我正在尝试使用python在pygame中制作一个2048游戏。如果没有给出参数，我希望函数stackright和stackleft对self.data进行操作，如果在参数中给出了矩阵，我希望返回一个stackedighted或stacklefted矩阵。我会去做这个。我正在使用python 2.7 这是我的代码：
import random
def getnull(): # returns null matrix
data = [
[0,0,2,2],
[0,4,0,2],
[0,2,0,0],
[2,0,2,0]
]
return data
class Data:
def fillrandom(self): #updates data by adding 2/4 randomly
exit = False
while not exit:
y = random.randint(0,3)
x = random.randint(0,3)
if self.data[y][x] == 0:
if random.randint(1,10) == 1:
self.data[y][x] = 4
exit = True
else:
self.data[y][x] = 2
exit = True
def alignright(self): #
list1 = [[],[],[],[]]
for i in range(4): # per row loop
for count in range(4): # per column loop
if self.data[i][count] != 0:
list1[i] += [self.data[i][count]]
list1[i] = [0]*(4-len(list1[i])) + list1[i]
self.data = list1
def alignleft(self):
list1 = [[],[],[],[]]
for i in range(4): # per row loop
for count in range(4): # per column loop
if self.data[i][count] != 0:
list1[i] += [self.data[i][count]]
list1[i] = list1[i] + [0]*(4-len(list1[i]))
self.data = list1
def printstate(self):
print(self.data[0])
print(self.data[1])
print(self.data[2])
print(self.data[3])
print("-------")
def checkfilled(self):
for count in range(4):
for i in range(4):
if self.data[count][i] == 0:
return False
return True
def stackright(self):
for i in range(4):
if self.data[i][3] == self.data[i][2]:
if self.data[i][1] == self.data[i][0]:
self.data[i][3] = self.data[i][3] *2
self.score += self.data[i][3]
self.data[i][2] = self.data[i][1] *2
self.score += self.data[i][2]
self.data[i][0] , self.data[i][1] = 0,0
else:
self.data[i][3] = self.data[i][3] *2
self.score += self.data[i][3]
self.data[i][2] = self.data[i][1]
self.data[i][1] = self.data[i][0]
self.data[i][0] = 0
elif self.data[i][2] == self.data[i][1]:
self.data[i][2] = self.data[i][2] *2
self.score += self.data[i][2]
self.data[i][1] = self.data[i][0]
self.data[i][0] = 0
elif self.data[i][1] == self.data[i][0]:
self.data[i][1] = self.data[i][1] *2
self.score += self.data[i][1]
self.data[i][0] = 0
def stackleft(self):
for i in range(4):
if self.data[i][0] == self.data[i][1]:
if self.data[i][2] == self.data[i][3]:
self.data[i][0] = self.data[i][0] *2
self.score += self.data[i][0]
self.data[i][1] = self.data[i][2] *2
self.score += self.data[i][1]
self.data[i][3] , self.data[i][2] = 0,0
else:
self.data[i][0] = self.data[i][0]*2
self.score += self.data[i][0]
self.data[i][1] = self.data[i][2]
self.data[i][2] = self.data[i][3]
self.data[i][3] = 0
elif self.data[i][1] == self.data[i][2]:
self.data[i][1] = self.data[i][1] *2
self.score += self.data[i][1]
self.data[i][2] = self.data[i][3]
self.data[i][3] = 0
elif self.data[i][2] == self.data[i][3]:
self.data[i][2] = self.data[i][2] *2
self.score += self.data[i][2]
self.data[i][3] = 0
def alignup(self):
col = [[],[],[],[]]
for i in range(4): #per col loop
for count in range(4): #per row loop
if self.data[count][i] != 0:
col[i] += [self.data[count][i]]
col[i] = col[i] + [0]*(4-len(col[i]))
for i in range(4): # convert column to row
for count in range(4):
self.data[count][i] = col[i][count]
def aligndown(self):
col = [[],[],[],[]]
for i in range(4): #per col loop
for count in range(4): #per row loop
if self.data[count][i] != 0:
col[i] += [self.data[count][i]]
col[i] = [0]*(4-len(col[i])) + col[i]
for i in range(4): # convert column to row
for count in range(4):
self.data[count][i] = col[i][count]
def __init__(self):
self.data = getnull()
self.score = 0
data = Data()
data.aligndown()
data.printstate()
print(data.score)
while True:
pass
答案 0 :(得分：3)
您可以拥有默认参数。例如：
def func(self, matrix=None):
if matrix is None:
#do stuff with self.data
else:
#do stuff
这样，如果没有给出参数，则默认值为None
那么你知道如果matrix的值为None，那么调用者没有指定一个值，所以你用self.data做了一些事情。但如果他确实指定了一个值（else），则表示调用者指定了一个值，你可以对矩阵做一些事情。
或者，如果您想将它们用作相同的值，您可以执行以下操作：
def func(self, matrix=None):
if matrix is None: matrix = self.data
#do stuff with the variable 'data'
现在data是您想要的任何内容
答案 1 :(得分：1)
这样的事情可以解决问题：
def stackright(self, *args):
if len(args) == 0:
#do things with no arguments
else:
#do things with arguments
print(args[0], args[1], ...)
如果您希望能够按名称调用参数，也可以将*args替换为**kwargs。你甚至可以同时使用两者：
def f(self, madatoryArgument1, mandatoryArgument2, *args, **kwargs)
这比使用None默认值传递参数的优点是，当参数数量增加时，它会简化它。
答案 2 :(得分：0)
看起来你要问的是，你的功能看起来像这样：
def stackright(self, data=None):
if data is None: data = self.data
for i in range(4):
if data[i][3] == data[i][2]:
if data[i][1] == data[i][0]:
data[i][3] = data[i][3] *2
self.score += data[i][3]
data[i][2] = data[i][1] *2
self.score += data[i][2]
data[i][0] , data[i][1] = 0,0
else:
data[i][3] = data[i][3] *2
self.score += data[i][3]
data[i][2] = data[i][1]
data[i][1] = data[i][0]
data[i][0] = 0
elif data[i][2] == data[i][1]:
data[i][2] = data[i][2] *2
self.score += data[i][2]
data[i][1] = data[i][0]
data[i][0] = 0
elif data[i][1] == data[i][0]:
data[i][1] = data[i][1] *2
self.score += data[i][1]
data[i][0] = 0
return data
def stackleft(self, data=None):
if data == None: data = self.data
for i in range(4):
if data[i][0] == data[i][1]:
if data[i][2] == data[i][3]:
data[i][0] = data[i][0] *2
self.score += data[i][0]
data[i][1] = data[i][2] *2
self.score += data[i][1]
data[i][3] , data[i][2] = 0,0
else:
data[i][0] = data[i][0]*2
self.score += data[i][0]
data[i][1] = data[i][2]
data[i][2] = data[i][3]
data[i][3] = 0
elif data[i][1] == data[i][2]:
data[i][1] = data[i][1] *2
self.score += data[i][1]
data[i][2] = data[i][3]
data[i][3] = 0
elif data[i][2] == data[i][3]:
data[i][2] = data[i][2] *2
self.score += data[i][2]
data[i][3] = 0
return data
然后调用data.stackright（）将对自身数据进行操作，但调用data.stackright（matrix2）将对matrix2进行操作（同时将分数添加到数据中，除非您更改上面的部分）