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Python Python Basics (2015) Shopping List App Continue

Jeffery Austin
Jeffery Austin
8,128 Points
Posted by Jeffery Austin
Jeffery Austin
8,128 Points

Having issues with challenge. Can't get this code to pass the challenge but the code does work in python shell.

def loopy(items):
    if items[0] == "a":
        for item in items[1:]:
            print(item)
    else:
        for item in items:
            print(item)

Not sure if this code works, but I also tried:

def loopy(items):
    for item in items:
        if items[0] == "a":
            continue
    print(item)

2 Answers

Seth Reece
Seth Reece
32,867 Points
Seth Reece
Seth Reece
32,867 Points

Hi Jeffery,

Your second one is real close. Since your loop is item in items your want to check for "a" in item, not items. e.g.

def loopy(items):
    for item in items:
      if item[0] == "a":
        continue
      print(item)

p.s. Notice that the print statement is indented to keep it inside the loop.

Jeffery Austin
Jeffery Austin
8,128 Points
Jeffery Austin
Jeffery Austin
8,128 Points

That worked! It wasn't at first because I was using:

if items[0] == "a":

instead of: 

if item[0] == "a":

Although I don't understand why items[0] doesn't work as items is the list being evaluated???

Evan Demaris
Evan Demaris
64,262 Points
Evan Demaris
Evan Demaris
64,262 Points

Hi Jeffery,

You original question was answered, but for your second question - why items[0] doesn't work - it's because that code would check if the first word in the list items is "a" (instead of checking index 0 of each word), and you're just checking that first word again as many times as there are indices in the list.

Jeffery Austin
Jeffery Austin
8,128 Points
Jeffery Austin
Jeffery Austin
8,128 Points

Aww okay I miss read the challenge thinking it was talking about the letter "a" at the index 0 of items. But if I had a list of ["a", "b", "c"]. Wouldn't this run?:

list = ["a", "b", "c"]
def loopy(items):
    for item in items:
      if items[0] == "a":
        continue
      print(item)
loopy(list)

Because I ran this in the python shell and it returns nothing, is it because the for loop is only evaluating each item and I can't check items[0] in the for loop?

Evan Demaris
Evan Demaris
64,262 Points

It would run, but I don't think that's what you want. Assuming full code of:

def loopy(items):
    #  For each index of your list;
    for item in items:
      #  If the first index of your list is "a" 
      #  (which it will always be, for a list of ["a", "b", "c"]);
      if items[0] == "a":
        #  At this point it is always true, 
        #  so 'continue' through the entire list without printing.
        continue
      print(item)
Jeffery Austin
Jeffery Austin
8,128 Points
Jeffery Austin
Jeffery Austin
8,128 Points

Aww so because 

items[0] == "a"

is always true it continues without printing the rest of the items?

Evan Demaris
Evan Demaris
64,262 Points

Right, it never even looks at the other items to see what they are; they are only used to tell how many times to check if items[0] == "a".

Jeffery Austin
Jeffery Austin
8,128 Points
Jeffery Austin
Jeffery Austin
8,128 Points

Okay, thanks for that clarification. That makes a lot of sense. I should have thought of that. I had it in my head that it would work, but now that makes sense. It can't keep evaluating to True! :)