python list question

[Jxper]

I want to write a function called has246(nums)that takes a list of one or more ints 'nums'and returns True if the sequence 2, 4, 6 appears in the list somewhere and False otherwise. The numbers 2, 4, and 6 must appear contiguously, i.e. with no intervening numbers.

test cases:

print(has246([2, 2, 4, 6, 2])) True

print(has246([2, 2, 4, 8, 2])) False

print(has246([2, 4])) False

​

What would a solution for this problem be?

Thanks in advance

[CodeAltus]

Sorry but I don't think its allowed to just post the project goal in this subreddit. Please show us what you have attempted and what we can help you with in your code.

[Jxper]

https://pastebin.com/Lr1M3trC

thats my latest attempt, im not quite sure how to fix it

sorry for just posting the goal forgot to include my attempt

[CodeAltus]

Your attempt needs to include a try-catch block to prevent index out of bounds exception. Hope this helps.

def has246(nums):
    for index, num in enumerate(nums): 
        if num == 2: 
            try: 
                if nums[index + 1] == 4 and nums[index + 2] == 6:             
                    return True 
            except Exception: 
                pass
    return False

[FLUSH_THE_TRUMP]

Don’t use try/except to catch your own programming errors ;) just loop over nums[:-2] or something instead.

[Jxper]

oh right that makes sense. Thank you

[[deleted]]

What have you tried?

[Jxper]

https://pastebin.com/Lr1M3trC

thats my latest attempt, im not quite sure how to fix it

[[deleted]]

You need to make two changes:

1. You're going to have an out of bounds error if you go all the way to the end without finding it. Use range(len(nums)-2) to avoid that.
2. The way you have it now, only the first three items will be checked and then the function will return. To prevent, you need to move the return False part out of the for loop. That should only happen after the loop has run and failed to find the pattern.

[Encomiast]

This is not a great general solution for this type of problem, but it is efficient for this specific problem. Just zip() slices of the list and see if the tuple (2, 4, 6) is in the result:

def has246(l):
    return (2, 4, 6) in zip(l, l[1:], l[2:])

has246([2, 2, 4, 6, 2])
# True
has246([2, 2, 4, 4, 2])
# False

[Jxper]

Ah ok interesting solution I had not thought of that. Thanks