Python速度测试-时差-毫秒-Python 实用宝典

Python速度测试-时差-毫秒

为了快速测试一段代码,在Python中进行两次比较的正确方法是什么?我尝试阅读API文档。我不确定我是否了解timedelta。 到目前为止,我有以下代码: from datetime import datetime tstart = datetime.now() print t1 # code to speed test tend = datetime.now() print t2 # what am I missing? # I'd like to print the time diff here

问题:Python速度测试-时差-毫秒

为了快速测试一段代码,在Python中进行两次比较的正确方法是什么?我尝试阅读API文档。我不确定我是否了解timedelta。

到目前为止,我有以下代码:

from datetime import datetime

tstart = datetime.now()
print t1

# code to speed test

tend = datetime.now()
print t2
# what am I missing?
# I'd like to print the time diff here

What is the proper way to compare 2 times in Python in order to speed test a section of code? I tried reading the API docs. I'm not sure I understand the timedelta thing.

So far I have this code:

from datetime import datetime

tstart = datetime.now()
print t1

# code to speed test

tend = datetime.now()
print t2
# what am I missing?
# I'd like to print the time diff here

回答 0

datetime.timedelta 只是两个日期时间之间的差...所以就像一段时间,以天/秒/微秒为单位

>>> import datetime
>>> a = datetime.datetime.now()
>>> b = datetime.datetime.now()
>>> c = b - a

>>> c
datetime.timedelta(0, 4, 316543)
>>> c.days
0
>>> c.seconds
4
>>> c.microseconds
316543

请注意,它c.microseconds仅返回timedelta的微秒部分!出于计时目的,请始终使用c.total_seconds()

您可以使用datetime.timedelta进行各种数学运算,例如:

>>> c / 10
datetime.timedelta(0, 0, 431654)

不过,查看CPU时间而不是墙上时钟时间可能更有用……虽然这取决于操作系统,但在类Unix系统下,请检查“ time”命令。

datetime.timedelta is just the difference between two datetimes ... so it's like a period of time, in days / seconds / microseconds

>>> import datetime
>>> a = datetime.datetime.now()
>>> b = datetime.datetime.now()
>>> c = b - a

>>> c
datetime.timedelta(0, 4, 316543)
>>> c.days
0
>>> c.seconds
4
>>> c.microseconds
316543

Be aware that c.microseconds only returns the microseconds portion of the timedelta! For timing purposes always use c.total_seconds().

You can do all sorts of maths with datetime.timedelta, eg:

>>> c / 10
datetime.timedelta(0, 0, 431654)

It might be more useful to look at CPU time instead of wallclock time though ... that's operating system dependant though ... under Unix-like systems, check out the 'time' command.


回答 1

从Python 2.7开始,有了timedelta.total_seconds()方法。因此,要获得经过的毫秒数:

>>> import datetime
>>> a = datetime.datetime.now()
>>> b = datetime.datetime.now()
>>> delta = b - a
>>> print delta
0:00:05.077263
>>> int(delta.total_seconds() * 1000) # milliseconds
5077

Since Python 2.7 there's the timedelta.total_seconds() method. So, to get the elapsed milliseconds:

>>> import datetime
>>> a = datetime.datetime.now()
>>> b = datetime.datetime.now()
>>> delta = b - a
>>> print delta
0:00:05.077263
>>> int(delta.total_seconds() * 1000) # milliseconds
5077

回答 2

您可能要改用timeit模块

You might want to use the timeit module instead.


回答 3

您还可以使用:

import time

start = time.clock()
do_something()
end = time.clock()
print "%.2gs" % (end-start)

或者您可以使用python分析器

You could also use:

import time

start = time.clock()
do_something()
end = time.clock()
print "%.2gs" % (end-start)

Or you could use the python profilers.


回答 4

我知道这很晚了,但实际上我真的很喜欢使用:

import time
start = time.time()

##### your timed code here ... #####

print "Process time: " + (time.time() - start)

time.time()从纪元开始,您可以得到秒数。因为这是标准时间(以秒为单位),所以您可以简单地从结束时间中减去开始时间来获得处理时间(以秒为单位)。time.clock()对基准测试非常有用,但是如果您想知道过程花费了多长时间,我发现它毫无用处。例如,说“我的过程需要10秒”比说“我的过程需要10个处理器时钟单位”要直观得多。

>>> start = time.time(); sum([each**8.3 for each in range(1,100000)]) ; print (time.time() - start)
3.4001404476250935e+45
0.0637760162354
>>> start = time.clock(); sum([each**8.3 for each in range(1,100000)]) ; print (time.clock() - start)
3.4001404476250935e+45
0.05

在上面的第一个示例中,显示的时间time.clock()为0.05,而time.time()为0.06377

>>> start = time.clock(); time.sleep(1) ; print "process time: " + (time.clock() - start)
process time: 0.0
>>> start = time.time(); time.sleep(1) ; print "process time: " + (time.time() - start)
process time: 1.00111794472

在第二个示例中,即使进程睡眠了一秒钟,处理器时间也以某种方式显示为“ 0”。time.time()正确显示多于1秒。

I know this is late, but I actually really like using:

import time
start = time.time()

##### your timed code here ... #####

print "Process time: " + (time.time() - start)

time.time() gives you seconds since the epoch. Because this is a standardized time in seconds, you can simply subtract the start time from the end time to get the process time (in seconds). time.clock() is good for benchmarking, but I have found it kind of useless if you want to know how long your process took. For example, it's much more intuitive to say "my process takes 10 seconds" than it is to say "my process takes 10 processor clock units"

>>> start = time.time(); sum([each**8.3 for each in range(1,100000)]) ; print (time.time() - start)
3.4001404476250935e+45
0.0637760162354
>>> start = time.clock(); sum([each**8.3 for each in range(1,100000)]) ; print (time.clock() - start)
3.4001404476250935e+45
0.05

In the first example above, you are shown a time of 0.05 for time.clock() vs 0.06377 for time.time()

>>> start = time.clock(); time.sleep(1) ; print "process time: " + (time.clock() - start)
process time: 0.0
>>> start = time.time(); time.sleep(1) ; print "process time: " + (time.time() - start)
process time: 1.00111794472

In the second example, somehow the processor time shows "0" even though the process slept for a second. time.time() correctly shows a little more than 1 second.


回答 5

以下代码应显示时间说明...

from datetime import datetime

tstart = datetime.now()

# code to speed test

tend = datetime.now()
print tend - tstart

The following code should display the time detla...

from datetime import datetime

tstart = datetime.now()

# code to speed test

tend = datetime.now()
print tend - tstart

回答 6

您可以简单地打印出差异:

print tend - tstart

You could simply print the difference:

print tend - tstart

回答 7

我不是Python程序员,但我确实知道如何使用Google,这就是我发现的内容:您使用“-”运算符。要完成您的代码:

from datetime import datetime

tstart = datetime.now()

# code to speed test

tend = datetime.now()
print tend - tstart

此外,看起来您可以使用strftime()函数格式化时间跨度计算以呈现时间,但是这会让您感到高兴。

I am not a Python programmer, but I do know how to use Google and here's what I found: you use the "-" operator. To complete your code:

from datetime import datetime

tstart = datetime.now()

# code to speed test

tend = datetime.now()
print tend - tstart

Additionally, it looks like you can use the strftime() function to format the timespan calculation in order to render the time however makes you happy.


回答 8

time.time()/ datetime可以快速使用,但并不总是100%精确。出于这个原因,我喜欢使用其中一个std lib 分析器(尤其是hotshot)来找出问题所在。

time.time() / datetime is good for quick use, but is not always 100% precise. For that reason, I like to use one of the std lib profilers (especially hotshot) to find out what's what.


回答 9

您可能需要研究配置文件模块。您会更好地了解减速的位置,并且大部分工作将完全自动化。

You may want to look into the profile modules. You'll get a better read out of where your slowdowns are, and much of your work will be full-on automated.


回答 10

这是一个模仿Matlab / Octave tic toc函数的自定义函数。

使用示例:

time_var = time_me(); # get a variable with the current timestamp

... run operation ...

time_me(time_var); # print the time difference (e.g. '5 seconds 821.12314 ms')

功能:

def time_me(*arg):
    if len(arg) != 0: 
        elapsedTime = time.time() - arg[0];
        #print(elapsedTime);
        hours = math.floor(elapsedTime / (60*60))
        elapsedTime = elapsedTime - hours * (60*60);
        minutes = math.floor(elapsedTime / 60)
        elapsedTime = elapsedTime - minutes * (60);
        seconds = math.floor(elapsedTime);
        elapsedTime = elapsedTime - seconds;
        ms = elapsedTime * 1000;
        if(hours != 0):
            print ("%d hours %d minutes %d seconds" % (hours, minutes, seconds)) 
        elif(minutes != 0):
            print ("%d minutes %d seconds" % (minutes, seconds))
        else :
            print ("%d seconds %f ms" % (seconds, ms))
    else:
        #print ('does not exist. here you go.');
        return time.time()

Here is a custom function that mimic's Matlab's/Octave's tic toc functions.

Example of use:

time_var = time_me(); # get a variable with the current timestamp

... run operation ...

time_me(time_var); # print the time difference (e.g. '5 seconds 821.12314 ms')

Function :

def time_me(*arg):
    if len(arg) != 0: 
        elapsedTime = time.time() - arg[0];
        #print(elapsedTime);
        hours = math.floor(elapsedTime / (60*60))
        elapsedTime = elapsedTime - hours * (60*60);
        minutes = math.floor(elapsedTime / 60)
        elapsedTime = elapsedTime - minutes * (60);
        seconds = math.floor(elapsedTime);
        elapsedTime = elapsedTime - seconds;
        ms = elapsedTime * 1000;
        if(hours != 0):
            print ("%d hours %d minutes %d seconds" % (hours, minutes, seconds)) 
        elif(minutes != 0):
            print ("%d minutes %d seconds" % (minutes, seconds))
        else :
            print ("%d seconds %f ms" % (seconds, ms))
    else:
        #print ('does not exist. here you go.');
        return time.time()

回答 11

您可以像这样使用timeit测试名为module.py的脚本。

$ python -mtimeit -s 'import module'

You could use timeit like this to test a script named module.py

$ python -mtimeit -s 'import module'

回答 12

《箭头》:Python的更好日期和时间

import arrow
start_time = arrow.utcnow()
end_time = arrow.utcnow()
(end_time - start_time).total_seconds()  # senconds
(end_time - start_time).total_seconds() * 1000  # milliseconds

Arrow: Better dates & times for Python

import arrow
start_time = arrow.utcnow()
end_time = arrow.utcnow()
(end_time - start_time).total_seconds()  # senconds
(end_time - start_time).total_seconds() * 1000  # milliseconds

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