Same idea as the last one. My loopy function needs to skip an item this time, though. Loop through each item in items ag

The challange is asking you to only print items that do not start with 'a'. So an implementation in psuedo code:

for each item in items:
    if that item does not start with a:
        print the item

Here is an example using something a bit easier to understand than the next:

def loopy(items):
    final_lyst = []    
    for item in items:
        if not item.startswith('a'):
            final_lyst.append(item)
    print(final_lyst)

Here is an example using list comprehension:

def loopy(items):
    items = [ item for item in items if not item.startswith('a')]
    print(items)

I changed my question could you check it again please?

Problem 1

• It would be wise for you to start testing your code in the python emulator, if you have quetions as to why something isn't working there is a built in help() function that will give you insight.

# Inside the interpreter
>>> help(list.index)

# From help()
index(...) method of builtins.list instance
    L.index(value, [start, [stop]]) -> integer -- return first index of value.
    Raises ValueError if the value is not present.

Problem 2

• So a common pitfall I believe every new programmer has is performing unecessary checks that makes there code redundant.

def loopy(items): 
    for item in items: 
        # ----- This block -------     
        if .index(0) = "a" : 
            continue 
        # ----- End Block -----
        else: 
            print(item)

From the block of code I have marked above.... what if instead of checking to see if the first index is an 'a', you check to see if it isn't.

When you are finding that the value is an 'a' you are continueing ... seems unecessary. Instead look at this.

def loop(items):
    for item in items:
        if 'a' != item[0]:
            print(item)