Shouldn't this work?

Question

def loopy(items):
    for item in items:
       if items[0] == 'a':
           continue
print(items[1:6:1])
loopy("abcdef")
```breaks.py
def loopy(items):
    for item in items:
       if items[0] == 'a':
           continue
       # else:
    print(items[1:6:1])
    ## print(items)
loopy("abcdef")
```

Brian Boring · Accepted Answer

So the question/challenge is: "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"."
Currently, the loop in your code doesn't have any effect on the output. All your code is doing right now is printing out characters 1-5 of the argument passed to it (in this case "abcdef"). But the challenge states that if you send it a list (["abc", "xyz"]), then it should loop through the list and only return those items that do not start with "a". So, what needs to be done here is:
create a new list that is empty
new_items = []
run your for loop, but you should add an ELSE. The ELSE needs to add anything that doesn't start with "a" into the new list.
else: new_items.append(item)
Then just return/print your new list
print (new_items)
And you should use Loopy(["abc", "xyz"]) so you know it is producing the correct result.
Hope that helps!

John Skelton · Answer

This worked perfectly:  Thank you!
def loopy(items):
    newitems = []
    for item in items:
        if "a" in item:
            continue
        else: newitems.append(item)
        print(new_items)
loopy(["abc","xyz"])

John Skelton · Answer

I got it to work using Brian's suggestion.

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John Skelton

John Skelton

Shouldn't this work?

3 Answers

Brian Boring

Brian Boring

create a new list that is empty

run your for loop, but you should add an ELSE. The ELSE needs to add anything that doesn't start with "a" into the new list.

Then just return/print your new list

John Skelton

John Skelton

John Skelton

John Skelton