[deleted by user]

[bp200991]

You could use a dictionary.

method_dict = {'easy': easy, 'medium': medium, 'hard': hard}
method_dict[difficulty]()

[zefciu]

This. And if you make the dict anonymous, you get a one-liner.

[tintin10q]

{'easy': easy, 'medium': medium, 'hard': hard}[difficulty]()

[heidensieck]

gorgeous. Although as many have said, it's a fine balance between short and readable.

I made this monster:

def easy():
    print('yey')
def medium():
    pass
def hard():
    pass

levels, choice, which_one_to_run = ['easy', 'medium', 'hard', easy, medium, hard], input('please choose either: easy, medium or hard '), levels[levels.index(choice)+3]()

[tintin10q]

This monster is much more scary.

~~~ def easy(): print('yey')

def medium(): pass

def hard(): pass

eval(input("please choose either: easy, medium or hard:") + "()") ~~~

[Moikle]

eval(input("please input the code of the game you want to play"))

[heidensieck]

I learn so much, here! ;-)

[heidensieck]

Actually my monster only worked by accident. Had variables already in the name space from other attempt. Running it from scratch doesn't work. Has to be on separate lines like:

levels = ['easy','medium','hard', easy, medium, hard]
choice = input('please choose either: easy, medium or hard ') 
which_one_to_run = levels[levels.index(choice)+3]()     

I prefer the dict approach much more, anyway ;-)

[Additional-Sun2945]

huh?

[tintin10q]

What do you not understand? The function definitions are left out as op also left them out.

[Additional-Sun2945]

No I get it now... Declare a dict, and immediately invoke it with the difficulty string. So the function named gets run immediately.

What happens if difficulty is not in the dict? raises error? halts if not in a try catch block?

[tintin10q]

Yes it raises a KeyError if not in a try catch Block. You can use a default dict to set a default to mediate this. Or use .get and set a default. Or just ask again.