Using variables in a dictionary

Question

I'm a beginner in Python and I started messing around with it. I have this simple code:

def guess_number(Name,Gender):
    if Gender=='Male':
        Title='Mr.'
    else:
        Title='Ms.'
    number=int(raw_input("Hello " + Title + Name + ", guess what my favorite number is between 1-10"))
    if number==4:
        print number
        print "That's my favorite number!"
    else:
        print number
        print "Try Again!"
    return number

choice_dict=dict([(1,0),(2,0),(3,0),(4,0),(5,0),(6,0),(7,0),(8,0),(9,0),(10,0)])

for i in range(10):
    guess_number("Noam","Male")
    choice_dict[number]=choice_dict[number]+1

print choice_dict[1], choice_dict[2], choice_dict[3], choice_dict[4], choice_dict[5], choice_dict[6], choice_dict[7], choice_dict[8], choice_dict[9], choice_dict[10]

It's a simple process where in a function called "guess_number" it asks a person for a certain number between 1-10. It repeats the function 10 times and for each number chosen it will add +1 to the number in a dictionary, at the end it prints how many times was each number chosen.

For some reason I don't understand it tells me that "number" is not defined even though I returned the variable "number" in the function "guess_number".

What is missing?

score 4 · Answer 1 · answered Jan 10 '12 at 17:58

number is a variable local to the function guess_number. It does not exist outside of it. What you return is the value - not the variable itself, but its value - which you need to use, or assign to something. You can even assign it to a variable of the same name - as long as you understand it's another variable that just happens to have the same name:

number = guess_number("Noam", "Male")

score 1 · Answer 2 · edited May 23 '17 at 10:34

Change

guess_number("Noam","Male")

to

number = guess_number("Noam","Male")

number is defined in the (local) scope of the guess_number function, but not in the global scope. To reference number in the global scope, number first needs to be bound to a value accessible from the global scope, and (local variables from a function are not accessible. See the LEGB rules, below.)

Reference:

Python's LEGB rule regarding scope

score 0 · Answer 3 · answered Jan 10 '12 at 18:02

The variable "number" in your function "guess_number" is only local to that function. To fix your problem, you have to assign a variable "number" to the result from "guess_number". Thus, change

guess_number("Noam","Male")

to

number = guess_number("Noam","Male")

Using variables in a dictionary

3 Answers3