How do I make a define function in Python that cloud check a list?

Question

Hi,

I wanted to make a define function in Python that could check for a 7 in a list. If the list has a 7, I want to function to return a print statement that says that the item is in the list. I set the parameter to the variable that represented the number in the list. E.g. this is what I did for number 1 in the list:

def check(seven):
    if seven == 7:
        print("Seven is in the list.")
    else:
        print("Seven is not in the list.")
print(check(num1))

Thank you,

@YC72909 Scratch & Online GDB

Answer 1

def check(seven):
    if 7 in seven:
        print('Seven is in the list')
    else:
        print('Seven is not in the list')

Comments

Thank you for your help.

Answer 2

While using a return and print statement in the same function can be done, it's a bit confusing, particularly in the same if/else control block. I'm not sure if the seven parameter is supposed to represent your list but I would approach the solution like this.

llist = [1,2,3,4,5,6,7,8,9]

def check(llist):
    for x in llist:
        if(x == 7):
            return "Seven is in the List"
    return "Seven is not in the list"
    
seven = check(llist)

print(seven)

- I use a for loop inside the function, where x takes on each successive value in the llist, which is passed as a parameter.

- The if condition is inside the loop and the return string for finding seven is in that scope as well. Once a return statement is executed, it will automatically exit the function, so we don't need to worry about the 2nd return statement running in this case.

- In case 7 is not found, the second return string, which is outside the for loop scope, is sent out of the function.

- Whichever return statement is sent out of the function and stored in the variable seven, can then be passed to print and the string will be output.

 

Comments

sure, I'll try.

---

Thank you, it worked. Also thank you for your explanation.

YC72909 Scratch & Online GDB