Adding fruits to a basket, allowing selection from ID and name

Question

I'm writing a tool which users can select multiple choices at once. Whilst I can't provide the actual content, I've replicated it with fruit. I'm replicating some functionality from Nikto where it allows different flags like "-mutation abcd".

for example here we have three fruits:

1- apple
2- orange
3- kiwi

I wanna let users to type "12" or "appleorange" or "21" or "orangeapple" for selecting apple and orange, or "123" or "appleorangekiwi" or vice versa to select apple, orange and kiwi and etc.

Assume that there are more than 3 choices, so it would be pain in the neck if I want to write elif statement for all multiple selections. How can I make my code simpler and shorter?

Here is my code:

fruits_ls = []


def add_fruits():
    while True:
        print("choose your favorite fruits:\n"
            "1- apple\n"
            "2- orange\n"
            "3- kiwi\n"
            "4- exit\n")

        my_fruits = input()
        if my_fruits == str(1) or my_fruits == "apple":
            fruits_ls.append("Apple")
        elif my_fruits == str(2) or my_fruits == "orange":
            fruits_ls.append("Orange")
        elif my_fruits == str(3) or my_fruits == "kiwi":
            fruits_ls.append("Kiwi")
        elif my_fruits == str(12) or my_fruits == "appleorange" or my_fruits == str(21) or my_fruits == "orangeapple":
            fruits_ls.append("Apple")
            fruits_ls.append("Orange")

        elif my_fruits == str(13) or my_fruits == "applekiwi" or my_fruits == str(31) or my_fruits == "kiwiapple":
            fruits_ls.append("Apple")
            fruits_ls.append("Kiwi")

        elif my_fruits == str(23) or my_fruits == "orangekiwi" or my_fruits == str(32) or my_fruits == "kiwiorange":
            fruits_ls.append("Orange")
            fruits_ls.append("Kiwi")

        elif my_fruits == str(4) or my_fruits == "exit":
            break


add_fruits()

print(fruits_ls)

Answer 1

Creating a mapping from input to output can be very useful here. A dictionary is the ideal data structure for this

VALID_INPUTS = {
    '1': 'Apple',
    'apple': 'Apple',
    '2': 'Orange',
    'orange': 'Orange',
    '3': 'Kiwi',
    'kiwi': 'Kiwi',
}
EXIT_INPUTS = ['4', 'exit']

Then you can loop over this mapping and append the matching inputs

if my_fruits in EXIT_INPUTS:
    break
for valid_input, fruit in VALID_INPUTS.items():
    if valid_input in my_fruits:
         fruits_ls.append(fruit)

Answer 2

The first thing I would like to do is applaud you for not immediately turning to back-to-back if statements. It's a trap I commonly see when tackling these kinds of problems. However, in this case, I'd ask a question that is counter to my previous point: are your options mutually exclusive? In other words, is it a problem if a person has multiple options in a loop accepted? Given you are accepting inputs like 31 or 12, my expectation is no. As such, instead of:

if my_fruits == str(1) or my_fruits == "apple":
    fruits_ls.append("Apple")
elif my_fruits == str(2) or my_fruits == "orange":
    fruits_ls.append("Orange")
elif my_fruits == str(3) or my_fruits == "kiwi":
    fruits_ls.append("Kiwi")
elif my_fruits == str(12) or my_fruits == "appleorange" or my_fruits == str(21) or my_fruits == "orangeapple":
    fruits_ls.append("Apple")
    fruits_ls.append("Orange")

elif my_fruits == str(13) or my_fruits == "applekiwi" or my_fruits == str(31) or my_fruits == "kiwiapple":
    fruits_ls.append("Apple")
    fruits_ls.append("Kiwi")

elif my_fruits == str(23) or my_fruits == "orangekiwi" or my_fruits == str(32) or my_fruits == "kiwiorange":
    fruits_ls.append("Orange")
    fruits_ls.append("Kiwi")

elif my_fruits == str(4) or my_fruits == "exit":
    break

A possibility would be checking if an instance of an option is inside the input:

if '1' in my_fruits or 'apple' in my_fruits:
    fruits_ls.append('Apple')
if '2' in my_fruits or 'orange' in my_fruits:
    fruits_ls.append('Orange')
if '3' in my_fruits or 'kiwi' in my_fruits:
    fruits_ls.append('Kiwi')
if '4' == my_fruits:
    break

There are some small problems with the above solution that may or may not be in scope. For example, "121" will only give "AppleOrange" when "AppleOrangeApple" may be desired. I can provide further insight if this is the case.

Hopefully this helps. Assuming the need for elifs is always a safe bet at first, but when refactoring code, elifs can be ditched if the answers can coexist without hindering the code's performance. Let me know if I can help further.

Answer 3

This could be made into a generic function that accepts a list of items. It displays a menu based on the list and lets the user select one or more items from the menu. Rather than keep a global list, the function returns the list of select items.

def select_items(prompt, items):
    selected_items = set()

    while True:
        print(f"{prompt}:")
        for n, item in enumerate(items, 1):
            print(f"    {n:2} - {item}")
        exit_no = n+1
        print(f"    {exit_no:2} - exit")

        raw_selection = input("> ")
        selection = raw_selection.lower()

        for n, item in enumerate(items, 1):
            if item in selection or str(n) in selection:
                selected_items.add(item)

        if 'exit' in selection or str(exit_no) in selection:
            return selected_items


select_items("Choose your favorite fruits", "apple orange kiwi banana".split())