Function Lists Python

Question

I am trying to figure out this problem on Codecademy, but I can not figure it out. Ok so I need to make this so it will return the number of times a word is said in a list.

Here is what it says to do:

Write a function that counts how many times the string "fizz" appears in a list.

1.Write a function called fizz_count that takes a list x as input.

2.Create a variable count to hold the ongoing count. Initialize it to zero.

3.for each item in x:, if that item is equal to the string "fizz" then increment the count variable.

4.After the loop, please return the count variable.

For example, fizz_count(["fizz","cat","fizz"]) should return 2.

Then here is what I have written:

def fizz_count(x):
    count = 0
    for item in x:
        if item == "fizz":
            count = count + 1
            return count

To look at this lesson the number is A Day at the Supermarket 4/13

Just for completeness: In Python you have the method .count() on lists, so your function could return just x.count('fizz'). Also, instead of count = count + 1 you can use count += 1. This is shorted and avoids doubling the code for the lvalue (count) in the idiom. — Alfe
– Alfe, Commented Mar 30, 2014 at 0:39

Two-Bit Alchemist · Accepted Answer · 2014-03-30 00:36:14Z

8

You're so close! You've just got the indentation wrong.

What you have:

for item in x:
    if item == 'fizz':
        count = count + 1
        return count # Returns any time we hit "fizz"!

What you need:

for item in x:
    if item == 'fizz':
        count += 1  # how we normally write this
return count   # after the loop, at the base indent level of the function

answered Mar 30, 2014 at 0:36

Two-Bit Alchemist

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Comments

Jason Baker · Accepted Answer · 2014-03-30 00:34:33Z

4

You're returning count inside your for loop; you should finish iterating over your list, and then return.

answered Mar 30, 2014 at 0:34

Jason Baker

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Comments

The Dude · Accepted Answer · 2014-03-30 00:36:04Z

4

You're almost there. The indentation is wrong. Your code should be:

def fizz_count(x):
    count = 0
    for item in x:
        if item == "fizz":
            count = count + 1
    return count

answered Mar 30, 2014 at 0:36

The Dude

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2 Comments

user3448743 Over a year ago

TY IT WORKED!! :D Could I also do this with like "Count += 1"?

The Dude Over a year ago

Yes, this will also work. Without the capital though: count += 1

Christian Tapia · Accepted Answer · 2014-03-30 00:36:25Z

3

What happens is that when a return statement is reached, the function will be exited. What you should do instead is to return the count after the for loop ends:

def fizz_count(x):
    count = 0
    for item in x:
        if item == "fizz":
            count = count + 1
    return count

Note:

Indentation is extremely important in Python since that's the way blocks are grouped.

answered Mar 30, 2014 at 0:36

Christian Tapia

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Comments

Yaw Boakye · Accepted Answer · 2014-03-30 00:35:19Z

2

def fizz_count(words_list):
    return len([word for word in words_list if word == 'fizz'])

OR

def fizz_count(words_list):
    return words_list.count('fizz')

answered Mar 30, 2014 at 0:35

Yaw Boakye

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1 Comment

Yaw Boakye Over a year ago

I know it doesn't answer your question, really, but give it a thought when you advance your Python (which you can start now).

Daniele Pantaleone · Accepted Answer · 2014-03-30 00:35:29Z

2

You need to place the return statement after the for loop

def fizz_count(x):
    count = 0
    for item in x:
        if item == "fizz":
            count = count + 1
    return count

answered Mar 30, 2014 at 0:35

Daniele Pantaleone

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2 Comments

Two-Bit Alchemist Over a year ago

Now you've got it. :)

Daniele Pantaleone Over a year ago

Yeah, I'm not used to this syntax editor and I failed the indentation when I submitted the solution at first :P

Snoop Dogg · Accepted Answer · 2014-03-30 00:35:45Z

2

put your return at the end of your for loop. so it would be:

    for item in x:
        if item == "fizz":
            count+= 1
    return count

answered Mar 30, 2014 at 0:35

Snoop Dogg

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3 Comments

Two-Bit Alchemist Over a year ago

Hint: Use leading four-space indentation to format code blocks.

Snoop Dogg Over a year ago

@Two-BitAlchemist thanks I'm new and I wasn't sure, i got it now though!

Two-Bit Alchemist Over a year ago

NP. Markdown is really fun and useful (it's used here, git, and other places). There's also a wonderful utility called pandoc that can do all kinds of things with it. I encourage you to learn it!

user3286261 · Accepted Answer · 2014-03-30 00:37:36Z

1

return is part of the if block. Do not return until the for block has finished:

def fizz_count(x):
count = 0
for item in x:
    if item == "fizz":
        count = count + 1
return count

answered Mar 30, 2014 at 0:37

user3286261

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Comments

spinlok · Accepted Answer · 2014-03-30 01:32:44Z

1

I'm aware that the assignment requires you to write an explicit for loop for counting, but this kind of error due to misindentation is exactly why I personally prefer using functional idioms in python. You can implement this like:

def fizz_count(x):
    return len([item for item in x if item == "fizz"])

answered Mar 30, 2014 at 1:32

spinlok

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