python - 高效统计 NumPy 中唯一子数组的出现次数？

Question

我有一个形状为 (128, 36, 8) 的数组，我想找出最后一个维度中长度为 8 的唯一子数组的出现次数。

我知道 np.unique 和 np.bincount，但它们似乎是针对元素而不是子数组。我看过this question但它是关于查找特定子数组的第一次出现，而不是所有唯一子数组的计数。

Answer 1

问题指出输入数组的形状为 (128, 36, 8)，我们有兴趣在最后一个维度中找到长度为 8 的唯一子数组。所以，我假设唯一性是沿着前两个维度合并在一起的。让我们假设 A 作为输入 3D 数组。

获取唯一子数组的个数

# Reshape the 3D array to a 2D array merging the first two dimensions
Ar = A.reshape(-1,A.shape[2])

# Perform lex sort and get the sorted indices and xy pairs
sorted_idx = np.lexsort(Ar.T)
sorted_Ar =  Ar[sorted_idx,:]

# Get the count of rows that have at least one TRUE value 
# indicating presence of unique subarray there
unq_out = np.any(np.diff(sorted_Ar,axis=0),1).sum()+1

sample 运行-

In [159]: A # A is (2,2,3)
Out[159]: 
array([[[0, 0, 0],
        [0, 0, 2]],

       [[0, 0, 2],
        [2, 0, 1]]])

In [160]: unq_out
Out[160]: 3

获取唯一子数组的出现次数

# Reshape the 3D array to a 2D array merging the first two dimensions
Ar = A.reshape(-1,A.shape[2])

# Perform lex sort and get the sorted indices and xy pairs
sorted_idx = np.lexsort(Ar.T)
sorted_Ar =  Ar[sorted_idx,:]

# Get IDs for each element based on their uniqueness
id = np.append([0],np.any(np.diff(sorted_Ar,axis=0),1).cumsum())

# Get counts for each ID as the final output
unq_count = np.bincount(id) 

sample 运行-

In [64]: A
Out[64]: 
array([[[0, 0, 2],
        [1, 1, 1]],

       [[1, 1, 1],
        [1, 2, 0]]])

In [65]: unq_count
Out[65]: array([1, 2, 1], dtype=int64)