours = temp//3600
temp = temp - 3600*hours
minutes = temp//60
seconds = temp - 60*minutes
print('%d:%d:%d' %(hours,minutes,seconds))
If you want one line of code without import any other library, that worked for me (Python v3.7.11):
print("Elapsed time: " + time.strftime("%H:%M:%S.{}".format(str(elapsed % 1)[2:])[:15], time.gmtime(elapsed)))
Output:
Elapsed time: 00:38:01.357318
You can control the milliseconds to be displayed by modifying "[:15]" to "[:11]", then you will get the desired result:
Elapsed time: 00:45:18.65
I have a while loop, and I want it to keep running through for 15 minutes. it is currently:
while True:
#blah blah blah
(this runs through, and then restarts. I need it to continue doing this except after 15 minutes it exits the loop)
Thanks!
Try this:
import time
t_end = time.time() + 60 * 15
while time.time() < t_end:
# do whatever you do
This will run for 15 min x 60 s = 900 seconds.
Function time.time returns the current time in seconds since 1st Jan 1970. The value is in floating point, so you can even use it with sub-second precision. In the beginning the value t_end is calculated to be "now" + 15 minutes. The loop will run until the current time exceeds this preset ending time.
If I understand you, you can do it with a datetime.timedelta -
import datetime
endTime = datetime.datetime.now() + datetime.timedelta(minutes=15)
while True:
if datetime.datetime.now() >= endTime:
break
# Blah
# Blah
Simply You can do it
import time
delay=60*15 ###for 15 minutes delay
close_time=time.time()+delay
while True:
##bla bla
###bla bla
if time.time()>close_time
break
For those using asyncio, an easy way is to use asyncio.wait_for():
async def my_loop():
res = False
while not res:
res = await do_something()
await asyncio.wait_for(my_loop(), 10)
I was looking for an easier-to-read time-loop when I encountered this question here. Something like:
for sec in max_seconds(10):
do_something()
So I created this helper:
# allow easy time-boxing: 'for sec in max_seconds(42): do_something()'
def max_seconds(max_seconds, *, interval=1):
interval = int(interval)
start_time = time.time()
end_time = start_time + max_seconds
yield 0
while time.time() < end_time:
if interval > 0:
next_time = start_time
while next_time < time.time():
next_time += interval
time.sleep(int(round(next_time - time.time())))
yield int(round(time.time() - start_time))
if int(round(time.time() + interval)) > int(round(end_time)):
return
It only works with full seconds which was OK for my use-case.
Examples:
for sec in max_seconds(10) # -> 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
for sec in max_seconds(10, interval=3) # -> 0, 3, 6, 9
for sec in max_seconds(7): sleep(1.5) # -> 0, 2, 4, 6
for sec in max_seconds(8): sleep(1.5) # -> 0, 2, 4, 6, 8
Be aware that interval isn't that accurate, as I only wait full seconds (sleep never was any good for me with times < 1 sec). So if your job takes 500 ms and you ask for an interval of 1 sec, you'll get called at: 0, 500ms, 2000ms, 2500ms, 4000ms and so on. One could fix this by measuring time in a loop rather than sleep() ...
The best solution for best performance is to use #DrV answer and the suggestion from #jfs to use time.monotonic():
import time
from datetime import datetime, timedelta
count = 0
end_time = time.monotonic() + 10
while time.monotonic() < end_time:
count += 1
print(f'10 second result: {count=:,}')
# 10 second result: count=185,519,745
count = 0
end_time = time.time() + 10
while time.time() < end_time:
count += 1
print(f'10 second result: {count=:,}')
# 10 second result: count=158,219,172
count = 0
end_time = datetime.now() + timedelta(seconds=10)
while datetime.now() < end_time:
count += 1
print(f'10 second result: {count=:,}')
# 10 second result: count=39,168,578
try this:
import time
import os
n = 0
for x in range(10): #enter your value here
print(n)
time.sleep(1) #to wait a second
os.system('cls') #to clear previous number
#use ('clear') if you are using linux or mac!
n = n + 1