我正在尝试以该面板为中心，这是我尝试使用％* s的代码，但没有成功。
任何的想法？
board = [["1","2","2"],["8*1","2@3","5*6"],["9","5","8"],["2","2","2"],
["5*6","6*8","0@2"],["1","2","8"],["1","9","8"],["2*7","7*5","4@2"],["1","3","3"]]
counter = 0
print("--------------+---------------+-------------------")
for row in board:
counter += 1
print("|" "%s" "|") % (" | ".join(row).center(47))
if counter == 3 or counter == 6 or counter == 9:
print("---------------+---------------------+--------------")
每个盒子的输出应该像这样
+-----------+
| 1 | 1 | 9 |
|2@3|1*6|7*2|
| 4 | 1 | 2 |
+-----------+
最佳答案
我修改了您的代码，使其产生的内容接近您的要求：
board = [["1","2","2"],["8*1","2@3","5*6"],["9","5","8"],
["2","2","2"],["5*6","6*8","0@2"],["1","2","8"],
["1","9","8"],["2*7","7*5","4@2"],["1","3","3"]]
counter = 0
print("+-----------+")
for row in board:
counter += 1
print("|%s|" % "|".join(c.center(3) for c in row))
if counter == 3 or counter == 6 or counter == 9:
print("+-----------+")
产生：
+-----------+
| 1 | 2 | 2 |
|8*1|2@3|5*6|
| 9 | 5 | 8 |
+-----------+
| 2 | 2 | 2 |
|5*6|6*8|0@2|
| 1 | 2 | 8 |
+-----------+
| 1 | 9 | 8 |
|2*7|7*5|4@2|
| 1 | 3 | 3 |
+-----------+
关于python - 居中矩阵，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/11497485/