python – How to use timeit module

The Question :

382 people think this question is useful

I understand the concept of what timeit does but I am not sure how to implement it in my code.

How can I compare two functions, say insertion_sort and tim_sort, with timeit?

The Question Comments :

The Answer 1

282 people think this answer is useful

The way timeit works is to run setup code once and then make repeated calls to a series of statements. So, if you want to test sorting, some care is required so that one pass at an in-place sort doesn’t affect the next pass with already sorted data (that, of course, would make the Timsort really shine because it performs best when the data already partially ordered).

Here is an example of how to set up a test for sorting:

>>> import timeit

>>> setup = '''
import random

s = [random.random() for i in range(1000)]
timsort = list.sort

>>> print min(timeit.Timer('a=s[:]; timsort(a)', setup=setup).repeat(7, 1000))

Note that the series of statements makes a fresh copy of the unsorted data on every pass.

Also, note the timing technique of running the measurement suite seven times and keeping only the best time — this can really help reduce measurement distortions due to other processes running on your system.

Those are my tips for using timeit correctly. Hope this helps 🙂

The Answer 2

290 people think this answer is useful

If you want to use timeit in an interactive Python session, there are two convenient options:

  1. Use the IPython shell. It features the convenient %timeit special function:

    In [1]: def f(x):
       ...:     return x*x
    In [2]: %timeit for x in range(100): f(x)
    100000 loops, best of 3: 20.3 us per loop
  2. In a standard Python interpreter, you can access functions and other names you defined earlier during the interactive session by importing them from __main__ in the setup statement:

    >>> def f(x):
    ...     return x * x 
    >>> import timeit
    >>> timeit.repeat("for x in range(100): f(x)", "from __main__ import f",
    [2.0640320777893066, 2.0876040458679199, 2.0520210266113281]

The Answer 3

150 people think this answer is useful

I’ll let you in on a secret: the best way to use timeit is on the command line.

On the command line, timeit does proper statistical analysis: it tells you how long the shortest run took. This is good because all error in timing is positive. So the shortest time has the least error in it. There’s no way to get negative error because a computer can’t ever compute faster than it can compute!

So, the command-line interface:

%~> python -m timeit "1 + 2"
10000000 loops, best of 3: 0.0468 usec per loop

That’s quite simple, eh?

You can set stuff up:

%~> python -m timeit -s "x = range(10000)" "sum(x)"
1000 loops, best of 3: 543 usec per loop

which is useful, too!

If you want multiple lines, you can either use the shell’s automatic continuation or use separate arguments:

%~> python -m timeit -s "x = range(10000)" -s "y = range(100)" "sum(x)" "min(y)"
1000 loops, best of 3: 554 usec per loop

That gives a setup of

x = range(1000)
y = range(100)

and times


If you want to have longer scripts you might be tempted to move to timeit inside a Python script. I suggest avoiding that because the analysis and timing is simply better on the command line. Instead, I tend to make shell scripts:


 ... # lots of stuff


 echo Minmod arr1
 python -m timeit -s "$SETUP" "Minmod(arr1)"

 echo pure_minmod arr1
 python -m timeit -s "$SETUP" "pure_minmod(arr1)"

 echo better_minmod arr1
 python -m timeit -s "$SETUP" "better_minmod(arr1)"

 ... etc

This can take a bit longer due to the multiple initialisations, but normally that’s not a big deal.

But what if you want to use timeit inside your module?

Well, the simple way is to do:

def function(...):


and that gives you cumulative (not minimum!) time to run that number of times.

To get a good analysis, use .repeat and take the minimum:

min(timeit.Timer(function).repeat(repeat=REPEATS, number=NUMBER))

You should normally combine this with functools.partial instead of lambda: ... to lower overhead. Thus you could have something like:

from functools import partial

def to_time(items):

test_items = [1, 2, 3] * 100
times = timeit.Timer(partial(to_time, test_items)).repeat(3, 1000)

# Divide by the number of repeats
time_taken = min(times) / 1000

You can also do:

timeit.timeit("...", setup="from __main__ import ...", number=NUMBER)

which would give you something closer to the interface from the command-line, but in a much less cool manner. The "from __main__ import ..." lets you use code from your main module inside the artificial environment created by timeit.

It’s worth noting that this is a convenience wrapper for Timer(...).timeit(...) and so isn’t particularly good at timing. I personally far prefer using Timer(...).repeat(...) as I’ve shown above.


There are a few caveats with timeit that hold everywhere.

  • Overhead is not accounted for. Say you want to time x += 1, to find out how long addition takes:

    >>> python -m timeit -s "x = 0" "x += 1"
    10000000 loops, best of 3: 0.0476 usec per loop

    Well, it’s not 0.0476 µs. You only know that it’s less than that. All error is positive.

    So try and find pure overhead:

    >>> python -m timeit -s "x = 0" ""      
    100000000 loops, best of 3: 0.014 usec per loop

    That’s a good 30% overhead just from timing! This can massively skew relative timings. But you only really cared about the adding timings; the look-up timings for x also need to be included in overhead:

    >>> python -m timeit -s "x = 0" "x"
    100000000 loops, best of 3: 0.0166 usec per loop

    The difference isn’t much larger, but it’s there.

  • Mutating methods are dangerous.

    >>> python -m timeit -s "x = [0]*100000" "while x: x.pop()"
    10000000 loops, best of 3: 0.0436 usec per loop

    But that’s completely wrong! x is the empty list after the first iteration. You’ll need to reinitialize:

    >>> python -m timeit "x = [0]*100000" "while x: x.pop()"
    100 loops, best of 3: 9.79 msec per loop

    But then you have lots of overhead. Account for that separately.

    >>> python -m timeit "x = [0]*100000"                   
    1000 loops, best of 3: 261 usec per loop

    Note that subtracting the overhead is reasonable here only because the overhead is a small-ish fraction of the time.

    For your example, it’s worth noting that both Insertion Sort and Tim Sort have completely unusual timing behaviours for already-sorted lists. This means you will require a random.shuffle between sorts if you want to avoid wrecking your timings.

The Answer 4

109 people think this answer is useful

If you want to compare two blocks of code / functions quickly you could do:

import timeit

start_time = timeit.default_timer()
print(timeit.default_timer() - start_time)

start_time = timeit.default_timer()
print(timeit.default_timer() - start_time)

The Answer 5

47 people think this answer is useful

I find the easiest way to use timeit is from the command line:


def InsertionSort(): ...
def TimSort(): ...

run timeit like this:

% python -mtimeit -s'import test' 'test.InsertionSort()'
% python -mtimeit -s'import test' 'test.TimSort()'

The Answer 6

24 people think this answer is useful

for me, this is the fastest way:

import timeit
def foo():
    print("here is my code to time...")

timeit.timeit(stmt=foo, number=1234567)

The Answer 7

12 people think this answer is useful
# Генерация целых чисел

def gen_prime(x):
    multiples = []
    results = []
    for i in range(2, x+1):
        if i not in multiples:
            for j in range(i*i, x+1, i):

    return results

import timeit

# Засекаем время

start_time = timeit.default_timer()
print(timeit.default_timer() - start_time)

# start_time = timeit.default_timer()
# gen_prime(1001)
# print(timeit.default_timer() - start_time)

The Answer 8

9 people think this answer is useful

This works great:

  python -m timeit -c "$(cat"

The Answer 9

6 people think this answer is useful

simply pass your entire code as an argument of timeit:

import timeit


limit = 10000
prime_list = [i for i in range(2, limit+1)]

for prime in prime_list:
    for elem in range(prime*2, max(prime_list)+1, prime):
        if elem in prime_list:
, number=10))

The Answer 10

3 people think this answer is useful

lets setup the same dictionary in each of the following and test the execution time.

The setup argument is basically setting up the dictionary

Number is to run the code 1000000 times. Not the setup but the stmt

When you run this you can see that index is way faster than get. You can run it multiple times to see.

The code basically tries to get the value of c in the dictionary.

import timeit

print('Getting value of C by index:', timeit.timeit(stmt="mydict['c']", setup="mydict={'a':5, 'b':6, 'c':7}", number=1000000))
print('Getting value of C by get:', timeit.timeit(stmt="mydict.get('c')", setup="mydict={'a':5, 'b':6, 'c':7}", number=1000000))

Here are my results, yours will differ.

by index: 0.20900007452246427

by get: 0.54841166886888

The Answer 11

1 people think this answer is useful

Example of how to use Python REPL interpreter with function that accepts parameters.

>>> import timeit                                                                                         

>>> def naive_func(x):                                                                                    
...     a = 0                                                                                             
...     for i in range(a):                                                                                
...         a += i                                                                                        
...     return a                                                                                          

>>> def wrapper(func, *args, **kwargs):                                                                   
...     def wrapper():                                                                                    
...         return func(*args, **kwargs)                                                                  
...     return wrapper                                                                                    

>>> wrapped = wrapper(naive_func, 1_000)                                                                  

>>> timeit.timeit(wrapped, number=1_000_000)                                                              

The Answer 12

1 people think this answer is useful
import timeit

def oct(x):
   return x*x

timeit.Timer("for x in range(100): oct(x)", "gc.enable()").timeit()

The Answer 13

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The built-in timeit module works best from the IPython command line.

To time functions from within a module:

from timeit import default_timer as timer
import sys

def timefunc(func, *args, **kwargs):
    """Time a function. 


    Usage example:
        timeit(myfunc, 1, b=2)
        iterations = kwargs.pop('iterations')
    except KeyError:
        iterations = 3
    elapsed = sys.maxsize
    for _ in range(iterations):
        start = timer()
        result = func(*args, **kwargs)
        elapsed = min(timer() - start, elapsed)
    print(('Best of {} {}(): {:.9f}'.format(iterations, func.__name__, elapsed)))
    return result

The Answer 14

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You would create two functions and then run something similar to this. Notice, you want to choose the same number of execution/run to compare apple to apple.
This was tested under Python 3.7.

enter image description here Here is the code for ease of copying it

import timeit

def fibonacci(n):
    Returns the n-th Fibonacci number.
    if(n == 0):
        result = 0
    elif(n == 1):
        result = 1
        result = fibonacci(n-1) + fibonacci(n-2)
    return result

if __name__ == '__main__':
    import timeit
    t1 = timeit.Timer("fibonacci(13)", "from __main__ import fibonacci")
    print("fibonacci ran:",t1.timeit(number=1000), "milliseconds")

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