Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Basics (2015) Shopping List App Continue

omri zakai
omri zakai
289 Points
Posted by omri zakai
omri zakai
289 Points

Help with basic Challenge

Q: Same idea as the last one. My loopy function needs to skip an item this time, though. Loop through every item in items. If the current item's index 0 is the letter "a", continue to the next one. Otherwise, print out the current member.

My answare: 

def loopy(items):
    for item in items:
       if len(item) == "a"
         continue
       else print(len(item))

Need your help :)

1 Answer

Alexander Davison
Alexander Davison
65,469 Points
Alexander Davison
Alexander Davison
65,469 Points

Good job! There's some mistakes here, though, take a look:

  • You are missing colons for your if/else statement
  • There's no need for using the "len()" function
  • And the list goes on.

Instead, try this: (yeah you can remove the comments, they are just for reference.)

# Make the loopy() function as you'd expect
def loopy(items):
    for item in items:
        # If the first letter of the word is "a", then don't print the word.
        if item[0] == "a":
            continue
        else:
            # If the word DOESN'T start with "a", then print the word.
            print(word)

Hope it helps and Good luck on your amazing journey of programming! :) ~Alex