Question

请问我对Python有点新鲜并且它很好，我可以评论说python非常性感，直到我需要移动4x4矩阵的内容，我想用它来构建2048游戏演示游戏是here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]
          ])

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]]

我需要有人以numpy的方式帮助我这样做，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{cover_up {1}}，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]]

提前致谢。

Answer 1

这是一个灵感来自this other post的向量化方法，并概括为涵盖所有四个方向的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]])

Answer 2

感谢所有这些我后来使用

def justify(a, direction):
    mask = a>0
    justified_mask = numpy.sort(mask,0) if direction == 'up' or direction =='down' else numpy.sort(mask, 1)
    if direction == 'up':
        justified_mask = justified_mask[::-1]
    if direction =='left':
        justified_mask = justified_mask[:,::-1]
    if direction =='right':
        justified_mask = justified_mask[::-1, :]    
    out = numpy.zeros_like(a) 
    out.T[justified_mask.T] = a.T[mask.T]
    return out

Python：证明NumPy数组的正当性

2 个答案: