我有一个Python命令行程序,需要一段时间才能完成。我想知道完成跑步所需的确切时间。
我看过timeit模块,但它似乎只适用于小代码片段。我想给整个节目计时。
我有一个Python命令行程序,需要一段时间才能完成。我想知道完成跑步所需的确切时间。
我看过timeit模块,但它似乎只适用于小代码片段。我想给整个节目计时。
当前回答
如果您想以微秒为单位测量时间,那么可以使用以下版本,完全基于Paul McGuire和Nicojo的答案——这是Python 3代码。我还为它添加了一些颜色:
import atexit
from time import time
from datetime import timedelta, datetime
def seconds_to_str(elapsed=None):
if elapsed is None:
return datetime.now().strftime("%Y-%m-%d %H:%M:%S.%f")
else:
return str(timedelta(seconds=elapsed))
def log(txt, elapsed=None):
colour_cyan = '\033[36m'
colour_reset = '\033[0;0;39m'
colour_red = '\033[31m'
print('\n ' + colour_cyan + ' [TIMING]> [' + seconds_to_str() + '] ----> ' + txt + '\n' + colour_reset)
if elapsed:
print("\n " + colour_red + " [TIMING]> Elapsed time ==> " + elapsed + "\n" + colour_reset)
def end_log():
end = time()
elapsed = end-start
log("End Program", seconds_to_str(elapsed))
start = time()
atexit.register(end_log)
log("Start Program")
log()=>打印定时信息的函数。
txt==>要记录的第一个参数及其用于标记计时的字符串。
atexit==>Python模块,用于注册程序退出时可以调用的函数。
其他回答
这是保罗·麦奎尔的回答,对我来说很有用。以防有人在运行这个问题时遇到问题。
import atexit
from time import clock
def reduce(function, iterable, initializer=None):
it = iter(iterable)
if initializer is None:
value = next(it)
else:
value = initializer
for element in it:
value = function(value, element)
return value
def secondsToStr(t):
return "%d:%02d:%02d.%03d" % \
reduce(lambda ll,b : divmod(ll[0],b) + ll[1:],
[(t*1000,),1000,60,60])
line = "="*40
def log(s, elapsed=None):
print (line)
print (secondsToStr(clock()), '-', s)
if elapsed:
print ("Elapsed time:", elapsed)
print (line)
def endlog():
end = clock()
elapsed = end-start
log("End Program", secondsToStr(elapsed))
def now():
return secondsToStr(clock())
def main():
start = clock()
atexit.register(endlog)
log("Start Program")
导入文件后,从程序中调用timing.main()。
以下代码段以可读的<HH:MM:SS>格式打印经过的时间。
import time
from datetime import timedelta
start_time = time.time()
#
# Perform lots of computations.
#
elapsed_time_secs = time.time() - start_time
msg = "Execution took: %s secs (Wall clock time)" % timedelta(seconds=round(elapsed_time_secs))
print(msg)
我也喜欢Paul McGuire的回答,并提出了一个更符合我需求的上下文管理器表单。
import datetime as dt
import timeit
class TimingManager(object):
"""Context Manager used with the statement 'with' to time some execution.
Example:
with TimingManager() as t:
# Code to time
"""
clock = timeit.default_timer
def __enter__(self):
"""
"""
self.start = self.clock()
self.log('\n=> Start Timing: {}')
return self
def __exit__(self, exc_type, exc_val, exc_tb):
"""
"""
self.endlog()
return False
def log(self, s, elapsed=None):
"""Log current time and elapsed time if present.
:param s: Text to display, use '{}' to format the text with
the current time.
:param elapsed: Elapsed time to display. Dafault: None, no display.
"""
print s.format(self._secondsToStr(self.clock()))
if(elapsed is not None):
print 'Elapsed time: {}\n'.format(elapsed)
def endlog(self):
"""Log time for the end of execution with elapsed time.
"""
self.log('=> End Timing: {}', self.now())
def now(self):
"""Return current elapsed time as hh:mm:ss string.
:return: String.
"""
return str(dt.timedelta(seconds = self.clock() - self.start))
def _secondsToStr(self, sec):
"""Convert timestamp to h:mm:ss string.
:param sec: Timestamp.
"""
return str(dt.datetime.fromtimestamp(sec))
只需使用timeit模块。它同时适用于Python 2和Python 3。
import timeit
start = timeit.default_timer()
# All the program statements
stop = timeit.default_timer()
execution_time = stop - start
print("Program Executed in "+str(execution_time)) # It returns time in seconds
它在几秒钟内返回,您可以获得执行时间。这很简单,但您应该在启动程序执行的主函数中编写这些。如果您想获得执行时间,即使在出现错误时,也可以将参数“Start”设置为它,并在那里进行如下计算:
def sample_function(start,**kwargs):
try:
# Your statements
except:
# except statements run when your statements raise an exception
stop = timeit.default_timer()
execution_time = stop - start
print("Program executed in " + str(execution_time))
Python程序执行度量的时间可能不一致,具体取决于:
可以使用不同的算法评估相同的程序运行时间因算法而异运行时间因实现而异运行时间因计算机而异基于小输入,运行时间不可预测
这是因为最有效的方法是使用“增长顺序”,并学习大“O”符号来正确地执行。
无论如何,您可以尝试使用以下简单算法来评估任何Python程序在每秒特定机器计数步骤中的性能:使其适应您想要评估的计划
import time
now = time.time()
future = now + 10
step = 4 # Why 4 steps? Because until here already four operations executed
while time.time() < future:
step += 3 # Why 3 again? Because a while loop executes one comparison and one plus equal statement
step += 4 # Why 3 more? Because one comparison starting while when time is over plus the final assignment of step + 1 and print statement
print(str(int(step / 10)) + " steps per second")