所以說我有一個二維陣列，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]
]
我想將 3 周圍的所有值 2 級別設定為特定數字，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]
]
請注意，3 可以在串列中的任何位置，我正在使用隨機生成器來獲取它。那么我怎么能做到這一點呢？也許使用 for 回圈？
uj5u.com熱心網友回復：
繼續評論 - 我發現 numpy 對于這樣的切片非常有用；
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
顯然也可以采用純 python/串列形式，但你可能會在雙重迭代中陷入困境
uj5u.com熱心網友回復：
這是一個純 Python 方法（雖然相當笨拙）：
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熱心網友回復：
純蟒蛇
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
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