Challenge: Make loopy function skip an item.

Question

I cannot figure out this challenge. I've tried so many iterations of my code posted below, but nothing works and I cannot seem to figure out how to accomplish it.
"Same idea as the last one. My loopy function needs to skip an item this time, though.
Loop through each item in items again. If the character at index 0 of the current item is the letter "a", continue to the next one. Otherwise, print out the current member.
Example: ["abc", "xyz"] will just print "xyz"."
breaks.py
def loopy(items):
    for item in items:
        if item == "a".index(0):
            continue
        else:
            print(item)

nakalkucing · Accepted Answer

You need your code to say if the item in the oneth place is equal to "a" continue.  What your code says right now is: if "a" is at the index of zero continue. Hope this helps, Nakal

Paul Walker · Answer

Thank you so much, @nakalkucing (https://teamtreehouse.com/nakalkucing)! You explanation cleared it up. I was going by specific index on the list instead of thinking about how many items are in the list. Here was the end result:
```python
def loopy(items):
    for item in items:
        if item[0] == "a":
            continue
        else:
            print(item)
```

Welcome to the Treehouse Community

Looking to learn something new?

Paul Walker

Paul Walker

Challenge: Make loopy function skip an item.

2 Answers

nakalkucing

nakalkucing

Paul Walker

Paul Walker

nakalkucing

nakalkucing