Python 如何循环Pyspark RDD中的每一项并将其转换为键？使用地图功能？

首先，我有一些类似的输入：

A:<phone1,phone2>,<location1>,<email1>
B:<phone1>,<location2>,<email1,email2>

phone1: A:<phone1,phone2>,<location1>,<email1>
phone1: B:<phone1>,<location2>,<email1,email2>
phone2: A:<phone1,phone2>,<location1>,<email1>
location1: A:<phone1,phone2>,<location1>,<email1>
location2: B:<phone1>,<location2>,<email1,email2>
email1: A:<phone1,phone2>,<location1>,<email1>
email1: B:<phone1>,<location2>,<email1,email2>
email2: B:<phone1>,<location2>,<email1,email2>

A:，，
B:，，

我想使用Pyspark.rdd.map（）函数在行中每次循环，并将它们转换为键值对，如下所示：

A:<phone1,phone2>,<location1>,<email1>
B:<phone1>,<location2>,<email1,email2>

phone1: A:<phone1,phone2>,<location1>,<email1>
phone1: B:<phone1>,<location2>,<email1,email2>
phone2: A:<phone1,phone2>,<location1>,<email1>
location1: A:<phone1,phone2>,<location1>,<email1>
location2: B:<phone1>,<location2>,<email1,email2>
email1: A:<phone1,phone2>,<location1>,<email1>
email1: B:<phone1>,<location2>,<email1,email2>
email2: B:<phone1>,<location2>,<email1,email2>

电话1:A:，， 电话1:B:，， 电话2:A:，， 地点1:A:，， 地点二:乙:,，， 电子邮件1:A:，， 电子邮件1:B:，， 电邮2:B:，， 在我以前的尝试中，我试图在map函数内部的lambda函数上添加一个循环，但它不支持。还有别的办法吗

scala>val-rdd=sc.parallelize（顺序（“A:，，，，，，，，，，”））
    scala> val rdd =  sc.parallelize(Seq("A:<phone1,phone2>,<location1>,<email1>", "B:<phone1>,<location2>,<email1,email2>"))

    scala> rdd.foreach(println)
    A:<phone1,phone2>,<location1>,<email1>
    B:<phone1>,<location2>,<email1,email2>

    scala> case class dataclass(c0:String, c1:String)

    scala> val df = rdd.map(x => x.split(":")).map(y => dataclass(y(0), y(1))).toDF

    scala> df.show(false)
    +---+------------------------------------+
    |c0 |c1                                  |
    +---+------------------------------------+
    |A  |<phone1,phone2>,<location1>,<email1>|
    |B  |<phone1>,<location2>,<email1,email2>|
    +---+------------------------------------+


    scala> val df1 = df.withColumn("tempCol",regexp_replace(regexp_replace(col("c1"), "<", ""),">", ""))
                       .withColumn("tempCol", explode(split(col("tempCol"), ",")))
                       .withColumn("out", concat(col("tempCol"), lit(":"), col("c0"), lit(":"), col("c1")))
                       .drop("c0", "c1", "tempCol")

    scala> df1.show(false)
    +------------------------------------------------+
    |out                                             |
    +------------------------------------------------+
    |phone1:A:<phone1,phone2>,<location1>,<email1>   |
    |phone2:A:<phone1,phone2>,<location1>,<email1>   |
    |location1:A:<phone1,phone2>,<location1>,<email1>|
    |email1:A:<phone1,phone2>,<location1>,<email1>   |
    |phone1:B:<phone1>,<location2>,<email1,email2>   |
    |location2:B:<phone1>,<location2>,<email1,email2>|
    |email1:B:<phone1>,<location2>,<email1,email2>   |
    |email2:B:<phone1>,<location2>,<email1,email2>   |
    +------------------------------------------------+

    scala> val rdd2 = df1.rdd.map(_(0))
    scala> rdd2.foreach(println)
    phone1:A:<phone1,phone2>,<location1>,<email1>
    phone2:A:<phone1,phone2>,<location1>,<email1>
    location1:A:<phone1,phone2>,<location1>,<email1>
    email1:A:<phone1,phone2>,<location1>,<email1>
    phone1:B:<phone1>,<location2>,<email1,email2>
    location2:B:<phone1>,<location2>,<email1,email2>
    email1:B:<phone1>,<location2>,<email1,email2>
    email2:B:<phone1>,<location2>,<email1,email2>

scala>rdd.foreach（println）
A:，，
B:，，
scala>案例类dataclass（c0:String，c1:String）
scala>valdf=rdd.map（x=>x.split（“：”）.map（y=>dataclass（y（0），y（1））.toDF
scala>df.show（假）
+---+------------------------------------+
|c0 | c1|
+---+------------------------------------+
|A ||
|B ||
+---+------------------------------------+
scala>val df1=df.withColumn（“tempCol”、regexp\u replace（regexp\u replace（col（“c1”）、“”、“”）
.withColumn（“tempCol”），explode（拆分（col（“tempCol”），“，”））
.带列（“out”、concat（col（“tempCol”）、亮（“：”）列（“c0”）、亮（“：”）列（“c1”））
.drop（“c0”、“c1”、“tempCol”）
scala>df1.show（false）
+------------------------------------------------+
|出去|
+------------------------------------------------+
|电话1:A:|
|电话2:A:|
|地点1:A:|
|电子邮件1:A:|
|电话1:B:|
|地点二:乙:,|
|电子邮件1:B:|
|电邮2:B:|
+------------------------------------------------+
scala>val rdd2=df1.rdd.map（0））
scala>rdd2.foreach（println）
电话1:A:，，
电话2:A:，，
地点1:A:，，
电子邮件1:A:，，
电话1:B:，，
地点二:乙:,，，
电子邮件1:B:，，
电邮2:B:，，