clock

why same python code has different clock time?

你说的曾经没有我的故事 提交于 2021-02-19 04:15:43
问题 I'm benchmarking my server using following python code: import time initial_clock = time.clock() res = 0 for i in range(1, 10000000): res += i * i print (time.clock() - initial_clock) When I run it multiple times I get different execution times from 2.163377 seconds to 2.970836 seconds. I know same code may have different execution time due to variation in CPU load but as is said in time.clock documentation it only considers current process clocks, so it should have same execution time using

why same python code has different clock time?

纵饮孤独 提交于 2021-02-19 04:15:30
问题 I'm benchmarking my server using following python code: import time initial_clock = time.clock() res = 0 for i in range(1, 10000000): res += i * i print (time.clock() - initial_clock) When I run it multiple times I get different execution times from 2.163377 seconds to 2.970836 seconds. I know same code may have different execution time due to variation in CPU load but as is said in time.clock documentation it only considers current process clocks, so it should have same execution time using

why same python code has different clock time?

牧云@^-^@ 提交于 2021-02-19 04:15:04
问题 I'm benchmarking my server using following python code: import time initial_clock = time.clock() res = 0 for i in range(1, 10000000): res += i * i print (time.clock() - initial_clock) When I run it multiple times I get different execution times from 2.163377 seconds to 2.970836 seconds. I know same code may have different execution time due to variation in CPU load but as is said in time.clock documentation it only considers current process clocks, so it should have same execution time using

Changing brightness depending on sound (Processing)

。_饼干妹妹 提交于 2021-02-08 06:14:47
问题 I am learning processing right now and I am trying to make a sketch that could change colour when the sound changes. (When Amplitude + , Then Brightness+ ) Because changing colour does not need to change as rapid as the draw() function. So how could I build a clock so that the color would not change in every draw? This is the code I am using right now: import ddf.minim.*; import ddf.minim.signals.*; import ddf.minim.analysis.*; import ddf.minim.effects.*; Minim minim; AudioPlayer song; FFT

Changing brightness depending on sound (Processing)

房东的猫 提交于 2021-02-08 06:14:33
问题 I am learning processing right now and I am trying to make a sketch that could change colour when the sound changes. (When Amplitude + , Then Brightness+ ) Because changing colour does not need to change as rapid as the draw() function. So how could I build a clock so that the color would not change in every draw? This is the code I am using right now: import ddf.minim.*; import ddf.minim.signals.*; import ddf.minim.analysis.*; import ddf.minim.effects.*; Minim minim; AudioPlayer song; FFT

Digital clock in status bar in python 3 and tkinter

若如初见. 提交于 2021-02-07 10:49:20
问题 I want to put this digital clock: import sys from tkinter import * import time root = Tk() time1 = '' clock = Label(root, font=('times', 20, 'bold'), bg='green') clock.pack(fill=BOTH, expand=1) def tick(): global time1 # get the current local time from the PC time2 = time.strftime('%H:%M:%S') # if time string has changed, update it if time2 != time1: time1 = time2 clock.config(text=time2) # calls itself every 200 milliseconds # to update the time display as needed # could use >200 ms, but

How do I calculate the angle between the hour and minutes hands?

痞子三分冷 提交于 2021-02-06 19:59:23
问题 I'm trying to workout this problem, but I am still struggling to understand the logic to solve this problem. hour degree = 360 / 12 = 30 minutes degree = 360 / 12 / 60 = 0.5 So, according to this, I thought I could formulate the following function in python: def clockangles(hour, min): return (hour * 30) + (min * 0.5) For the hour, it works fine, as it appears to have a 1=1 mapping. But for the minute there is one problem at least. When it's 0 minutes, the minutes hand points to 12. For

How do I calculate the angle between the hour and minutes hands?

≯℡__Kan透↙ 提交于 2021-02-06 19:58:13
问题 I'm trying to workout this problem, but I am still struggling to understand the logic to solve this problem. hour degree = 360 / 12 = 30 minutes degree = 360 / 12 / 60 = 0.5 So, according to this, I thought I could formulate the following function in python: def clockangles(hour, min): return (hour * 30) + (min * 0.5) For the hour, it works fine, as it appears to have a 1=1 mapping. But for the minute there is one problem at least. When it's 0 minutes, the minutes hand points to 12. For

Python sched.scheduler exceeds max recursion depth

情到浓时终转凉″ 提交于 2021-02-06 12:44:27
问题 I have recently started learning Python and part of the simple app I am making includes a timer with a hh:mm:ss display running in its own thread. Looking around the web I found two ways of implementing this: Using sched.scheduler Using threading.Timer The way I did it looks similar for both implementations: sched: def tick(self, display, alarm_time): # Schedule this function to run every minute s = sched.scheduler(time.time, time.sleep) s.enter(1, 1, self.tick, ([display, alarm_time])) #

Getting Negative Values Using clock_gettime

江枫思渺然 提交于 2021-02-04 13:50:16
问题 In the following program, I have tried to measure the execution time of a job(for loop). Most of the time it works fine, however, sometimes, it returns negative values!! My first guess is that the variable may get overflowed. Can anyone please let me whether I am right or not? How can I solve the problem? Thanks int main(int argc, char **argv) { long int ST; long int ET; struct timespec gettime_now; clock_gettime(CLOCK_REALTIME, &gettime_now); ST= gettime_now.tv_nsec; for (i=0; i < 1000; i++)