Python如何修改代码将二维列表board中所有'1'移动到每列底部位置

需求说明 #

将二维列表board中所有的'1'移动到每一列的最底部位置，示例如下：
输入参考：

board = [[" "," ","1"," "],
         [" "," ","1"," "],
         ["1","1"," "," "],
         ["1"," "," ","1"]]

预期输出：

[[" "," "," "," "],
 [" "," "," "," "],
 ["1"," ","1"," "],
 ["1","1","1","1"]] 

原代码问题分析 #

现有代码存在两个核心问题：

1. 没有保留原二维列表的行列结构，直接把所有元素打平为一维列表处理
2. 没有按列维度统计'1'的数量并单独处理每列，而是把所有'1'直接拼接在打平后的列表末尾，完全不符合按列下沉的要求

修改后的可运行代码 #

board = [[" "," ","1"," "],
         [" "," ","1"," "],
         ["1","1"," "," "],
         ["1"," "," ","1"]]

def sink_ones_to_col_bottom(board):
    row_num = len(board)
    if not row_num:
        return []
    col_num = len(board[0])
    # 先转置二维列表，把列转为行方便处理
    transposed_cols = list(zip(*board))
    processed_cols = []
    for col in transposed_cols:
        # 统计当前列的'1'总数
        one_cnt = col.count('1')
        # 构造新列：前面填充空格，末尾放对应数量的'1'
        new_col = [' ']*(row_num - one_cnt) + ['1']*one_cnt
        processed_cols.append(new_col)
    # 转置回原行列结构，将元组转为列表保证格式一致
    return [list(row) for row in zip(*processed_cols)]

# 输出结果
result = sink_ones_to_col_bottom(board)
print(result)

运行输出：

[[' ', ' ', ' ', ' '], [' ', ' ', ' ', ' '], ['1', ' ', '1', ' '], ['1', '1', '1', '1']]

完全匹配预期要求。

内容的提问来源于stack exchange，提问作者Joshua Bradley