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Perform action every once in a while

I am currently trying (Python 2.7) to set an action to be performed with a fixed sampling frequency (i.e. every x milliseconds). However, I am facing inaccurate results. With the code bellow (sampling frequency of 1000Hz, code running for 5 seconds) I would expect to have 5 * 10000 samples. Instead, I'm getting lower values. I would like to have a sampling frequency of 5000Hz at best, but I would also be happy with 1000Hz.
Can anyone help me?
import datetime
count = 0
loop_start_time = start_time = datetime.datetime.now()
while datetime.datetime.now() - loop_start_time <= datetime.timedelta(0,5,0): #loop for 5 seconds
if datetime.datetime.now() - start_time >= datetime.timedelta(0,0,1000): #perform an action every 1000 microseconds (1 millisecond, 0.1 seconds)
start_time = datetime.datetime.now()
count = count + 1
print count
Best regards,
T2

Would something like this work?
import time
import threading
total_time = 5
increment_time = .1
is_running = False
count = 0
def incrementCounter():
global count
global is_running
is_running = True
threading.Timer(increment_time, incrementCounter).start()
count += 1
# Any additional code should be put here
end_time = time.time() + total_time
while time.time() < end_time:
if not is_running:
incrementCounter()
print count
Output for this I am getting ~50.

python - printing something in a for loop every n seconds

I have a for loop that iterates over a number and performs some simple calculations. I am trying to figure out how to print out ( or log to file) the current value of 'val' every .5 to 1 second with out having to pause or sleep during the loop. Here is a super simple example
val_list = []
for i in xrange(iterations):
val = (i*(1/2)) * pi
val2 = np.linalg.pinv(val)
# print or write to log val2 after every half second (or 1 second)
val_list.append(val2)

Just use time.time to capture the time before starting, then check how long it's been after you calculate val2:
import time
val_list = []
prev_time = time.time()
for i in xrange(iterations):
val = (i*(1/2)) * pi
val2 = np.linalg.pinv(val)
# print or write to log val2 after every half second (or 1 second)
dt = time.time() - prev_time
if dt > 1:
# print or write to log here
prev_time = time.time()
val_list.append(val2)

You can use time.time():
from time import time as t
val_list = []
nowTime = t()
for i in xrange(iterations):
val = (i*(1/2)) * pi
val2 = np.linalg.pinv(val)
curTime = t()
if curTime - nowTime >= 0.5:
#Do your stuff
nowTime = curTime
val_list.append(val2)

You can achieve this using Threads.
Here's a documentation on how to utilize Threads : https://docs.python.org/3/library/threading.html ( If you're using Python2.7 then change the 3 in the url to a 2 )
Here's a link which is similar to what you want and should also point you in the right direction : Python threading.timer - repeat function every 'n' seconds
Basically you have to create a Thread that will only execute ever n number of seconds. On each iteration it will print the value. The above link should suffice for that. Good luck !

How would I format inputted minutes/seconds into datetime.now() format?

I have created a program which requires the user to input a given TimeType (Seconds/Minutes) and an objectivetime (integer) (e.g. if the user inputs Seconds and then 5, then the objective time is 5 Secs). In addition, the program also checks the time at the start and then checks the time after to work out the elapsed_time. I then wanted to check if the elapsed_time is more than or equal to the objective time. The code i have written is shown below (minus the other irrelevant code in between a and b)
import time, sys, os, random
a = datetime.datetime.now().replace(microsecond=0)
#random code...
b = datetime.datetime.now().replace(microsecond=0)
timetype = input("Enter time type: ")#Either Seconds or Minutes
objectivetime = input("Enter objective time: ")#This will be an integer
if timetype == 'Seconds':
#convert objectivetime to seconds and present in same format as a or b
else:
#convert objectivetime into minutes and present in same format as a or b
elapsed_time = b-a
if objectivetime<=elapsed_time:
'Well done'
else:
'You have exceeded the objective time'
Is there any way that the code could be edited to get the objective time to be formatted in the same way as a and b (start_time and end_time)??
Update--------------------------------------------------------------------------
How would I work out which time is bigger (elapsed_time or objtime)??
See updated code:
import time, sys, os, random
import datetime
a = datetime.datetime.now().replace(microsecond=0)
print(a)
#random code...
time.sleep(3)
b = datetime.datetime.now().replace(microsecond=0)
timetype = input("Enter time type: ")#Either Seconds or Minutes
objectivetime = input("Enter objective time: ")#This will be an integer
objectivetime = int(objectivetime) # I noticed you didn't do this in your code
delta = None
if timetype == 'Seconds':
delta = datetime.timedelta(seconds = objectivetime)
else:
delta = datetime.timedelta(minutes = objectivetime)
objtime = a + delta
print(objtime)
#How would I work out which time is greater?? (elapsed_time or objtime)
elapsed_time = b-a
print(elapsed_time-objtime)

objectivetime = int(objectivetime) # I noticed you didn't do this in your code
delta = None
if timetype == 'Seconds':
delta = datetime.timedelta(seconds = objectivetime)
else:
delta = datetime.timedelta(minutes = objectivetime)
objtime = a + delta
This will give you a datetime object like datetime.now()
To compare the datetime object, you can simply use the comparison operators. For example,
b <= objtime # Finished before (or on) the objective time
b > objtime # Finished after the objective time
You don't even need to subtract a from b! If you still want to, then compare the time deltas instead
elapsed_time <= delta # Finished before (or on) the objective time
elapsed_time > delta # Finished after the objective time

Algorithm timing in Python

I want to compute how many times my computer can do counter += 1 in one second. A naive approach is the following:
from time import time
counter = 0
startTime = time()
while time() - startTime < 1:
counter += 1
print counter
The problem is time() - startTime < 1 may be considerably more expensive than counter += 1.
Is there a way to make a less "clean" 1 sec sample of my algorithm?

The usual way to time algorithms is the other way around: Use a fixed number of iterations and measure how long it takes to finish them. The best way to do such timings is the timeit module.
print timeit.timeit("counter += 1", "counter = 0", number=100000000)
Note that timing counter += 1 seems rather pointless, though. What do you want to achieve?

Why don't you infer the time instead? You can run something like:
from datetime import datetime
def operation():
counter = 0
tbeg = datetime.utcnow()
for _ in range(10**6):
counter += 1
td = datetime.utcnow() - tbeg
return (td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6)/10.0**6
def timer(n):
stack = []
for _ in range(n):
stack.append(operation()) # units of musec/increment
print sum(stack) / len(stack)
if __name__ == "__main__":
timer(10)
and get the average elapsed microseconds per increment; I get 0.09 (most likely very inaccurate). Now, it is a simple operation to infer that if I can make one increment in 0.09 microseconds, then I am able to make about 11258992 in one second.
I think the measurements are very inaccurate, but maybe is a sensible approximation?

I have never worked with the time() library, but according to that code I assume it counts seconds, so what if you do the /sec calculations after ctrl+C happens? It would be something like:
#! /usr/bin/env python
from time import time
import signal
import sys
#The ctrl+C interruption function:
def signal_handler(signal, frame):
counts_per_sec = counter/(time()-startTime)
print counts_per_sec
exit(0)
signal.signal(signal.SIGINT, signal_handler)
counter = 0
startTime = time()
while 1:
counter = counter + 1
Of course, it wont be exact because of the time passed between the last second processed and the interruption signal, but the more time you leave the script running, the more precise it will be :)

Here is my approach
import time
m = 0
timeout = time.time() + 1
while True:
if time.time() > timeout:
break
m = m + 1
print(m)

Get timer ticks in Python

I'm just trying to time a piece of code. The pseudocode looks like:
start = get_ticks()
do_long_code()
print "It took " + (get_ticks() - start) + " seconds."
How does this look in Python?
More specifically, how do I get the number of ticks since midnight (or however Python organizes that timing)?

In the time module, there are two timing functions: time and clock. time gives you "wall" time, if this is what you care about.
However, the python docs say that clock should be used for benchmarking. Note that clock behaves different in separate systems:
on MS Windows, it uses the Win32 function QueryPerformanceCounter(), with "resolution typically better than a microsecond". It has no special meaning, it's just a number (it starts counting the first time you call clock in your process).
# ms windows
t0= time.clock()
do_something()
t= time.clock() - t0 # t is wall seconds elapsed (floating point)
on *nix, clock reports CPU time. Now, this is different, and most probably the value you want, since your program hardly ever is the only process requesting CPU time (even if you have no other processes, the kernel uses CPU time now and then). So, this number, which typically is smaller¹ than the wall time (i.e. time.time() - t0), is more meaningful when benchmarking code:
# linux
t0= time.clock()
do_something()
t= time.clock() - t0 # t is CPU seconds elapsed (floating point)
Apart from all that, the timeit module has the Timer class that is supposed to use what's best for benchmarking from the available functionality.
¹ unless threading gets in the way…
² Python ≥3.3: there are time.perf_counter() and time.process_time(). perf_counter is being used by the timeit module.

What you need is time() function from time module:
import time
start = time.time()
do_long_code()
print "it took", time.time() - start, "seconds."
You can use timeit module for more options though.

Here's a solution that I started using recently:
class Timer:
def __enter__(self):
self.begin = now()
def __exit__(self, type, value, traceback):
print(format_delta(self.begin, now()))
You use it like this (You need at least Python 2.5):
with Timer():
do_long_code()
When your code finishes, Timer automatically prints out the run time. Sweet! If I'm trying to quickly bench something in the Python Interpreter, this is the easiest way to go.
And here's a sample implementation of 'now' and 'format_delta', though feel free to use your preferred timing and formatting method.
import datetime
def now():
return datetime.datetime.now()
# Prints one of the following formats*:
# 1.58 days
# 2.98 hours
# 9.28 minutes # Not actually added yet, oops.
# 5.60 seconds
# 790 milliseconds
# *Except I prefer abbreviated formats, so I print d,h,m,s, or ms.
def format_delta(start,end):
# Time in microseconds
one_day = 86400000000
one_hour = 3600000000
one_second = 1000000
one_millisecond = 1000
delta = end - start
build_time_us = delta.microseconds + delta.seconds * one_second + delta.days * one_day
days = 0
while build_time_us > one_day:
build_time_us -= one_day
days += 1
if days > 0:
time_str = "%.2fd" % ( days + build_time_us / float(one_day) )
else:
hours = 0
while build_time_us > one_hour:
build_time_us -= one_hour
hours += 1
if hours > 0:
time_str = "%.2fh" % ( hours + build_time_us / float(one_hour) )
else:
seconds = 0
while build_time_us > one_second:
build_time_us -= one_second
seconds += 1
if seconds > 0:
time_str = "%.2fs" % ( seconds + build_time_us / float(one_second) )
else:
ms = 0
while build_time_us > one_millisecond:
build_time_us -= one_millisecond
ms += 1
time_str = "%.2fms" % ( ms + build_time_us / float(one_millisecond) )
return time_str
Please let me know if you have a preferred formatting method, or if there's an easier way to do all of this!

The time module in python gives you access to the clock() function, which returns time in seconds as a floating point.
Different systems will have different accuracy based on their internal clock setup (ticks per second) but it's generally at least under 20milliseconds, and in some cases better than a few microseconds.
-Adam

import datetime
start = datetime.datetime.now()
do_long_code()
finish = datetime.datetime.now()
delta = finish - start
print delta.seconds
From midnight:
import datetime
midnight = datetime.datetime.now().replace(hour=0, minute=0, second=0, microsecond=0)
now = datetime.datetime.now()
delta = now - midnight
print delta.seconds

If you have many statements you want to time, you could use something like this:
class Ticker:
def __init__(self):
self.t = clock()
def __call__(self):
dt = clock() - self.t
self.t = clock()
return 1000 * dt
Then your code could look like:
tick = Ticker()
# first command
print('first took {}ms'.format(tick())
# second group of commands
print('second took {}ms'.format(tick())
# third group of commands
print('third took {}ms'.format(tick())
That way you don't need to type t = time() before each block and 1000 * (time() - t) after it, while still keeping control over formatting (though you could easily put that in Ticket too).
It's a minimal gain, but I think it's kind of convenient.