I'm working on a tic-tac-toe game, and run into a problem.

My problem lies in lines 24/25 and 30/31.
I want to take the returned dictionary from the convert_to_yx() function and assign the keys and values as the x and y of the game_grid I have on line 1. I get a TypeError that says the int is not iterable.

I'm guessing it's a rather simple problem with the syntax somewhere or I'm trying to do something impossible. If there is another way I can take 1 value and return 2 values, I'd be pleased to know.

Thanks for any help. I'll be available to answer any questions about my code should they arise.

``````game_grid = [{0:None, 1:None, 2:None},
{0:None, 1:None, 2:None},
{0:None, 1:None, 2:None}]

reference_grid = [{0:'1', 1:'2', 2:'3'},
{0:'4', 1:'5', 2:'6'},
{0:'7', 1:'8', 2:'9'}]

def display_grid(grid_type=game_grid):

for x in range(3):
for y in range(3):
print grid_type[y][x],
if y in range(2):
print "|",
if x in range(2):
print "\n--+---+--\n",

#takes the user's choice 1-9 and returns y, x values
def convert_to_yx(choice):
convert_table = {1:{0:0}, 2:{0:1}, 3:{0:2},
4:{1:0}, 5:{1:1}, 6:{1:2},
7:{2:0}, 8:{2:1}, 9:{2:2}}
yx = convert_table[choice]
return yx

#accepts cell choice, sends to convert_to_yx() and applies to grid
def choose_cell():
choice = input("\nWhich spot would you like to place your X? >>> ")
for y, x in convert_to_yx(choice):
game_grid[y][x]='X'
display_grid()

if __name__ == "__main__":
display_grid(reference_grid)
choose_cell()``````
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Last Post by woooee

Add some print statements so you can tell what is going on, such as

``````convert_table = {1:{0:0}, 2:{0:1}, 3:{0:2},
4:{1:0}, 5:{1:1}, 6:{1:2},
7:{2:0}, 8:{2:1}, 9:{2:2}}

yx = convert_table[3]
print yx, type(yx)``````

There isn't a y, x in yx (convert_table[choice]). It's a dictionary not a list. You'll have to use something else to get this data. Also add more print statements so you know what is going on. Test each function individually, especially the values passed to the functions and the values it returns.

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