range() vs enumerate()

I use the interactive Python Terminal iPython. It has color coating and many useful built in methods. You can install it here.

Mind blown. Thank you.

Hi James, I installed iPython and ran it via command line but I don’t see similar interface like your have. Can you please tell me how I can have the similar setup as you? Thanks

Hi @Abu I am also using VSCode in the video with “Dainty – Monokai” theme. I have a split screen setup with iPython + Terminal on one side and code on the other side.

I actually realized it after posting the comment :-) Thanks a lot though.

Hey @James Thanks for the video and mentioning ipython. I didn’t knew about ipython before, so I tried it after watching the view. But when i get into ipython it was surprising the interface was very familiar. I was using it daily for the past couple o months in django shell, didn’t noticed I was using ipython. LOL.

Hello James, great material! My question is does the order of the if statements matter? I had thought that the if statement for those numbers divisible by 5 and 3 should come first, because it is the strictest. I am a newb, so I apologize if this is a dumb question. Thanks, Mark

@Agil Yea IPython creeps into everything! I just learned about the ipdb debugger which expands on pdb. import ipdb; ipdb.set_trace() pypi.org/project/ipdb/

@Mark No dumb questions at all! You are right there is another way to write the code:

if num % 5 == 0 and num % 3 ==0:
    numbers[i] = "fizzbuzz"
elif num % 5 == 0:
    numbers[i] = "buzz"
elif num % 3 == 0:
    numbers[i] = "fizz"

But be careful you need to use if-elif-elif or else the following conditions will override the first one (3 if statements). Try it out yourself!

Thanks, James!

One More version

"""
numbers = [45, 22, 14, 65, 97, 72]
Replace all integers divisible by 3 with "fizz"
Replace all integers divisible by 5 with "buzz"
Replace all integers divisible by 3 and 5 with "fizzbuzz"
>>> my_list_1 = [45, 22, 14, 65, 97, 72]
>>> my_list_2 = [45, 22, 14, 65, 97, 72]
>>> use_range(my_list_1)
Values: ['fizzbuzz', 22, 14, 'buzz', 97, 'fizz']
>>> use_enumerate(my_list_2)
Values: ['fizzbuzz', 22, 14, 'buzz', 97, 'fizz']
"""
from functools import lru_cache


@lru_cache
def is_true_div_by(num, div_num):
    val = True if num % div_num == 0 else False
    return val


def use_range(my_list: list) -> None:
    for idx in range(len(my_list)):
        num = my_list[idx]
        if is_true_div_by(num, 3) and is_true_div_by(num, 5):
            my_list[idx] = "fizzbuzz"
        elif is_true_div_by(num, 3):
            my_list[idx] = "fizz"
        elif is_true_div_by(num, 5):
            my_list[idx] = "buzz"
    print(f"Values: {my_list}")


def use_enumerate(my_list: list) -> None:
    for idx, num in enumerate(my_list):
        if is_true_div_by(num, 3) and is_true_div_by(num, 5):
            my_list[idx] = "fizzbuzz"
        elif is_true_div_by(num, 3):
            my_list[idx] = "fizz"
        elif is_true_div_by(num, 5):
            my_list[idx] = "buzz"
    print(f"Values: {my_list}")

You wrote this:

if num % 3 == 0 and num % 5 == 0:

We can rewrite this line like this:

if num % 15 == 0:

If the number should be divisible by 3 and by 5, then it should be divisible by 3 * 5 = 15.

I think it looks more concise. :)

Awesome.

I believe y’all know it already, but I can say that this logic wouldn’t work if the question asks us to print. Can’t I?

  for i in range(len(numbers)):
    num = numbers[i]

    if num%3 == 0:
        print("fizz")
    if num%5 == 0:
        print("buzz")
    if num%3 == 0 and num%5 == 0:
        print("fizzbuzz")

Say, number[0] = 15

Output would be

fizz
buzz
fizzbuzz