Python 应用指定的规则构建数组

由于b包含了所有的对，你可以有效地将其重新塑造为正方形，并按其行数/列数索引，然后用sliding_window_view形成索引对，并索引正方形中间:

from numpy.lib.stride_tricks import sliding_window_view as swv

s = np.full((b[:, 0].max()+1, b[:, 1].max()+1), -1)
s[b[:, 0], b[:, 1]] = b[:, 2]

v = swv(a, 2, axis=0)
out = s[v[..., 0], v[..., 1]]

具有基于0的索引的变体(产生稍微更紧凑的中间部分):

s = np.full((b[:, 0].max(), b[:, 1].max()), -1)
s[b[:, 0]-1, b[:, 1]-1] = b[:, 2]

v = swv(a, 2, axis=0)-1
out = s[v[..., 0], v[..., 1]]

输出:

array([[4, 0, 6, 8],
       [1, 6, 6, 0],
       [6, 9, 0, 4],
       [9, 8, 6, 8],
       [0, 1, 9, 0]])

中级s:

array([[-1, -1, -1, -1],
       [-1,  6,  0,  9],
       [-1,  6,  0,  4],
       [-1,  1,  0,  8]])

# variant
array([[6, 0, 9],
       [6, 0, 4],
       [1, 0, 8]])

If you had arbitrary values making it difficult to generate a dense square intermediate, you could use pandas's merge:

from numpy.lib.stride_tricks import sliding_window_view as swv
import pandas as pd

out = (pd.DataFrame(swv(a, 2, axis=0).reshape(-1, 2))
         .merge(pd.DataFrame(b), how='left')
         [2].to_numpy()
         .reshape(-1, a.shape[1])
       )

输出:

array([[4, 0, 6, 8],
       [1, 6, 6, 0],
       [6, 9, 0, 4],
       [9, 8, 6, 8],
       [0, 1, 9, 0]])

timings

对于密集输入(即索引为0—n(或1—n)的形式而没有缺失索引)，具有1->N个索引的正方形NxN a(N = a0):>>

# numpy
16.5 ms ± 941 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# pandas
173 ms ± 17 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

使用稀疏输入(索引是从更大的集合中 Select 的N个值；这里是从50_000个可能性中 Select 的N0个值；～2%密度；～0.04%密度以正方形形式表示):

# numpy
5.04 s ± 2.5 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

# pandas
192 ms ± 13.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)