# Help

#1

``````x = ["fizz","buzz","fizz","fizz"]
def fizz_count(x):
count = 0
for x in x:
if x == "fizz":
count = count + 1
return count
elif x != "fizz":
count = count
return count``````

WHAT DO YOU WANT FROM ME CODECADEMY!?!?!?!??!?!?!!?!??!?!?!?!?!?!??!?!?!?!?!?!?!?!?!??!?!!??!?!?

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#2

Oops, try again. fizz_count(['fizz', 'fizz', 1, 'fizz', 7, 'fuzz', 10, 'fIzZ']) returned 1 instead of the correct answer: 3

this is what it tells me.

#3

a function ends the moment a return keyword is reached, if a return keyword is reached in the loop, the loop will break

count the number of fizz inside the loop, then after the loop return the count

#4

Try using a discrete iterator variable...

``for n in x:``

#5

ive tried that - i took the "return" out of the variable, but it gave me an error message saying it has to be inside a function: File "python", line 12 SyntaxError: 'return' outside function

#6

you have to place the return inside the function but outside the loop? Can i see how you did this?

#7

``````x = ["fizz","buzz","fizz","fizz"]
def fizz_count(x):
count = 0
for item in x:
if item == "fizz":
count = count + 1

elif item != "fizz":
count = count

return count``````

#8

I have been on this one for like a month now, i am honestly considering breaking my computer now.

#9

the indent of return count should match the indention level of your for loop. This way, its inside the function but outside the loop

#10

x = ["fizz","buzz","fizz","fizz"]
def fizz_count(x):
count = 0
for item in x:
if item == "fizz":
count = count + 1

``````    elif item != "fizz":
count = count

return count``````

like this?

#11

Nvm my friend just helped me. Im kind of mad now cause the error was so simple...

#12