您可以使用索引：

# output with the shape of a
c = np.zeros_like(a)

# non-zero positions
m = a != 0

# assign values
c[m] = b[a[m]-1]

numpy.where更短的替代方案（但如果有很多零，效率可能会降低）：

c = np.where(a!=0, b[a-1], 0)

输出：

array([[ 9,  1,  8,  0,  0],
       [ 6,  4,  0,  0,  0],
       [ 4,  4, 14, 11,  0],
       [ 6,  0,  0,  0,  0],
       [19,  5,  0,  0,  0],
       [33,  9,  4,  0,  0],
       [ 0,  0,  0,  0,  0],
       [ 5,  4,  0,  0,  0],
       [ 5,  0,  0,  0,  0],
       [14,  3,  6,  4,  0]])

可重现输入：

a = np.array([[15,  2,  1,  0,  0],
              [ 8, 12,  0,  0,  0],
              [ 4, 12, 10,  9,  0],
              [ 3,  0,  0,  0,  0],
              [19,  7,  0,  0,  0],
              [13, 15,  4,  0,  0],
              [ 0,  0,  0,  0,  0],
              [11,  4,  0,  0,  0],
              [ 7,  0,  0,  0,  0],
              [10,  6,  8,  4,  0]])

b =  np.array([ 8,  1,  6,  4,  9,  3,  5,  6, 11, 14,  5,  4, 33,  7,  9, 15,  7,
                3, 19,  3])

计时

在（10000，5）上，大部分为非零

# indexing
203 µs ± 6 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

# numpy.where
97.2 µs ± 2.02 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

On（10000，10005），只有前5列为非零

# indexing
2.2 s ± 384 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

# numpy.where
4.73 s ± 680 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)