在python中使用sleep做一个计时器。 如何计算时间？

Question

I'm trying to make a simple timer which prints the time remaining every second.

for k in range(100):
    print(100-k)
    t.sleep(1)
#output
#100
#99
#98
#...
#1

However, this will take slightly longer than 100 seconds, because there will be a delay added when print() is used. For long periods, this is slightly noticeable. Is there a way to account for this, and accurately display the time every second? I know I could just sleep(100) , but this wouldn't let the time left be printed.

Answer 1

import time
start_time=time.time()
for k in range(25):
    print(25-k)
    time.sleep(1)
print("it took  "+str(float(time.time()-start_time)*1000)+" Milliseconds")

the output with print is: it took 26412.75382041931 Milliseconds

the output without print : it took 25053.035020828247 Milliseconds

it should have been just 25000 milliseconds but it is not

printing will take time, even reading the code takes time point is don't expect accurate timing with time.sleep() !!!

Answer 2

You can use time.time() to measure elapsed time.

import time
start_time = time.time()

for k in range(100):
    # k seconds SHOULD be elapsed at this point
    print(100 - k)
    slept_time = time.time() - start_time
    time.sleep(1 + k-slept_time)

Answer 3

Using time.sleep will never give you the accurate time for your timer, since the time it takes is the one second sleep time + printing time, you can use threading.Timer to get more accurate results. https://repl.it/Hwkt :

import threading, time

start_time=time.time()
def count_loop(counter):
  if counter <= 0: 
    print("it took  "+str(float(time.time()-start_time)*1000)+" Milliseconds")
    return

  threading.Timer(1.0, count_loop, args=[counter-1]).start()
  print(counter)      

count_loop(100)

This is still not accurate, but with only very minimum offset, only 45 ms. However, when using time.sleep from legendisback's example, there is apparently 81 ms delay. https://repl.it/HwlK