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
根据这个答案,创建了一个简单但方便的工具。
import time
from datetime import timedelta
def start_time_measure(message=None):
if message:
print(message)
return time.monotonic()
def end_time_measure(start_time, print_prefix=None):
end_time = time.monotonic()
if print_prefix:
print(print_prefix + str(timedelta(seconds=end_time - start_time)))
return end_time
用法:
total_start_time = start_time_measure()
start_time = start_time_measure('Doing something...')
# Do something
end_time_measure(start_time, 'Done in: ')
start_time = start_time_measure('Doing something else...')
# Do something else
end_time_measure(start_time, 'Done in: ')
end_time_measure(total_start_time, 'Total time: ')
输出:
Doing something...
Done in: 0:00:01.218000
Doing something else...
Done in: 0:00:01.313000
Total time: 0:00:02.672000
其他回答
我在查找两种不同方法的运行时间时遇到的问题,这两种方法用于查找所有<=一个数的素数。当在程序中进行用户输入时。
错误的方法
#Sample input for a number 20
#Sample output [2, 3, 5, 7, 11, 13, 17, 19]
#Total Running time = 0.634 seconds
import time
start_time = time.time()
#Method 1 to find all the prime numbers <= a Number
# Function to check whether a number is prime or not.
def prime_no(num):
if num<2:
return False
else:
for i in range(2, num//2+1):
if num % i == 0:
return False
return True
#To print all the values <= n
def Prime_under_num(n):
a = [2]
if n <2:
print("None")
elif n==2:
print(2)
else:
"Neglecting all even numbers as even numbers won't be prime in order to reduce the time complexity."
for i in range(3, n+1, 2):
if prime_no(i):
a.append(i)
print(a)
"When Method 1 is only used outputs of running time for different inputs"
#Total Running time = 2.73761 seconds #n = 100
#Total Running time = 3.14781 seconds #n = 1000
#Total Running time = 8.69278 seconds #n = 10000
#Total Running time = 18.73701 seconds #n = 100000
#Method 2 to find all the prime numbers <= a Number
def Prime_under_num(n):
a = [2]
if n <2:
print("None")
elif n==2:
print(2)
else:
for i in range(3, n+1, 2):
if n%i ==0:
pass
else:
a.append(i)
print(a)
"When Method 2 is only used outputs of running time for different inputs"
# Total Running time = 2.75935 seconds #n = 100
# Total Running time = 2.86332 seconds #n = 1000
# Total Running time = 4.59884 seconds #n = 10000
# Total Running time = 8.55057 seconds #n = 100000
if __name__ == "__main__" :
n = int(input())
Prime_under_num(n)
print("Total Running time = {:.5f} seconds".format(time.time() - start_time))
上述所有情况下获得的不同运行时间都是错误的。对于我们正在接受输入的问题,我们必须在接受输入后才开始计时。这里,用户键入输入所花费的时间也与运行时间一起计算。
正确的方法
我们必须从开头删除start_time=time.time()并将其添加到主块中。
if __name__ == "__main__" :
n = int(input())
start_time = time.time()
Prime_under_num(n)
print("Total Running time = {:.3f} seconds".format(time.time() - start_time))
因此,两种方法单独使用时的输出如下:-
# Method 1
# Total Running time = 0.00159 seconds #n = 100
# Total Running time = 0.00506 seconds #n = 1000
# Total Running time = 0.22987 seconds #n = 10000
# Total Running time = 18.55819 seconds #n = 100000
# Method 2
# Total Running time = 0.00011 seconds #n = 100
# Total Running time = 0.00118 seconds #n = 1000
# Total Running time = 0.00302 seconds #n = 10000
# Total Running time = 0.01450 seconds #n = 100000
现在我们可以看到,与错误方法相比,总运行时间有显著差异。即使方法2在两种方法中的性能优于方法1,但第一种方法(错误方法)是错误的。
您可以使用Python分析器cProfile来测量CPU时间,以及每个函数内部花费的时间以及每个函数被调用的次数。如果您想在不知道从哪里开始的情况下提高脚本的性能,这非常有用。对另一个堆栈溢出问题的回答很好。查看文档总是很好的。
以下是如何从命令行使用cProfile评测脚本的示例:
$ python -m cProfile euler048.py
1007 function calls in 0.061 CPU seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 0.061 0.061 <string>:1(<module>)
1000 0.051 0.000 0.051 0.000 euler048.py:2(<lambda>)
1 0.005 0.005 0.061 0.061 euler048.py:2(<module>)
1 0.000 0.000 0.061 0.061 {execfile}
1 0.002 0.002 0.053 0.053 {map}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler objects}
1 0.000 0.000 0.000 0.000 {range}
1 0.003 0.003 0.003 0.003 {sum}
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))
在Linux或Unix中:
$ time python yourprogram.py
在Windows中,请参阅StackOverflow问题:如何在Windows命令行上测量命令的执行时间?
对于更详细的输出,
$ time -v python yourprogram.py
Command being timed: "python3 yourprogram.py"
User time (seconds): 0.08
System time (seconds): 0.02
Percent of CPU this job got: 98%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:00.10
Average shared text size (kbytes): 0
Average unshared data size (kbytes): 0
Average stack size (kbytes): 0
Average total size (kbytes): 0
Maximum resident set size (kbytes): 9480
Average resident set size (kbytes): 0
Major (requiring I/O) page faults: 0
Minor (reclaiming a frame) page faults: 1114
Voluntary context switches: 0
Involuntary context switches: 22
Swaps: 0
File system inputs: 0
File system outputs: 0
Socket messages sent: 0
Socket messages received: 0
Signals delivered: 0
Page size (bytes): 4096
Exit status: 0
from time import time
start_time = time()
...
end_time = time()
time_taken = end_time - start_time # time_taken is in seconds
hours, rest = divmod(time_taken,3600)
minutes, seconds = divmod(rest, 60)