我想测量执行一个函数所花费的时间。我没时间工作:

import timeit
start = timeit.timeit()
print("hello")
end = timeit.timeit()
print(end - start)

当前回答

使用timeit.default_timer而不是timeit.timeit。前者自动提供您的平台和Python版本上可用的最佳时钟:

from timeit import default_timer as timer

start = timer()
# ...
end = timer()
print(end - start) # Time in seconds, e.g. 5.38091952400282

timeit.default_timer被分配给time.time()或time.clock(),具体取决于操作系统。在Python 3.3+default_timer上,所有平台上都有time.perf_counter()。请参见Python-time.cclock()与time.time()-精度?

另请参见:

正在优化代码如何优化速度

其他回答

测量时间(秒):

from timeit import default_timer as timer
from datetime import timedelta

start = timer()

# ....
# (your code runs here)
# ...

end = timer()
print(timedelta(seconds=end-start))

输出:

0:00:01.946339

要深入了解递归调用的每个函数,请执行以下操作:

%load_ext snakeviz
%%snakeviz

它只需要在Jupyter笔记本中使用这两行代码,就可以生成一个很好的交互图。例如:

这是代码。同样,以%开头的2行是使用snakeviz所需的唯一额外代码行:

# !pip install snakeviz
%load_ext snakeviz
import glob
import hashlib

%%snakeviz

files = glob.glob('*.txt')
def print_files_hashed(files):
    for file in files:
        with open(file) as f:
            print(hashlib.md5(f.read().encode('utf-8')).hexdigest())
print_files_hashed(files)

在笔记本外运行snakeviz似乎也是可能的。更多信息请访问snakeviz网站。

下面是一个返回“hh:mm:ss”字符串的小型计时器类:

class Timer:
  def __init__(self):
    self.start = time.time()

  def restart(self):
    self.start = time.time()

  def get_time_hhmmss(self):
    end = time.time()
    m, s = divmod(end - self.start, 60)
    h, m = divmod(m, 60)
    time_str = "%02d:%02d:%02d" % (h, m, s)
    return time_str

用法:

# Start timer
my_timer = Timer()

# ... do something

# Get time string:
time_hhmmss = my_timer.get_time_hhmmss()
print("Time elapsed: %s" % time_hhmmss )

# ... use the timer again
my_timer.restart()

# ... do something

# Get time:
time_hhmmss = my_timer.get_time_hhmmss()

# ... etc
import time

def getElapsedTime(startTime, units):
    elapsedInSeconds = time.time() - startTime
    if units == 'sec':
        return elapsedInSeconds
    if units == 'min':
        return elapsedInSeconds/60
    if units == 'hour':
        return elapsedInSeconds/(60*60)

虽然问题中没有严格要求,但通常情况下,您需要一种简单、统一的方法来递增地测量几行代码之间的经过时间。

如果您使用的是Python 3.8或更高版本,则可以使用赋值表达式(也称为walrus运算符)以相当优雅的方式实现这一点:

import time

start, times = time.perf_counter(), {}

print("hello")
times["print"] = -start + (start := time.perf_counter())

time.sleep(1.42)
times["sleep"] = -start + (start := time.perf_counter())

a = [n**2 for n in range(10000)]
times["pow"] = -start + (start := time.perf_counter())

print(times)

=>

{'print': 2.193450927734375e-05, 'sleep': 1.4210970401763916, 'power': 0.005671024322509766}