Python program to manipulate a list based on user input

Question

I have this Python code:

def main() -> None:
    print("Welcome To UltiList!")

    lst = []
    len_of_list = int(input("Enter the len of the list: "))

    while len(lst) <= len_of_list:
        print(">> ", end="")
        args = input().strip().split(" ")

        if args[0] == "append":
            lst.append(int(args[1]))
        elif args[0] == "insert":
            lst.insert(int(args[1]), int(args[2]))
        elif args[0] == "remove":
            lst.remove(int(args[1]))
        elif args[0] == "pop":
            lst.pop()
        elif args[0] == "sort":
            lst.sort()
        elif args[0] == "reverse":
            lst.reverse()
        elif args[0] == "print":
            print(lst)
        elif args[0] == "exit":
            print("Bye!")
            exit()
        else:
            print("That's not a valid command!")


if __name__ == "__main__":
    main()

But I thinks is very repetitive,Eg: In JavaScript I could do something like:


const commands = {
  append: (arg)=>lst.append(arg),
  remove: (arg)=>lst.remove(arg),
  ...
}

// This would go in the while loop
commands[`${args[0]}`](args[1])

And I would no longer have to make any comparison, this will make it more readable and shorter in my opinion.

What can be improved?

Please see What to do when someone answers. I have rolled back Rev 2 → 1. — 200_success
– 200_success, Commented Jul 27, 2021 at 7:01

Peilonrayz · Accepted Answer · 2021-07-25 16:46:38Z

2

Your 'switch statement' is bad because you've chosen to not use the same algorithm as your JavaScript example:

const commands = {
  append: (arg)=>lst.append(arg),
  remove: (arg)=>lst.remove(arg),
  ...
}

Converted into Python:

commands = {
    "append": lambda arg: lst.append(arg),
    "remove": lambda arg: lst.remove(arg),
}

commands[f"{args[0]}"](args[1])

Python has no equivalent to JavaScript's () => {...} lambdas. So you can abuse class if you need them.

class Commands:
    def append(arg):
        lst.append(arg)

    def exit(arg):
        print("bye")
        exit()

getattr(Commands, f"{args[0]}")(args[1])

answered Jul 25, 2021 at 16:46

Peilonrayz♦

44.5k7 gold badges80 silver badges158 bronze badges

\$\begingroup\$ lambda arg:lst.append(arg) is just a closure lst.append. F-string is an overkill, non-str aguments would not be found in dict keys. Else/default option can be done with commands.get(args[0], None) and adding None:lambda print(...) to commands. \$\endgroup\$

Pavlo Slavynskyy
– Pavlo Slavynskyy

2021-07-25 18:38:03 +00:00
Commented Jul 25, 2021 at 18:38
1

\$\begingroup\$ @PavloSlavynskyy 1. The point is to show lambda exists in Python, not to cut characters. 2. I'm just copying the JS example in the question. 3. Yes, or you could just do default = lambda arg: print(...) commands.get(args[0], default)(args[1]). \$\endgroup\$

Peilonrayz
– Peilonrayz ♦

2021-07-25 18:54:23 +00:00
Commented Jul 25, 2021 at 18:54

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