How can I avoid a lot of if's and elif's in the options menu

Question

I want to build an options menu where the user types a number and according to the number he calls a function that does something .. For example:

def func1():
    pass
def func2():
    pass
def func3():
    pass

user_input = int(input("enter option: "))
if user_input == 1:
    func1()
elif user_input == 2:
    func2()
elif user_input == 3:
    func3()

So instead, build something in a few lines of code that will work in the same way and of course also on a function that gets parameters

score 1 · Answer 1 · answered Jun 08 '21 at 14:31

You can store the functions themselves in a dictionary and then access them using the user input

def func1():
    pass
def func2():
    pass
def func3():
    pass

funcs = {1:func1, 2:func2, 3:func2}

user_input = int(input("enter option: "))
funcs[user_input]()

score 0 · Answer 2 · answered Jun 08 '21 at 14:32

0

You can have a dict which can fetch the right function for you to execute:

def func1():
    pass
def func2():
    pass
def func3():
    pass

f = {1:func1, 2:func2, 3: func3}

user_input = int(input("enter option: "))
if user_input in f:
    f[user_input]()

answered Jun 08 '21 at 14:32

quamrana

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Okay sounds good, but let's say function 3 gets a parameter named name, like this: def func3(name) and the others don't. How do I call the function along with a parameter – FlyFox6767 Jun 08 '21 at 14:36
But where does `name` come from? – quamrana Jun 08 '21 at 14:37
Let's say it's also input from the user – FlyFox6767 Jun 08 '21 at 14:39
Perhaps you mean: `def func3(): func3_n(input(''))`? – quamrana Jun 08 '21 at 14:40
umm i mean like: `def func3(name): pass ` and in the main: `name = input("enter your name: )` and then we call the func with the param in your way but the other two still get nothing – FlyFox6767 Jun 08 '21 at 14:46

How can I avoid a lot of if's and elif's in the options menu

2 Answers2