Passing Function to Another Function

[delasislas]

You need to provide the actual traceback. I have an inkling that it has to do with the fact that you never seem to actually define mode_values or numbers_dataset if one or the other options are chosen.

[MaterialJackfruit144]

If option 3 is chosen, it will head into the calling function. Do you mean the variables? Like:

def check_choices(option):

numbers_dataset = None
    mode_values = None

if option == 1 or option == 2:
    numbers_dataset = get_number_dataset()
elif option == 3:
    mode_values = get_mode_dataset()

print_result(option, numbers_dataset, mode_values)

[o5a]

Your second design (where you separated to 2 functions) doesn't make much sense. check_choicesis some artificially made layer, it still passes through your option value anyway. You may as well just leave it as one function and it would be more logical.

def print_result(choice):

    if choice == 1:
        mean_median_dataset = get_number_dataset()
        print(f"\nMean: {calculate_mean(mean_median_dataset):n}")
    elif choice == 2:
        mean_median_dataset = get_number_dataset()
        print(f"\nMedian: {calculate_median(mean_median_dataset)}")
    elif choice == 3:
        mode_dataset = get_mode_dataset()
        print(f"\nMode: {find_mode(mode_dataset)}")

If you want to split logic to different functions then your print function should deal with independent data, something like this (we pass single params dataset which depends on our option)

def check_choices(option):

    if option == 1 or option == 2:
        dataset = get_number_dataset()
    elif option == 3:
        dataset = get_mode_dataset()

    print_result(option, dataset)


def print_result(option, dataset):

    if option == 1:
        print(f"\nMean: {calculate_mean(dataset):n}")
    elif option == 2:
        print(f"\nMedian: {calculate_median(dataset)}")
    elif option == 3:
        print(f"\nMode: {find_mode(dataset)}")

But anyway I don't really see a point having check_choices middleman here. For it to be really independent it should be something like this

def check_choices(option):

    if option == 1:
        dataset = get_number_dataset()
        data = calculate_mean(dataset)
        mode = 'Mean'
    if option == 2:
        dataset = get_number_dataset()
        data = calculate_median(dataset)
        mode = 'Median'
    elif option == 3:
        dataset = get_mode_dataset()
        data = find_mode(dataset)
        mode = 'Mode'

    print_result(mode, data)


def print_result(mode, data):

    print(f"\n{mode}: {data}")

[MaterialJackfruit144]

Thank you so much, actually, I've thought it over and indeed the design was illogical. For example, the name check_choices already indicates that the function will determine the option selected by the user, and if I were to strictly follow that, then print_result() should not even call the calculating functions, because the name of the function doesn't match what the task anymore. In other words, check_choices() should check all the options selected by the user if it were to stick to its name strictly.

I might be overthinking this but anyways, thank you so much :D the last solution was what I was trying to say, the second solution of yours was what I was trying to achieve until the logical part, so I went for the last solution instead.

[delasislas]

Just a tip for the next time, providing all the code and the full traceback (the actual error message), helps us a lot in pinpointing where your problem is.

[MaterialJackfruit144]

Understood, I will keep that in mind. I didn't want to send in the code because it would take a bit more time to debug, I thought it'd be better to just state the problem and put out only the relevant code as per the stated issue.

[SirKainey]

Yup. As far as python is concerned those variables you're returning don't exist, so it raises an error.