CODEWITHSUNDEEP
pythongenerator
Yasser Elbarbay
Yasser Elbarbay

Reputation: 133

How to use python generators with multiple if statements?

say something basic like:

def fizzBuzz(n: int) -> List[str]:
        l =[]
        for i in range(1,n+1):
            if i%3==0 and i%5==0:
                l.append("FizzBuzz")
            elif i%3==0:
                l.append("Fizz")
            elif i%5==0:
                l.append("Buzz")
            else:
                l.append(str(i))
        return l

where input: n=15.
output: [ "1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz", "11", "Fizz", "13", "14", "FizzBuzz" ]
I have started with something like:

["FizzBuzz" if x%3 ==0 and x%5==0 else str(x) for x in range(1, n+1)]

Upvotes: 0

Views: 1014

Answers (1)

Gulzar
Gulzar

Reputation: 28094

Since you asked for a generator, here is your function converted to one, along with usage.

Notice that fizz_buzz(n) returns a NEW generator, so if you want multiple iterations, you would have to generate a new one every time.

n = 15


def fizz_buzz(n: int):
    for i in range(1, n + 1):
        if i % 3 == 0 and i % 5 == 0:
            yield "FizzBuzz"
        elif i % 3 == 0:
            yield "Fizz"
        elif i % 5 == 0:
            yield "Buzz"
        else:
            yield str(i)


a = fizz_buzz(n)
for f in a:
    print(f)

Output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

However you may have wanted a list comprehension one liner, and didn't know what it is called and assumed the name was a "generator".

In that case,

b = ["FizzBuzz" if i%3==0 and i%5==0 else "Fizz" if i%3==0 else "Buzz" if i%5==0 else str(i) for i in range(n)]
print(b)

And, this can also be a one-liner-generator:

c = ("FizzBuzz" if i%3==0 and i%5==0 else "Fizz" if i%3==0 else "Buzz" if i%5==0 else str(i) for i in range(n))
for f in c:
    print(f)

Upvotes: 2

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