[Tut] List Difference | The Most Pythonic Way

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List Difference | The Most Pythonic Way

Short answer: The most Pythonic way to compute the difference between two lists

Code:

l1

and

Code:

l2

is the list comprehension statement

Code:

[x for x in l1 if x not in set(l2)]

. This works even if you have duplicate list entries, it maintains the original list ordering, and it’s efficient due to the constant runtime complexity of the set membership operation.



What’s the best way to compute the difference between two lists in Python?

Code:

a = [5, 4, 3, 2, 1] b = [4, 5, 6, 7] # a - b == [3, 2, 1] # b - a == [6, 7]

In Python, you always have multiple ways to solve the same (or a similar) problem. Let’s have an overview in the following interactive code shell:

https://trinket.io/embed/python/41397b54a0

Exercise: Run the code and think about your preferred way!

Let’s dive into each of the methods to find the most Pythonic one for your particular scenario.

Method 1: Set Difference

The naive approach to solve this problem is to convert both lists into sets and use the set minus (or set difference) operation.

Code:

# Method 1: Set Difference print(set(a) - set(b)) # {1, 2, 3} print(set(b) - set(a)) # {6, 7}

This approach is elegant because it’s readable, efficient, and concise.

However, there are some unique properties to this method which you should be aware of:

• The result is a setand not a list. You can convertit back to a list by using the
  Code:

  list(...)
  constructor.
  
  
• All duplicatedlist entries are removed in the process because sets cannot have duplicated elements.
  
  
• The order of the original list is lost because sets do not maintain the ordering of the elements.
  

If all three properties are acceptable to you, this is by far the most efficient approach as evaluated later in this article!

However, how can you maintain the order of the original list elements while also allow duplicates? Let’s dive into the list comprehension alternative!

Method 2: List Comprehension

List comprehension is a compact way of creating lists. The simple formula is

Code:

[expression + context]

.

• Expression: What to do with each list element?
  
  
• Context: What elements to select? The context consists of an arbitrary number of
  Code:

  for
  and
  Code:

  if
  statements.
  

You can use list comprehension to go over all elements in the first list but ignore them if they are in the second list:

Code:

# Method 2: List Comprehension print([x for x in a if x not in set(b)]) # [3, 2, 1]

We used a small but effective optimization of converting the second list

Code:

b

to a set first. The reason is that checking membership

Code:

x in b

is muchfaster for sets than for lists. However, semantically, both variants are identical.

Here are the distinctive properties of this approach:

• The result of the list comprehension statement is a list.
  
  
• The order of the original list is maintained.
  
  
• Duplicate elements are maintained.
  

If you rely on these more powerful guarantees, use the list comprehension approach because it’s the most Pythonic one.

Method 3: Simple For Loop

Surprisingly, some online tutorials recommend using a nested for loop (e.g.,those guys):

Code:

# Method 3: Nested For Loop d = [] for x in a: if x not in b: d.append(x) print(d) # [3, 2, 1]

In my opinion, this approach would only be used by absolute beginners or coders who come from other programming languages such as C++ or Java and don’t know essential Python features like list comprehension. You can optimize this method by converting the list

Code:

b

to a set first to accelerate the check

Code:

if x not in b

by a significant margin.

Performance Evaluation

Want to know the most performant one? In the following, I tested three different approaches:

Code:

import timeit init = 'l1 = list(range(100)); l2 = list(range(50))' # 1. Set Conversion print(timeit.timeit('list(set(l1) - set(l2))', init, number = 10000)) # 2. List Comprehension print(timeit.timeit('[x for x in l1 if x not in l2]', init, number = 10000)) # 3. List Comprehension + set print(timeit.timeit('s = set(l2);[x for x in l1 if x not in s]', init, number = 10000)) ''' 0.1620231000000001 0.5186101000000001 0.057180300000000184 '''

You can run the code in our interactive Python shell:

https://repl.it/@finxter/listdifferencespeed?lite=true

Exercise: Run the code. Which is fastest and why?

Although the first approach seems to be fastest, you now know that it has some disadvantages, too. (Loses duplicate info, loses ordering info.) From the two list comprehension approaches, the second one kills the first one in terms of runtime complexity and performance!

Where to Go From Here?

Enough theory, let’s get some practice!

To become successful in coding, you need to get out there and solve real problems for real people. That’s how you can become a six-figure earner easily. And that’s how you polish the skills you really need in practice. After all, what’s the use of learning theory that nobody ever needs?

Practice projects is how you sharpen your saw in coding!

Do you want to become a code master by focusing on practical code projects that actually earn you money and solve problems for people?

Then become a Python freelance developer! It’s the best way of approaching the task of improving your Python skills—even if you are a complete beginner.

Join my free webinar “How to Build Your High-Income Skill Python” and watch how I grew my coding business online and how you can, too—from the comfort of your own home.

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