The Most Pythonic Way to Check if Two Ordered Lists Are Identical

The most Pythonic way to check if two ordered lists l1 and l2 are identical, is to use the l1 == l2 operator for element-wise comparison. If all elements are equal and the length of the lists are the same, the return value is True.

Problem: Given are two lists l1 and l2. You want to perform Boolean Comparison: Compare the lists element-wise and return True if your comparison metric returns True for all pairs of elements, and otherwise False.

Examples:

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3]
# compare(l1, l2) --> False

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3, 5, 4]
# compare(l1, l2) --> False

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3, 4, 5]
# compare(l1, l2) --> True

Let’s discuss the most Pythonic ways of solving this problem. Here’s a quick interactive code overview:

Exercise: Glance over all methods and run the code. What questions come to mind? Do you understand each method?

Read on to learn about each method in detail!

Method 1: Simple Comparison

Not always is the simplest method the best one. But for this particular problem, it is! The equality operator == compares a list element-wise—many Python coders don’t know this!

# 1. Simple Comparison
def method_1(l1, l2):
    return l1 == l2

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3]
print(method_1(l1, l2))
# False

So, if you just want to learn about the most Pythonic way to solve this problem, look no further.

But if you want to dive into the wonderful world of Python, learning about different interesting and powerful Python functions, read on!

Method 2: Simple For Loop

The following method is what you’d see from a coder coming from another programming language or from a beginner who doesn’t know about the equality operator on lists (see Method 1).

# 2. Simple For Loop
def method_2(l1, l2):
    for i in range(min(len(l1), len(l2))):
        if l1[i] != l2[i]:
            return False
    return len(l1) == len(l2)

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3]
print(method_2(l1, l2))
# False

In the code, you iterate over all indices from 0 to the last position of the smallest list as determined by the part min(len(l1), len(l2)). You then check if both elements at the same position are different. If they are different, i.e., l1[i] != l2[i], you can immediately return False because the lists are also different.

If you went through the whole loop without returning False, the list elements are similar. But one list may still be longer! So, by returning len(l1) == len(l2), you ensure to only return True if (1) all elements are equal and (2) the lists have the same length.

A lot of code to accomplish such a simple thing! Let’s see how a better coder would leverage the zip() function to reduce the complexity of the code.

Method 3: zip() + For Loop

The zip function takes a number of iterables and aggregates them to a single one by combining the i-th values of each iterable into a tuple for every i.

Let’s see how you can use the function to make the previous code more concise:

# 3. Zip + For Loop
def method_3(l1, l2):
    for x, y in zip(l1, l2):
        if x != y:
            return False
    return len(l1) == len(l2)

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3]
print(method_3(l1, l2))
# False

Instead of iterating over indices, you now iterate over pairs of elements (the ones zipped together). If the lists have different sizes, the remaining elements from the longer list will be skipped. This way, element-wise comparison becomes simpler and no elaborate indexing schemes are required. Avoiding indices by means of the zip() function is a more Pythonic way for sure!

Method 4: sum() + zip() + len()

But true Python coders will often avoid a for loop and use a generator expression instead.

  • You first create an iterable of Boolean values using the generator expression x == y for x, y in zip(l1, l2).
  • Then, you sum up over the Boolean values (another trick of pro coders) to find the number of elements that are the same and store it in variable num_equal.
  • Finally, you compare this with the length of both lists. If all three values are the same, both lists have the same elements and their length is the same, too. They are equal!
# 4. Sum + Zip + Len
def method_4(l1, l2):
    num_equal = sum(x == y for x, y in zip(l1, l2))
    return num_equal == len(l1) == len(l2)

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3]
print(method_4(l1, l2))
# False

print(method_4([1, 2], [1, 2]))
# True

From the methods except the first one using the == operator, this is the most Pythonic way due to the use of efficient Python helper functions like zip(), len(), and sum() and generator expressions to make the code more concise and more readable.

You could also write this in a single line of code!

sum(x == y for x, y in zip(l1, l2)) == len(l1) == len(l2)

If you love Python one-liners, check out my new book Python One-Liners with internationally renowned publisher NoStarch press. (Amazon Link)

Method 5: map() + reduce() + len()

The last method is just to train your functional programming skills.

# 5. map() + reduce() + len()
from functools import reduce
def method_5(l1, l2):
    equal = map(lambda x, y: x == y, l1, l2)
    result = reduce(lambda x, y: x and y, equal)
    return result and len(l1) == len(l2)

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3]
print(method_5(l1, l2))
# False

print(method_5([1, 2, 3], [1, 2, 3]))
# True

The map() function combines all pairs of elements to Boolean values (are the two elements equal?). The reduce() function combines all Boolean values performing an and operation. Sure, you can also use the more concise variant using the all() function:

Method 6: map() + all()

This is the same as the previous method—but using the all() function instead of reduce() to combine all Boolean values in a global and operation.

# 6. map() + all()
def method_6(l1, l2):
    result = all(map(lambda x, y: x == y, l1, l2))
    return result and len(l1) == len(l2)

l1 = [1, 2, 3, 4, 5]
l2 = [1, 2, 3]
print(method_5(l1, l2))
# False

print(method_5([1, 2, 3], [1, 2, 3]))
# True

Thanks for reading this article to the end! I hope you learned something new today. If you want to learn something new every day, join my free Python email series for continuous improvement in Python and computer science.

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