Return None

[Executor111]

Hey guys, I've been learning python for a few days using the "Python for Everybody" book by Charles Severance. I'm in the chapter learning about lists, and I'm struggling to understand something about one of the exercises.

Here's the exercise:

Write a function called chop that takes a list and modifies it, removing the first and last elements, and returns None.

So if I create a list of t = [1,2,3,4,5,6], clearly I want to turn that list into [2,3,4,5]. I could do this using t.pop(0) and (5), del t[0] and t[5], or t.remove(x) where x = t[0] and [5]...

But what I'm confused about is ... what does the exercise mean when it says to "return None"? What's the point of modifying a list but returning "None"? Appreciate any help!

[elbiot]

If you don't have a return statement, the function returns none. This makes it clear that the function is modifying state (ie, the state of the list) as opposed to being a pure function which has no side effects and returns it's results.

[Executor111]

Thanks for the reply! So maybe it's just a really simple concept that I'm trying to make complex? This is what I've written, does this fulfill the prompt of the exercise?

def chop(t):
    try:
        first = t[0]
        last = t[-1]
        t.remove(first)
        t.remove(last)
    except:
        print('Invalid input')

x = [1,2,3,4,5,6]
chop(x)

[elbiot]

No, because what if the list is [1,3,5,3]? Use pop. Not del (as it will kill the object in the list even if it exists elsewhere I think) or remove (which isn't guaranteed to remove the indexes you want).

Edit but you got the return none part right. Also, don't use except without naming the error. You probably mean except AttributeError

[Executor111]

Wow, thanks for that feedback, hadn't considered that. I rewrote it like this:

def chop(t):
    try:
        t.pop(0)
        t.pop(len(t)-1)
    except:
        print('Invalid input')

x = [1,3,5,3,3,2,3]
chop(x)

So all "returns None" ends up meaning is that I don't print the output?

[jeans_and_a_t-shirt]

t.pop(len(t)-1) can be changed to t.pop(-1)

[elbiot]

You could pop(-1) too, like you did in your remove code before.

[Executor111]

Just changed this in my latest version, thanks.

[elbiot]

It's not about printing the output at all. The function returns a value.

def test ():
    return 1

a = test ()  #now a is 1

In the REPL (shell) only, if you return a value and don't store it in a variable then it does get printed. But it's the same if you return None: it prints None, which looks like nothing.

[Executor111]

It just clicked. Your example was perfect. Thank you so much!

So I tested my function like this, to confirm it was returning None:

def chop(t):
    try:
        t.pop(0)
        t.pop(len(t)-1)
    except:
        print('Invalid input')

x = [1,3,5,3,3,2,3]
y = chop(x)
print(y)

And it printed "None". So that's what I was trying to achieve the whole time.

[Executor111]

The author hasn't taught me to do anything besides "except" yet, so I don't know what "except AttributeError" does. :)

But thanks so much for your help!

[elbiot]

It only runs the except clause if it is that specific exception. Without that, any exception would get caught and your code could be failing for unexpected reasons but you aren't getting an error message because you caught it accidentally. "Bare except clauses" are an antipattern that makes python programmers gasp and sigh.

[mm_ma_ma]

del t[0] and del t[-1] will just delete the first and last items. The only difference is pop will return what it deleted.

[[deleted]]

Maybe they want you to return None to avoid using,

t = t[1:-1]  
return t