请我有点Python陌生，感觉很好，我可以说python很性感，直到我需要移动4x4矩阵的内容，我想在构建游戏的2048游戏演示时使用它，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
如果你这样调用它，这就是函数的作用
cover_left([ [1,0,2,0], [3,0,4,0], [5,0,6,0], [0,7,0,8] ])
它将覆盖左侧的零并产生
[ [1, 2, 0, 0], [3, 4, 0, 0], [5, 6, 0, 0], [7, 8, 0, 0]]
请让我帮助某人，以numpy达到更快的速度并且需要更少的代码（我在深度优先搜索算法中使用的代码），更重要的是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]]
这里有一个量化的方法，通过启发this other post和推广到覆盖non-zeros所有四个方向-
this other post
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
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]])
