a = []def make_squares(arr, length, nums): if not nums: print(arr) a.append(arr) return r = 0 while r < len(arr) and len(arr[r]) == length: r += 1 for i in nums: nums.remove(i) arr[r].append(i) make_squares(arr, length, nums) nums.append(i) arr[r] = arr[r][:-1]make_squares([[] for i in range(3)], 3, [i+1 for i in range(3**2)])print(a)我正在嘗試使用上面的代碼創建nxn矩陣，每個矩陣都有一個 numbers 的排列i+1...n^2。我已經打印了每個附加到 a 的矩陣，它們看起來是正確的，但是當我最后打印 a 時，我得到了[[[], [], []], [[], [], []], ...]。這對我來說沒有任何意義。預期的結果是[[[1, 2, 3], [4, 5, 6], [7, 8, 9]], [[1, 2, 3], [4, 5, 6], [7, 9, 8]], ...]
1 回答
紫衣仙女
TA貢獻1839條經驗 獲得超15個贊
順序可能與您預期的結果不同，但這對標準庫itertools模塊來說是個竅門：
import itertools
def make_squares(w):
size = w * w
for perm in itertools.permutations(range(1, size + 1)):
# "grouper" from the itertools recipe list
yield list(itertools.zip_longest(*[iter(perm)] * w))
for square in make_squares(3):
print(square)
印刷
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]
[(1, 2, 3), (4, 5, 6), (7, 9, 8)]
[(1, 2, 3), (4, 5, 6), (8, 7, 9)]
[(1, 2, 3), (4, 5, 6), (8, 9, 7)]
[(1, 2, 3), (4, 5, 6), (9, 7, 8)]
[(1, 2, 3), (4, 5, 6), (9, 8, 7)]
[(1, 2, 3), (4, 5, 7), (6, 8, 9)]
[(1, 2, 3), (4, 5, 7), (6, 9, 8)]
[(1, 2, 3), (4, 5, 7), (8, 6, 9)]
[(1, 2, 3), (4, 5, 7), (8, 9, 6)]
[(1, 2, 3), (4, 5, 7), (9, 6, 8)]
[(1, 2, 3), (4, 5, 7), (9, 8, 6)]
[(1, 2, 3), (4, 5, 8), (6, 7, 9)]
[(1, 2, 3), (4, 5, 8), (6, 9, 7)]
[(1, 2, 3), (4, 5, 8), (7, 6, 9)]
...
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