Python Error Syntax on equal symbol
0
$begingroup$
I'm correcting a program about a memory game, but in my click function there is a syntax error on equal symbol. Why?
About my def code, something to improve?. Perhaps, about using compressed structures?
Thanks
def clic(x, y):
global display, board, symbol, temporary1, temporary2
display.onclick(None)
[j, i] = [int(x), int(y)]
if 0 <= i < len(board) and 0 <= i < len(board[0]):
# check if the cell ist close
if board[i][j] == cell_close:
# operate it if you want
if temporary1:
temporary1 = [i, j]
board[i][j] = cell_partiallyOpen
else:
temporary2 = [i, j]
board[i][j] = cell_partiallyOpen
print_cell(board, symbol, i, j)
if temporary2 is not None:
if symbol[temporary1[0], [temporary1[1]]
== symbol[temporary2[0]][temporary2[1]]:
board[temporary1[0], temporary1[1]] = cell_open # error!
board[temporary2[0], temporary2[1]] = cell_open
else:
sleep(0.5) # time delay, when click on second cell
board[temporary1[0], [temporary1[1]] = cell_close
board[temporary2[0], [temporary2[1]] = cell_close
print_cell(board, symbol, temporary1[0], temporary1[1])
print_cell(board, symbol, temporary2[0], temporary2[1])
temporary1 = None
temporary2 = None
print_cell(board, symbol, temporary1[0], temporary1[1])
if all_open(board):
display.bye()
else:
display.onclick(clic)
Python 2.7.15
python-2.x
asked 5 mins ago
JRRJRR
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0
$begingroup$
I'm correcting a program about a memory game, but in my click function there is a syntax error on equal symbol. Why?
About my def code, something to improve?. Perhaps, about using compressed structures?
Thanks
def clic(x, y):
global display, board, symbol, temporary1, temporary2
display.onclick(None)
[j, i] = [int(x), int(y)]
if 0 <= i < len(board) and 0 <= i < len(board[0]):
# check if the cell ist close
if board[i][j] == cell_close:
# operate it if you want
if temporary1:
temporary1 = [i, j]
board[i][j] = cell_partiallyOpen
else:
temporary2 = [i, j]
board[i][j] = cell_partiallyOpen
print_cell(board, symbol, i, j)
if temporary2 is not None:
if symbol[temporary1[0], [temporary1[1]]
== symbol[temporary2[0]][temporary2[1]]:
board[temporary1[0], temporary1[1]] = cell_open # error!
board[temporary2[0], temporary2[1]] = cell_open
else:
sleep(0.5) # time delay, when click on second cell
board[temporary1[0], [temporary1[1]] = cell_close
board[temporary2[0], [temporary2[1]] = cell_close
print_cell(board, symbol, temporary1[0], temporary1[1])
print_cell(board, symbol, temporary2[0], temporary2[1])
temporary1 = None
temporary2 = None
print_cell(board, symbol, temporary1[0], temporary1[1])
if all_open(board):
display.bye()
else:
display.onclick(clic)
Python 2.7.15
python-2.x
asked 5 mins ago
JRRJRR
11
New contributor
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
0
0
$begingroup$
I'm correcting a program about a memory game, but in my click function there is a syntax error on equal symbol. Why?
About my def code, something to improve?. Perhaps, about using compressed structures?
Thanks
def clic(x, y):
global display, board, symbol, temporary1, temporary2
display.onclick(None)
[j, i] = [int(x), int(y)]
if 0 <= i < len(board) and 0 <= i < len(board[0]):
# check if the cell ist close
if board[i][j] == cell_close:
# operate it if you want
if temporary1:
temporary1 = [i, j]
board[i][j] = cell_partiallyOpen
else:
temporary2 = [i, j]
board[i][j] = cell_partiallyOpen
print_cell(board, symbol, i, j)
if temporary2 is not None:
if symbol[temporary1[0], [temporary1[1]]
== symbol[temporary2[0]][temporary2[1]]:
board[temporary1[0], temporary1[1]] = cell_open # error!
board[temporary2[0], temporary2[1]] = cell_open
else:
sleep(0.5) # time delay, when click on second cell
board[temporary1[0], [temporary1[1]] = cell_close
board[temporary2[0], [temporary2[1]] = cell_close
print_cell(board, symbol, temporary1[0], temporary1[1])
print_cell(board, symbol, temporary2[0], temporary2[1])
temporary1 = None
temporary2 = None
print_cell(board, symbol, temporary1[0], temporary1[1])
if all_open(board):
display.bye()
else:
display.onclick(clic)
Python 2.7.15
python-2.x
asked 5 mins ago
JRRJRR
11
New contributor
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm correcting a program about a memory game, but in my click function there is a syntax error on equal symbol. Why?
About my def code, something to improve?. Perhaps, about using compressed structures?
Thanks
def clic(x, y):
global display, board, symbol, temporary1, temporary2
display.onclick(None)
[j, i] = [int(x), int(y)]
if 0 <= i < len(board) and 0 <= i < len(board[0]):
# check if the cell ist close
if board[i][j] == cell_close:
# operate it if you want
if temporary1:
temporary1 = [i, j]
board[i][j] = cell_partiallyOpen
else:
temporary2 = [i, j]
board[i][j] = cell_partiallyOpen
print_cell(board, symbol, i, j)
if temporary2 is not None:
if symbol[temporary1[0], [temporary1[1]]
== symbol[temporary2[0]][temporary2[1]]:
board[temporary1[0], temporary1[1]] = cell_open # error!
board[temporary2[0], temporary2[1]] = cell_open
else:
sleep(0.5) # time delay, when click on second cell
board[temporary1[0], [temporary1[1]] = cell_close
board[temporary2[0], [temporary2[1]] = cell_close
print_cell(board, symbol, temporary1[0], temporary1[1])
print_cell(board, symbol, temporary2[0], temporary2[1])
temporary1 = None
temporary2 = None
print_cell(board, symbol, temporary1[0], temporary1[1])
if all_open(board):
display.bye()
else:
display.onclick(clic)
Python 2.7.15
python-2.x
python-2.x
asked 5 mins ago
JRRJRR
11
New contributor
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
JRRJRR
11
New contributor
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
JRRJRR
11
New contributor
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
JRRJRR
11
asked 5 mins ago
JRRJRR
11
11
New contributor
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
JRR is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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