Aki*_*zri 9 python search artificial-intelligence numpy 2048
请我有点新鲜Python,它一直很好,我可以评论说python是非常性感的,直到我需要移动4x4矩阵的内容,我想用于构建游戏的2048游戏演示在这里我有这个功能
def cover_left(matrix):
new=[[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]]
for i in range(4):
count=0
for j in range(4):
if mat[i][j]!=0:
new[i][count]=mat[i][j]
count+=1
return new
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如果你像这样调用它,这就是这个函数的作用
cover_left([
[1,0,2,0],
[3,0,4,0],
[5,0,6,0],
[0,7,0,8]
])
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它会覆盖左边的零并产生
[ [1, 2, 0, 0],
[3, 4, 0, 0],
[5, 6, 0, 0],
[7, 8, 0, 0]]
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我需要有人帮助我numpy做一个这样做的方法,我相信会更快,需要更少的代码(我在深度优先搜索算法中使用),更重要的是实现cover_up,cover_down和
`cover_left`.
`cover_up`
[ [1, 7, 2, 8],
[3, 0, 4, 0],
[5, 0, 6, 0],
[0, 0, 0, 0]]
`cover_down`
[ [0, 0, 0, 0],
[1, 0, 2, 0],
[3, 0, 4, 0],
[5, 7, 6, 8]]
`cover_right`
[ [0, 0, 1, 2],
[0, 0, 3, 4],
[0, 0, 5, 6],
[0, 0, 7, 8]]
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提前致谢.
Div*_*kar 27
这是一个矢量化的方法,受到所有四个方向的启发this other post和概括non-zeros-
def justify(a, invalid_val=0, axis=1, side='left'):
"""
Justifies a 2D array
Parameters
----------
A : ndarray
Input array to be justified
axis : int
Axis along which justification is to be made
side : str
Direction of justification. It could be 'left', 'right', 'up', 'down'
It should be 'left' or 'right' for axis=1 and 'up' or 'down' for axis=0.
"""
if invalid_val is np.nan:
mask = ~np.isnan(a)
else:
mask = a!=invalid_val
justified_mask = np.sort(mask,axis=axis)
if (side=='up') | (side=='left'):
justified_mask = np.flip(justified_mask,axis=axis)
out = np.full(a.shape, invalid_val)
if axis==1:
out[justified_mask] = a[mask]
else:
out.T[justified_mask.T] = a.T[mask.T]
return out
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样品运行 -
In [473]: a # input array
Out[473]:
array([[1, 0, 2, 0],
[3, 0, 4, 0],
[5, 0, 6, 0],
[6, 7, 0, 8]])
In [474]: justify(a, axis=0, side='up')
Out[474]:
array([[1, 7, 2, 8],
[3, 0, 4, 0],
[5, 0, 6, 0],
[6, 0, 0, 0]])
In [475]: justify(a, axis=0, side='down')
Out[475]:
array([[1, 0, 0, 0],
[3, 0, 2, 0],
[5, 0, 4, 0],
[6, 7, 6, 8]])
In [476]: justify(a, axis=1, side='left')
Out[476]:
array([[1, 2, 0, 0],
[3, 4, 0, 0],
[5, 6, 0, 0],
[6, 7, 8, 0]])
In [477]: justify(a, axis=1, side='right')
Out[477]:
array([[0, 0, 1, 2],
[0, 0, 3, 4],
[0, 0, 5, 6],
[0, 6, 7, 8]])
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