Scala 动态重命名dataframe中的列，然后与另一个表联接

在dataframe中，我有一个如下所示的属性表

在要重命名的列中

我必须根据此输入重命名该列

如果cust_id标志为yes，我只想加入customer表

在最后的输出中，我想用实际的列名显示哈希列值

val maintab\u df=maintable

val cust_df=customertable

将主表列e重命名为a后连接主表和客户表

maintable.a=customertable.a

---

以下是一个如何操作的示例：

propertydf.show
+-----------------+------------+
|columns-to-rename|cust_id_flag|
+-----------------+------------+
|(e to a),(d to b)|           Y|
+-----------------+------------+

val columns_to_rename = propertydf.head(1)(0).getAs[String]("columns-to-rename")
val cust_id_flag = propertydf.head(1)(0).getAs[String]("cust_id_flag")

val parsed_columns = columns_to_rename.split(",")
    .map(c => c.replace("(", "").replace(")", "")
    .split(" to "))
// parsed_columns: Array[Array[String]] = Array(Array(e, a), Array(d, b))

val rename_columns = maintab_df.columns.map(c => {
    val matched = parsed_columns.filter(p => c == p(0))
    if (matched.size != 0)
        col(c).as(matched(0)(1).toString) 
    else 
        col(c)
})
// rename_columns: Array[org.apache.spark.sql.Column] = Array(e AS `a`, f, c, d AS `b`)

val select_columns = maintab_df.columns.map(c => {
    val matched = parsed_columns.filter(p => c == p(0))
    if (matched.size != 0) 
        col(matched(0)(1) + "_hash").as(matched(0)(1).toString) 
    else 
        col(c)
})
// select_columns: Array[org.apache.spark.sql.Column] = Array(a_hash AS `a`, f, c, b_hash AS `b`)

val join_cond = parsed_columns.map(_(1))
// join_cond: Array[String] = Array(a, b)

if (cust_id_flag == "Y") {
    val result = maintab_df.select(rename_columns:_*)
                           .join(cust_df, join_cond)
                           .select(select_columns:_*)
} else {
    val result = maintab_df
}

result.show
+------+---+---+--------+
|     a|  f|  c|       b|
+------+---+---+--------+
|*****!|  1| 11|    &&&&|
| ****%|  2| 12|;;;;;;;;|
|*****@|  3| 13|  \\\\\\|
+------+---+---+--------+

---

它不应该是从d到b而不是从f到b吗？哦，是的，对了，应该是从d到b！！！