python - 按值对矩阵进行分组和汇总

Question

我有两个矩阵，prob 和 totalHigh 都是 axbxcxd 形状。 a 和b 是坐标。以下是两个示例:

In [77]: prob[1,1,:]
Out[77]: 
array([[ 0.09,  0.01,  0.  ,  0.  ,  0.  ],
       [ 0.81,  0.09,  0.  ,  0.  ,  0.  ],
       [ 0.  ,  0.  ,  0.  ,  0.  ,  0.  ]])

In [78]: totalHigh[1,1,:]
Out[78]: 
array([[0, 1, 2, 3, 4],
       [1, 2, 3, 4, 5],
       [2, 3, 4, 5, 6]])

totalHigh 包含有关结果的信息，不幸的是在两个维度上。相应地，prob 包含这些结果的概率。例如，在坐标 1,1 处，结果 1 的总概率是 0.01+0.81。

如何去除多余的维度？

预期结果

simplifiedHigh[1,1,:]
array([0, 1, 2, 3, 4, 5, 6])
simplifiedProb[1,1,:]
array([0.09, 0.82, 0.09, 0, 0, 0, 0])

我如何以最有效的方式得到它？

Answer 1

您可以使用 np.bincount和 np.unique -

IDs = np.unique(totalHigh_sliced)
counts = np.bincount(totalHigh_sliced.ravel(),prob_sliced.ravel())

sample 运行-

In [215]: prob_sliced
Out[215]: 
array([[ 0.09,  0.01,  0.  ,  0.  ,  0.  ],
       [ 0.81,  0.09,  0.  ,  0.  ,  0.  ],
       [ 0.  ,  0.  ,  0.  ,  0.  ,  0.  ]])

In [216]: totalHigh_sliced
Out[216]: 
array([[0, 1, 2, 3, 4],
       [1, 2, 3, 4, 5],
       [2, 3, 4, 5, 6]])

In [217]: IDs = np.unique(totalHigh_sliced)
     ...: counts = np.bincount(totalHigh_sliced.ravel(),prob_sliced.ravel())
     ...: 

In [218]: IDs
Out[218]: array([0, 1, 2, 3, 4, 5, 6])

In [219]: counts
Out[219]: array([ 0.09,  0.82,  0.09,  0.  ,  0.  ,  0.  ,  0.  ])