Python:对齐NumPy数组 [英] Python: Justifying NumPy array

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问题描述

请问我对Python有点陌生,这很好,我可以评论说python非常性感,直到我需要移动4x4矩阵的内容为止,我想在构建4x4矩阵的2048游戏演示时使用该矩阵游戏是Gabriele Cirulli的此处我有此功能

Please I am a bit new to Python and it has been nice, I could comment that python is very sexy till I needed to shift content of a 4x4 matrix which I want to use in building a 2048 game demo of the game is here I have this function

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

如果您这样调用它,这就是函数的作用

This is what this function does if you call it like this

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

It will cover the zeros to the left and produce

[  [1, 2, 0, 0],
   [3, 4, 0, 0],
   [5, 6, 0, 0],
   [7, 8, 0, 0]]

请我需要有人帮助我实现numpy的方式,我相信这会更快并且需要更少的代码(我在深度优先搜索算法中使用),更重要的是cover_up的实现,cover_down

Please I need someone to help me with a numpy way of doing this which I believe will be faster and require less code (I am using in a depth-first search algo) and more importantly the implementation of cover_up, cover_down and

`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]]

提前谢谢.

推荐答案

这是一种受 this other post 启发的矢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]])

这篇关于Python:对齐NumPy数组的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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