如何使用类似于sudoku的行输出以下数据
p = [
[-8, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, -3, -6, 0, 0, 0, 0, 0],
[0, -7, 0, 0, -9, 0, -2, 0, 0],
[0, -5, 0, 0, 0, -7, 0, 0, 0],
[0, 0, 0, 0, -4, -5, -7, 0, 0],
[0, 0, 0, -1, 0, 0, 0, -3, 0],
[0, 0, -1, 0, 0, 0, 0, -6, -8],
[0, 0, -8, -5, 0, 0, 0, -1, 0],
[0, -9, 0, 0, 0, 0, -4, 0, 0]]
print ("Unsolved:")
w="\n".join(re.findall("(?p).{,9}", p))[:-1]
print (w)
以下列方式获取输出
0 | 0 | 3 | 0 | 2 | 0 | 6 | 0 | 0
9 | 0 | 0 | 3 | 0 | 5 | 0 | 0 | 1
0 | 0 | 1 | 8 | 0 | 6 | 4 | 0 | 0
0 | 0 | 8 | 1 | 0 | 2 | 9 | 0 | 0
7 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 8
0 | 0 | 6 | 7 | 0 | 8 | 2 | 0 | 0
0 | 0 | 2 | 6 | 0 | 9 | 5 | 0 | 0
8 | 0 | 0 | 2 | 0 | 3 | 0 | 0 | 9
0 | 0 | 5 | 0 | 1 | 0 | 3 | 0 | 0
答案 0 :(得分:3)
rows = ("|".join(map(" {:2d} ".format, row)) for row in p)
print("\n".join(rows))
-8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 0 | 0 | -3 | -6 | 0 | 0 | 0 | 0 | 0 0 | -7 | 0 | 0 | -9 | 0 | -2 | 0 | 0 0 | -5 | 0 | 0 | 0 | -7 | 0 | 0 | 0 0 | 0 | 0 | 0 | -4 | -5 | -7 | 0 | 0 0 | 0 | 0 | -1 | 0 | 0 | 0 | -3 | 0 0 | 0 | -1 | 0 | 0 | 0 | 0 | -6 | -8 0 | 0 | -8 | -5 | 0 | 0 | 0 | -1 | 0 0 | -9 | 0 | 0 | 0 | 0 | -4 | 0 | 0