Cannot solve this question. does anyone can help me out please? thanks

A couple things, first, putting quotes around 'available' makes it a string. In other words, you want the variable available and not simpley the word 'available' to split, i.e.: available.split()

also, split method takes in a string argument, or what you want it to split. in this case, available.split(";") note that the quotes are needed inside the method call.

finally, splitting the string available results in a list of outputs, i.e:

>>> available = "banana split;hot fudge;cherry;malted;black and white"
>>>
>>> available.split(";")
['banana split', 'hot fudge', 'cherry', 'malted', 'black and white']

so you can simply assign sundaes to available.split(";")

in otherwords:

available = "banana split;hot fudge;cherry;malted;black and white"
sundaes = available.split(";")

Moderator Edit: Moved response from Comments to Answers

Thank you again!!! :)

Hello Hamsternation, thank you again. You right I have to think in a much simpler way. Thank again

hamsternation thank you a lot for your help. :)

can't figure out this question. Can you help out please? thank you

banana.py available = "banana split;hot fudge;cherry;malted;black and white" available.split(';') sundaes = available.split(';') menu = "Our available flavors are: {}." ", ".join(sundaes) display_menu = sundaes menu.format(display_menu)

I steal have an error message, yet it works while using python shell

sundaes = available.split(";")

sundaes ['banana split', 'hot fudge', 'cherry', 'malted', 'black and white'] (", ").join(sundaes) 'banana split, hot fudge, cherry, malted, black and white' display_menu = sundaes menu.format(display_menu) "Our available flavors are: ['banana split', 'hot fudge', 'cherry', 'malted', 'black and white']"

Okay so when you type code on treehouse, use triple backticks (the button above your tab button) checkout the Markdown Cheatsheet for more information.

This makes the code a lot easier to read.

But here's the code I think you have:

available = "banana split;hot fudge;cherry;malted;black and white"
available.split(';')
sundaes = available.split(';')
menu = "Our available flavors are: {}."
", ".join(sundaes)
display_menu = sundaes
menu.format(display_menu)

so you don't actually need the second available.split(':') line, it doesn't really do anything since it's not assigned to any variable nor returned to anything. it just disappears, so you can get rid of it.

so I think what you're missing is that sundaes created a list of menu items, which looks like this:

['banana split', 'hot fudge', 'cherry', 'malted', 'black and white']

you then assigned sundaes to display_menu, which means the variable display_menu is just another copy of sundae, which as a reminder, still looks like this:

['banana split', 'hot fudge', 'cherry', 'malted', 'black and white'] so when you menu.format(display_menu), you're sticking literally the list into the bracket, making the answer wrongly formatted.

"Our available flavors are: ['banana split', 'hot fudge', 'cherry', 'malted', 'black and white']"

what you want is:

"Our available flavors are: banana split, hot fudge, cherry, malted, black and white."

you had the right idea with the ", ".join(sundaes) but you didn't assign it to anything. ", ".join(sundaes) returns the string: 'banana split, hot fudge, cherry, malted, black and white'

what you want to do is assign that to the display_menu as such:

display_menu = ", ".join(sundaes)

so now, display_menu is the string: 'banana split, hot fudge, cherry, malted, black and white'

now finally, use the menu.format(display_menu) and now you should have the right answer.

Good luck!

oh okay I see the challenge now.

First off, you made a syntax error with the

display_menu=",".join(sundaes)

part of your code. you're missing a space after the comma. Your code yields a list of sundaes with just comma and no space.

So about the challenge, you have to be very carefully with the instructions:

Alright, let's finish making our menu. Combine the sundaes list into a new variable named display_menu, where each item in the list is rejoined together by a comma and a space (", "). Then reassign the menu variable to use the existing variable and .format() to replace the placeholder with the new string in display_menu. If you're really brave, you can even accomplish this all on the same line where menu is currently being set.

The key word is that they want you to reassign the menu variable

That means the variable menu becomes the string "Our available flavors are: banana split, hot fudge, cherry, malted, black and white."

Basically you're missing the reassignment back to menu:

available = "banana split;hot fudge;cherry;malted;black and white" 
sundaes = available.split(";") 
menu ="Our available flavors are:{}." 
display_menu =", ".join(sundaes) 
menu = menu.format(display_menu)

A more "brave" version they're talking about is doing this in 1 step. It effectively assigns menu while formatting the join strings. phew.

available = "banana split;hot fudge;cherry;malted;black and white" 
sundaes = available.split(";") 
menu ="Our available flavors are:{}.".format(', '.join(sundaes))

Hope this helps! best of luck! :)

Hello Hamsternation,

I hope all is well. listen could you help me out again please. can pass the 2nd Q Challenge Task 1 of 2

I'm going to create a variable named age. I need you to make an if condition that sets admitted to True if age is 13 or more. Important: In each task of this code challenge, the code you write should be added to the code from the previous task. Restart Get Help Check Work conditions.py

True 1 admitted = None 2 age <= 13 3 if age <=13 and admitted: 4 admitted = False 5 else: 6 ` admitted = True

2Q:Oops! It looks like Task 1 is no longer passing. and I don't know why

admitted = None age <= 13 if age <=13 and admitted: admitted = False elif: age >=13 and admitted admitted = True else: admitted = False

Oops! It looks like Task 1 is no longer passing.

Hello! This task is actually a bit more straightforward than what you're thinking, let's break it down.

So let's start with your code for challenge 1:

admitted = None
age <= 13
if age <=13 and admitted:
    admitted = False
else:
    admitted = True

Let's see what your 2nd line of code does: age <= 13

The variable age is either greater than 13 or equal to 13 or less than 13, depending on what the challenge assigns, which we don't know.

Let's say the challenge makes "age = 15"

>>> age <= 13
False

If the challenge makes "age = 13"

>>> age <=13
True

I hope you see how line 2 of your code doesn't actually do anything in terms of the program other than returning a True or False that doesn't get used up for anything else, so we actually don't need line 2 of your code.

Now let's break down your if statement and why it passed challenge 1, but probably isn't for the reason you think it did:

admitted=None
if age <=13 and admitted:
    admitted = False
else:
    admitted = True

Let's say age = 15, then if age <=13 would return True and admitted (which is equal to "None") would return False as such:

if True and False:

True and False would be evaluated as False so it triggers the else, which sets admitted to True

it turns out that this challenge passes as long as you have an if statement and you've somehow set admitted to True

Which is a bug. So a meaningless code like this would pass this challenge:

admitted = None

if age:
    admitted = True

So let's figure out how to actually solve this:

The challenge asks:

I need you to make an if condition that sets admitted to True if age is 13 or more.

You basically build it as you read it:

1. create an if condition
2. if age is 13 or more, then
3. set admitted to True

so as follows:

if age >= 13:  #create an if condition that tests if age is 13 or more
    admitted = True  # set admitted to True

We don't actually need to worry about the else yet, but if you want to see the 2nd part:

OK, one more. Add an else to your if. In the else, set admitted to False.

if age >= 13:  #create an if condition that tests if age is 13 or more
    admitted = True  # set admitted to True
else:
    admitted = False # if it doesn't meet the if condition, set admitted to False

Ahhhh the zen of python.... :)

Hello Hamsternation, I hope all is well. I need you help again i f you can please.

here is the challenge: Challenge Task 1 of 1

Write a function named printer. The function should take a single argument, count, and should print "Hi " as many times as the count argument. Remember, you can multiply a string by an integer.

and my code is

def printer(count): count = 0 while count < 100: print("Hi") count += 1 printer(10)

I ran the code into the shell it worked but I have an incorrect answer "Bummer! Didn't find the right number of "Hi"s.

if you can help me out again please

thanks!!!