How To Check If A List Contains A Sublist With Python

While working through the CodingBat exercises today, I came across the following challenge:

Given an array of ints, return True if .. 1, 2, 3, .. appears in the array somewhere.

array123([1, 1, 2, 3, 1]) → True
array123([1, 1, 2, 4, 1]) → False
array123([1, 1, 2, 1, 2, 3]) → True

I personally challenged myself to refactor my solutions into one line of code, which is forcing me to look into some more advanced Python methods. Here’s my solution:

But first, take a look at the official CodeBat solution:

def array123(nums):
  # Note: iterate with length-2, so can use i+1 and i+2 in the loop
  for i in range(len(nums)-2):
    if nums[i]==1 and nums[i+1]==2 and nums[i+2]==3:
      return True
  return False

Not too bad, but it’s still seems more complicated than it needs to be.

After searching around for a better way, I stumbled upon this StackOverflow question.

Inspired by the first answer, I came up with the following solution:

def array123(nums):
  return set([1,2,3]) & set(nums) == set([1,2,3])

A set of list is a list of unique values in that list. For example, try this in your command line:

$ python
>>> arr = [1,1,2,2,4]
>>> set(arr)
>>> arr2 = [1,2,2]
>>> set(arr2)
>>> set(arr) & set(arr2)

Check out more on sets here. They definitely look useful for working with two sets of lists!

Enjoy the article? Join over 20,000+ Swift developers and enthusiasts who get my weekly updates.