calling functions python from outside

Question

I have small a python script looking something like this:

def number1():
    x = 1
    open_numbers = []
    open_numbers.append(x)
    return open_numbers

def myfunction(open_numbers):
    y = 1
    open_numbers.append(y)

I would like to call the myfunction in the the end of the script. Using

myfunction()

But it keeps telling me missing 1 required positional argument: 'open_numbers'

Tried passing the argument and got name 'open_numbers' is not defined

I plan to add more functions later and run them the same way

function(arg)
function2(arg)
function3(arg)

What data are you trying to give to myfunction? open_numbers only exists inside of number1. You need to use the return from that function if you want to use it. — Carcigenicate
– Carcigenicate, Commented Oct 21, 2019 at 18:17
Please fix the indentation. - Formatting help ... more Formatting ... Formatting sandbox — wwii
– wwii, Commented Oct 21, 2019 at 18:18
if you call the function myfunction() without any arguments, you'll get the first error: 1 required positional argument: 'open_numbers'. If you call it with the argument open_numbers, you need to define open_numbers out of the scope of the function or define it as a new local variable inside the function — Pranjal Aswani
– Pranjal Aswani, Commented Oct 21, 2019 at 18:19
I understand what you're asking, but in the future you really need to make a minimal reproducible example including how you're calling myfunction with open_numbers, and the full error message with traceback. See How to Ask for more tips. — wjandrea
– wjandrea, Commented Oct 21, 2019 at 18:25

CypherX · Accepted Answer · 2020-05-11 10:25:10Z

1

Solution

First of all, your code was not properly indented. I have corrected that.
The function myfunction takes in a list (open_numbers) as input and should return it as well.

I have passed in the output of number1() as the input to myfunction(). This should create a list: [1, 1]. And that's what it did.

def number1():
    x = 1
    open_numbers = []
    open_numbers.append(x)
    return open_numbers


def myfunction(open_numbers):
    y = 1
    open_numbers.append(y)
    return open_numbers

myfunction(number1())

Output:

[1, 1]

edited May 11, 2020 at 10:25

answered Oct 21, 2019 at 18:21

CypherX

7,4034 gold badges29 silver badges39 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sadrach Pierre · Accepted Answer · 2019-10-21 18:22:26Z

1

you need to pass in an object to your function. you can call your function with an empty list if you want:

a = []
myfunction(a)

answered Oct 21, 2019 at 18:22

Sadrach Pierre

1214 bronze badges

Comments

Jacek Rojek · Accepted Answer · 2019-10-21 18:22:31Z

0

You might want to define default parameter, and return the updated value

def myfunction(open_numbers = []):
    y = 1
    open_numbers.append(y)
    return open_numbers

Then you can call it with passing parameter myfunction([1]) or without myfunction()

answered Oct 21, 2019 at 18:22

Jacek Rojek

1,1328 silver badges16 bronze badges

1 Comment

wjandrea Over a year ago

Using a mutable default argument is probably a bad idea. For example try running for _ in range(2): print(myfunction())

Collectives™ on Stack Overflow

calling functions python from outside

3 Answers 3

Solution

Comments

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Solution

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related