显示并返回函数上重复字符串的数量？

Question

I'm already learning Python on a online free course. Now I have and exercise that I've been hours trying to do but.... I cannot see why it doesn't works..

The instructions are here: Write a function called fizz_count that takes a list x as input and returns the count of the string “fizz” in that list.

For example, fizz_count(["fizz","buzz","fizz"]) should return 2 . (Make sure your function return s the number instead of print ing it.) Check out the Hint if you need help!

Hint: Try making a counter variable (for example, count ) in your function. The counter variable could initially be set to zero. After that, you can loop through the list that you take as input and increase count by one every time an item in the list is equal to the string "fizz".

At the end, don't forget to return the number of "fizz" s!

Then here's my code:

x=["fizz","bear","fizz"]

def fizz_count(*x):
    count=str(0)
    for count in x:
        if x=='fizz':
            count=count+str(1)
        return count

print fizz_count(x) #It prints "fizz","bear","fizz" but the thing that I want is to                 
                    #print "2" because "fizz" is two times in that string..

Answer 1

To fix your specific issue:

x=["fizz","bear","fizz"]

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

print fizz_count(x)

A few issues:

1. No need of *x in the parameter in python
2. Use a different variable name when you are iterating through the elements.
3. count=0 is sufficient, no need to cast it as a string
4. if x=='fizz' should be if e=='fizz' - Check the element of the list, rather than the entire list.
5. The return statement had to be after the loop executes. Notice the indentation.

Ofcourse, there are better ways of achieving what you are looking for, but I shall leave it here as it seems like you are learning.

Hope this helps.

Answer 2

Few things about your script:

def fizz_count(*x)

This probably isn't the syntax you want here. What you're doing here is expanding the list so it takes its own argument slot. You probably want to pass it like this:

def fizz_count(x)

Next:

count = str(0)

You can simply initiate your counter like this:

count = 0

Next:

for count in x:

This now makes your count variable an entry in x , but since you expanded the list this probably isn't what you're seeing. Regardless, again, you want something like this to assign each member to a variable:

for item in x:

Next:

count=count+str(1)

Again, you don't need to convert this to a string, count=count+1 would suffice

Finally, not sure if it's just a formatting issue, but your return statement should be outside your for-loop.

Answer 3

First of all, you shouldn't really have x=["fizz","bear","fizz"] before the function, you should have it after as it makes more sense to have it together with the print function. I've tested it, this should work:

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

x=["fizz","bear","fizz"]
print (fizz_count(x))

What I don't understand is why you had count as a string. You don't need to. Just make it an ordinary integer and the variable i in automatically created when I declare this for loop. Also, you shouldn't have count=count+1 as although it still works, it isn't really the pythonic way of doing this and there's a reason the += operator was built in( it does the same thing but makes your life that little bit easier)