所以说我有一个二维阵列，例如：

[
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]

我想将 3 周围的所有值 2 级别设定为特定数字，例如：

[
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 3, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]

请注意，3 可以在串列中的任何位置，我正在使用随机生成器来获取它。那么我怎么能做到这一点呢？也许使用 for 回圈？

uj5u.com热心网友回复：

继续评论 - 我发现 numpy 对于这样的切片非常有用；

import numpy as np

arr = np.array([
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
])

xs, ys = np.where(arr == 3)
arr[xs[0] - 2: xs[0]   3, ys[0] - 2: ys[0]   3] = 1
arr[xs[0], ys[0]] = 3

显然也可以采用纯 python/串列形式，但你可能会在双重迭代中陷入困境

uj5u.com热心网友回复：

这是一个纯 Python 方法（虽然相当笨拙）：

A = [
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]

for j, a in enumerate(A):
    try:
        i = a.index(3)
        P = A[j-1] if j > 0 else None
        Q = A[j 1] if j < len(a) - 1 else None
        for k in range(max(0, i-2), min(i 3, len(a))):
            if P:
                P[k] = 1
            if Q:
                Q[k] = 1
            a[k] = 1
        a[i] = 3
        break
    except ValueError:
        pass
print(A)

uj5u.com热心网友回复：

纯蟒蛇

A = [
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]


location = [(i, j) for i, row in enumerate(A) for j, item in enumerate(row) if item == 3][0]
for i, row in enumerate(A):
    for j, item in enumerate(row):
        if (abs(i - location[0]) <= 2) and (abs(j - location[1]) <= 2) and not ((i, j) == location):
            A[i][j] = 1