如何测量python中代码行之间的时间？

So in Java, we can do How to measure time taken by a function to execute

But how is it done in python? To measure the time start and end time between lines of codes? Something that does this:

import some_time_library

starttime = some_time_library.some_module()
code_tobe_measured() 
endtime = some_time_library.some_module()

time_taken = endtime - starttime

If you want to measure CPU time, can use time.process_time() for Python 3.3 and above:

import time
start = time.process_time()
# your code here    
print(time.process_time() - start)

First call turns the timer on, and second call tells you how many seconds have elapsed.

There is also a function time.clock(), but it is deprecated since Python 3.3 and will be removed in Python 3.8.

There are better profiling tools like timeit and profile, however time.process_time() will measure the CPU time and this is what you’re are asking about.

If you want to measure wall clock time instead, use time.time().

You can also use time library:

import time

start = time.time()

# your code

# end

print(f'Time: {time.time() - start}')

With a help of a small convenience class, you can measure time spent in indented lines like this:

with CodeTimer():
   line_to_measure()
   another_line()
   # etc...

Which will show the following after the indented line(s) finishes executing:

Code block took: x.xxx ms

UPDATE: You can now get the class with pip install linetimer and then from linetimer import CodeTimer. See this GitHub project.

The code for above class:

import timeit

class CodeTimer:
    def __init__(self, name=None):
        self.name = " '"  + name + "'" if name else ''

    def __enter__(self):
        self.start = timeit.default_timer()

    def __exit__(self, exc_type, exc_value, traceback):
        self.took = (timeit.default_timer() - self.start) * 1000.0
        print('Code block' + self.name + ' took: ' + str(self.took) + ' ms')

You could then name the code blocks you want to measure:

with CodeTimer('loop 1'):
   for i in range(100000):
      pass

with CodeTimer('loop 2'):
   for i in range(100000):
      pass

Code block 'loop 1' took: 4.991 ms
Code block 'loop 2' took: 3.666 ms

And nest them:

with CodeTimer('Outer'):
   for i in range(100000):
      pass

   with CodeTimer('Inner'):
      for i in range(100000):
         pass

   for i in range(100000):
      pass

Code block 'Inner' took: 2.382 ms
Code block 'Outer' took: 10.466 ms

Regarding timeit.default_timer(), it uses the best timer based on OS and Python version, see this answer.

I always prefer to check time in hours, minutes and seconds (%H:%M:%S) format:

from datetime import datetime
start = datetime.now()
# your code
end = datetime.now()
time_taken = end - start
print('Time: ',time_taken) 

output:

Time:  0:00:00.000019

I was looking for a way how to output a formatted time with minimal code, so here is my solution. Many people use Pandas anyway, so in some cases this can save from additional library imports.

import pandas as pd
start = pd.Timestamp.now()
# code
print(pd.Timestamp.now()-start)

Output:

0 days 00:05:32.541600

I would recommend using this if time precision is not the most important, otherwise use time library:

%timeit pd.Timestamp.now() outputs 3.29 µs ± 214 ns per loop

%timeit time.time() outputs 154 ns ± 13.3 ns per loop

Putting the code in a function, then using a decorator for timing is another option. (Source) The advantage of this method is that you define timer once and use it with a simple additional line for every function.

First, define timer decorator:

import functools
import time

def timer(func):
    @functools.wraps(func)
    def wrapper(*args, **kwargs):
        start_time = time.perf_counter()
        value = func(*args, **kwargs)
        end_time = time.perf_counter()
        run_time = end_time - start_time
        print("Finished {} in {} secs".format(repr(func.__name__), round(run_time, 3)))
        return value

    return wrapper

Then, use the decorator while defining the function:

@timer
def doubled_and_add(num):
    res = sum([i*2 for i in range(num)])
    print("Result : {}".format(res))

Let’s try:

doubled_and_add(100000)
doubled_and_add(1000000)

Output:

Result : 9999900000
Finished 'doubled_and_add' in 0.0119 secs
Result : 999999000000
Finished 'doubled_and_add' in 0.0897 secs

Note: I’m not sure why to use time.perf_counter instead of time.time. Comments are welcome.

You can try this as well:

from time import perf_counter

t0 = perf_counter()

...

t1 = perf_counter()
time_taken = t1 - t0