How do I get time of a Python program's execution?

I put this timing.py module into my own site-packages directory, and just insert import timing at the top of my module:

import atexit
from time import clock

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
    print

def endlog():
    end = clock()
    elapsed = end-start
    log("End Program", secondsToStr(elapsed))

def now():
    return secondsToStr(clock())

start = clock()
atexit.register(endlog)
log("Start Program")

I can also call timing.log from within my program if there are significant stages within the program I want to show. But just including import timing will print the start and end times, and overall elapsed time. (Forgive my obscure secondsToStr function, it just formats a floating point number of seconds to hh:mm:ss.sss form.)

Note: A Python 3 version of the above code can be found here or here.

In Linux or Unix:

$ time python yourprogram.py

In Windows, see this StackOverflow question: How do I measure execution time of a command on the Windows command line?

For more verbose output,

$ 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

I really like Paul McGuire's answer, but I use Python 3. So for those who are interested: here's a modification of his answer that works with Python 3 on *nix (I imagine, under Windows, that clock() should be used instead of time()):

#python3
import atexit
from time import time, strftime, localtime
from datetime import timedelta

def secondsToStr(elapsed=None):
    if elapsed is None:
        return strftime("%Y-%m-%d %H:%M:%S", localtime())
    else:
        return str(timedelta(seconds=elapsed))

def log(s, elapsed=None):
    line = "="*40
    print(line)
    print(secondsToStr(), '-', s)
    if elapsed:
        print("Elapsed time:", elapsed)
    print(line)
    print()

def endlog():
    end = time()
    elapsed = end-start
    log("End Program", secondsToStr(elapsed))

start = time()
atexit.register(endlog)
log("Start Program")

If you find this useful, you should still up-vote his answer instead of this one, as he did most of the work ;).

import time

start_time = time.clock()
main()
print time.clock() - start_time, "seconds"

time.clock() returns the processor time, which allows us to calculate only the time used by this process (on Unix anyway). The documentation says "in any case, this is the function to use for benchmarking Python or timing algorithms"

I like the output the datetime module provides, where time delta objects show days, hours, minutes, etc. as necessary in a human-readable way.

For example:

from datetime import datetime
start_time = datetime.now()
# do your work here
end_time = datetime.now()
print('Duration: {}'.format(end_time - start_time))

Sample output e.g.

Duration: 0:00:08.309267

or

Duration: 1 day, 1:51:24.269711

As J.F. Sebastian mentioned, this approach might encounter some tricky cases with local time, so it's safer to use:

import time
from datetime import timedelta
start_time = time.monotonic()
end_time = time.monotonic()
print(timedelta(seconds=end_time - start_time))