How to use timeit module

Question

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?

Answer 1

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",
                     number=100000)
   [2.0640320777893066, 2.0876040458679199, 2.0520210266113281]
   

Answer 2

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

sum(x)
min(y)

---

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:

 SETUP="

 ... # 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(...):
    ...

timeit.Timer(function).timeit(number=NUMBER)

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.

---

Warnings

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.

Answer 3

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)                                                              
0.4458435332577161                                                                                        

Answer 4

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

import timeit

start_time = timeit.default_timer()
func1()
print(timeit.default_timer() - start_time)

start_time = timeit.default_timer()
func2()
print(timeit.default_timer() - start_time)

Answer 5

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

Answer 6

simply pass your entire code as an argument of timeit:

import timeit

print(timeit.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:
            prime_list.remove(elem)"""

, number=10))