CODEWITHSUNDEEP
pythonpython-2.7python-3.x
user3448743
user3448743

Reputation: 43

Function Lists Python

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

Upvotes: 1

Views: 4166

Answers (9)

spinlok
spinlok

Reputation: 3661

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"])

Upvotes: 1

user3286261
user3286261

Reputation: 501

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

Upvotes: 1

Christian Tapia
Christian Tapia

Reputation: 34146

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.

Upvotes: 3

Two-Bit Alchemist
Two-Bit Alchemist

Reputation: 18467

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

Upvotes: 8

The Dude
The Dude

Reputation: 4005

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

Upvotes: 4

Snoop Dogg
Snoop Dogg

Reputation: 60

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

Upvotes: 2

Daniele Pantaleone
Daniele Pantaleone

Reputation: 2733

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

Upvotes: 2

Yaw Boakye
Yaw Boakye

Reputation: 10742

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')

Upvotes: 2

Jason Baker
Jason Baker

Reputation: 2481

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

Upvotes: 4

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