A Stack?

A Stack is using the principle first-in-last-out.

It is like a stack of plates. The last one you put on the top is the first one you take.

How can you implement them in Python? Well, we are in luck, you can use a Stack, and if done correctly, you will have the same performance as an actual Stack implementation will have.

But first, how can you do it wrong?

Well, you might think that the first element of the list is the top of your stack, hence in you will insert the elements on the first position, and, hence, remove them from the first position as well.

# Create a list as a stack
s = []

# Insert into the first position.
element = 7
s.insert(0, element)

# Remove from the first position.
s.pop(0)

Sounds about right?

Let’s test that and compare it with a different approach. To add the newest element to the end of the list, and, hence, remove them from the end of the list.

# Create a list and use it as stack
s = []

# Insert element in last postion
element = 7
s.append(element)

# Remove from the last position
s.pop()

Let’s check the performance of those two approaches.

Comparing the performance of the two approaches

How do you compare. You can use cProfile library. It is easy to use and informative results

See the sample code below, which compares the two approaches by create a stack each and inserting n elements to it and removing them afterwards.

import cProfile


def profile_list_as_queue_wrong(n):
    s = []
    for i in range(n):
        s.insert(0, i)
    while len(s) > 0:
        s.pop(0)


def profile_list_as_queue_correct(n):
    s = []
    for i in range(n):
        s.append(i)
    while len(s) > 0:
        s.pop()


def profile(n):
    profile_list_as_queue_wrong(n)
    profile_list_as_queue_correct(n)


cProfile.run("profile(100000)")

The results are given here.

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    5.842    5.842 <string>:1(<module>)
        1    0.078    0.078    0.107    0.107 Stack.py:12(profile_list_as_queue_correct)
        1    0.000    0.000    5.842    5.842 Stack.py:20(profile)
        1    0.225    0.225    5.735    5.735 Stack.py:4(profile_list_as_queue_wrong)
   200002    0.017    0.000    0.017    0.000 {len}
   100000    0.007    0.000    0.007    0.000 {method 'append' of 'list' objects}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
   100000    3.547    0.000    3.547    0.000 {method 'insert' of 'list' objects}
   200000    1.954    0.000    1.954    0.000 {method 'pop' of 'list' objects}
        2    0.014    0.007    0.014    0.007 {range}

Observe that the “wrong” implementation takes over 5 seconds and the “correct” takes approximately 0.1 second. Over a factor 50 in difference.

Looking into the details

If we look at the complexities given by Python, it explains it all.

The Python lists amortised complexities are given on this page.

And you notice that the append and pop (last element) are O(1), which means constant time. Constant time, means that the operations are independent on the size of the lists. That means the correct implementation gives O(n) time complexity.

On the other hand, the insert and pop(0) have linear performance. That basically means that we with the wrong implementation end up with O(n^2) time complexity.