2048 game in Python

Question

As a beginner I started to code a 2048 game. I made the matrix and filled it with 0's. Then I wanted to write a function which loops trough the whole matrix and find all the 0 values. Then save the coordinates of the 0 values and later replace them with 2 or 4 values. I made a random variable to choose beetween 2 or 4. The problem is, that i don't really know how to push the x and y coordinates of the 0 values to and array and then read them.

table = [[0, 0, 0, 0],
 [0, 0, 0, 0],
 [0, 0, 0, 0],
 [0, 0, 0, 0]]
options = []
def saveOptions(i, j):
 options.append([{i , j}])
 return options
def optionsReaderandAdder(options):
 if len(options) > 0:
 spot = random.random(options)
 r = random.randint(0, 1)
 if r > 0.5:
 table <-------- THIS IS THE LINE WHERE I WOULD LIKE TO CHANGE THE VALUE OF THE 0 TO 2 OR 4.
def optionsFinder():
 for i in range(4):
 for j in range(4):
 if table[i][j] == 0:
 saveOptions(i, j)
 optionsReaderandAdder(options)
addNumber()
print('\n'.join([''.join(['{:4}'.format(item) for item in row])
 for row in table]))

• Open AI - Opting Out: {i , j} is an unordered set. You probably want a tuple, (i , j). Unless 2048 is symmetrical, I don't know.

Answer 1

You can iterate over the rows and columns of the table:

for row in table:
 for element in row:
 if element == 0:
 # now what?

We don't know what the coordinate is.

Python has a useful function named enumerate. We can iterate the same way as previously, but we also get the index, or position into the array.

zeroes = []
for i, row in enumerate(table):
 for j, element in enumerate(row):
 if element == 0:
 zeroes.append((i, j))

We can then set the values, for example all to 2:

for i, j in zeroes:
 table[i][j] = 2

Your random code looks fine.

• D.Tomi: Thank you very much! Gonna look at it home