我有一个下面的sparkDataframe，其中所有的列（除了主键列empïid）都由一个Map组成（其中键“from”和“to”可以有空值）。我想计算每一列的'from'和'to'（emp\u id除外），并向Map（名为'change'）添加一个新键，该Map的值为a）'insert'if'from'value为null，to'不为null b）'delete'if'to'value为null，from'不为null b）'update'if'from'和'to'不为null&'from'value不同于'to'value
注意：具有空值的列将保持不变。
我们如何在scala中实现这一点。

|emp_id|emp_city             |emp_name                    |emp_phone            |emp_sal                    |emp_site                          |

|1     |null                 |[from -> Will, to -> Watson]|null                 |[from -> 1000, to -> 8000]|[from ->, to -> Seattle]          |
|3     |null                 |[from -> Norman, to -> Nate]|null                 |[from -> 1000, to -> 8000]|[from -> CherryHill, to -> Newark]|
|4     |[from ->, to -> Iowa]|[from ->, to -> Ian]        |[from ->, to -> 1004]|[from ->, to -> 8000]     |[from ->, to -> Des Moines]       |

预期：

|emp_id|emp_city             |emp_name                    |emp_phone            |emp_sal                    |emp_site                          |

|1     |null                 |[from -> Will, to -> Watson, change -> update]|null                 |[from -> 1000, to -> 8000, change -> update]|[from ->, to -> Seattle, change -> insert]          |
|3     |null                 |[from -> Norman, to -> Nate, change -> update]|null                 |[from -> 1000, to -> 8000, change -> update]|[from -> CherryHill, to -> Newark, change -> update]|
|4     |[from ->, to -> Iowa, change -> insert]|[from ->, to -> Ian, change -> insert]        |[from ->, to -> 1004, change -> insert]|[from ->, to -> 8000, change -> insert]     |[from ->, to -> Des Moines, change -> insert]       |