Problem in a list with previously created variables. Error local variable 'Victory' refere-CodePudding

I have Victory, Draw, Lose variables for a hockey game. Obviously, based on the final result of a game, only one of these variables will be True. For this reason I created them inside a try and except: pass.

Now I would like to print only one of these variables (the True one, for example Victory, because Victory becomes True based on a scraping of the results of the hockey match), but I get the error UnboundLocalError: local variable 'Victory' referenced before assignment when I search to print the True variable, because Victory was already named before the assignment. Given the simplicity of the problem, I report only the part useful for the resolution:

Maybe I shouldn't use a list

#scraping and assignment of Point_Team_A and Point_Team_B
...

try:
    Victory = Point_Team_A > Point_Team_B 
    Draw = Point_Team_A == Point_Team_B
    Lose = Point_Team_A < Point_Team_B
except:
    pass

result=[Victory, Draw, Lose] #great problem is here

print (*filter(None, result), sep='\n') #and here

For example if the result is a Victory, I want to get the output: Victory

NOTE: I use try / except because only one (1) variable will be True between Victory, Draw, Lose. The two False variables will be recognized as an error, because I previously created Point_Team_A and Point_Team_B inside an if / else condition. Subsequently I use Point_Team_A and B to create the 3 variables above. For example, if the result of a hockey match is 7-3, it means that team A has won, because Victory = Point_Team_A> Point_Team_B. Consequently Draw and Lose will be recognized by Python as an error, for this use try / except.

CodePudding user response：

just do

if a_points > b_points:
    print('victory')
elif a_points == b_points:
    print('draw')
else:
    print('defeat')

your code as is, when it works, would literally only ever print True

if you want another way to reproduce this error you're seeing, try this:

def f():
    try:
        x = 1 / 0
    except ZeroDivisionError:
        pass

    print(x)

what's happening is that the python interpreter knows that the name x exists in scope but also that it's never bound to anything (because the exception occurs before the assignment). so you get UnboundLocalError

CodePudding user response：

When there is an exception in the try block the variables created in the try block will not be available in an outside try block, so create a variable and initialize it either in the outside try block or inside except

Victory = Draw = Lose = False
try:
    Victory = Point_Team_A > Point_Team_B 
    Draw = Point_Team_A == Point_Team_B
    Lose = Point_Team_A < Point_Team_B
except:
    pass

result=[Victory, Draw, Lose] 

print (*filter(None, result), sep='\n')