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 trialBapi Roy
14,237 PointsAm I missing something
def loopy(items):
# Code goes here
for idx, item in enumerate(items):
if idx == 0 and 'a' in item:
continue
else:
print(item)
Getting "Bummer! Try again!" in Challenge BUT working on my local shell
def loopy(items):
# Code goes here
for idx, item in enumerate(items):
if idx == 0 and 'a' in item:
continue
else:
print(item)
3 Answers
james south
Front End Web Development Techdegree Graduate 33,271 Pointsyour code does not do what the challenge asks. the enumeration gets the index of the list and each corresponding element, it does not check to see if the first letter of each element is an 'a'.
Prashant Nayak
2,300 Pointsitem[0]=='a'
Bapi Roy
14,237 PointsIt will not work when item is "abc"
Bapi Roy
14,237 Points@Kenneth Love Can you help me with this problem?
Bapi Roy
14,237 PointsBapi Roy
14,237 PointsIt is done in line
if idx == 0 and 'a' in item: