在Spark Dataframe中将一个列转换为多个列

Question

在Spark Dataframe中将一个列转换为多个列

scalaapache-sparkdataframehadoopapache-spark-sql

5

我有一个大的数据框（大约1.2GB），其结构如下：

+---------+--------------+------------------------------------------------------------------------------------------------------+
| country |  date_data   |                                                 text                                                 |
+---------+--------------+------------------------------------------------------------------------------------------------------+
| "EEUU"  | "2016-10-03" | "T_D: QQWE\nT_NAME: name_1\nT_IN: ind_1\nT_C: c1ws12\nT_ADD: Sec_1_P\n ...........\nT_R: 45ee"       |
| "EEUU"  | "2016-10-03" | "T_D: QQAA\nT_NAME: name_2\nT_IN: ind_2\nT_C: c1ws12\nT_ADD: Sec_1_P\n ...........\nT_R: 46ee"       |
| .       | .            | .                                                                                                    |
| .       | .            | .                                                                                                    |
| "EEUU"  | "2016-10-03" | "T_D: QQWE\nT_NAME: name_300000\nT_IN: ind_65\nT_C: c1ws12\nT_ADD: Sec_1_P\n ...........\nT_R: 47aa" |
+---------+--------------+------------------------------------------------------------------------------------------------------+

行数为300,000，"text"字段是一个大约5000个字符的字符串。

我想要将字段“text”分解成以下新字段：

+---------+------------+------+-------------+--------+--------+---------+--------+------+
| country | date_data  | t_d  |   t_name    |  t_in  |  t_c   |  t_add  | ...... | t_r  |
+---------+------------+------+-------------+--------+--------+---------+--------+------+
| EEUU    | 2016-10-03 | QQWE | name_1      | ind_1  | c1ws12 | Sec_1_P | ...... | 45ee |
| EEUU    | 2016-10-03 | QQAA | name_2      | ind_2  | c1ws12 | Sec_1_P | ...... | 45ee |
| .       | .          | .    | .           | .      | .      | .       | .      |      |
| .       | .          | .    | .           | .      | .      | .       | .      |      |
| .       | .          | .    | .           | .      | .      | .       | .      |      |
| EEUU    | 2016-10-03 | QQWE | name_300000 | ind_65 | c1ws12 | Sec_1_P | ...... | 47aa |
+---------+------------+------+-------------+--------+--------+---------+--------+------+

目前，我正在使用正则表达式来解决这个问题。首先，我编写正则表达式并创建一个从文本中提取单个字段的函数（总共有90个正则表达式）：

val D_text = "((?<=T_D: ).*?(?=\\\\n))".r
val NAME_text = "((?<=nT_NAME: ).*?(?=\\\\n))".r
val IN_text = "((?<=T_IN: ).*?(?=\\\\n))".r
val C_text = "((?<=T_C: ).*?(?=\\\\n))".r
val ADD_text = "((?<=T_ADD: ).*?(?=\\\\n))".r
        .
        .
        .
        .
val R_text = "((?<=T_R: ).*?(?=\\\\n))".r   

//UDF function:
 def getFirst(pattern2: scala.util.matching.Regex) = udf(
          (url: String) => pattern2.findFirstIn(url) match { 
              case Some(texst_new) => texst_new
              case None => "NULL"
              case null => "NULL"
          }
   )

然后，我使用正则表达式将函数应用于文本中的每个新字段，创建一个名为tbl_separate_fields的新数据框。

val tbl_separate_fields = hiveDF.select(
          hiveDF("country"),
          hiveDF("date_data"),   
          getFirst(D_text)(hiveDF("texst")).alias("t_d"),
          getFirst(NAME_text)(hiveDF("texst")).alias("t_name"),
          getFirst(IN_text)(hiveDF("texst")).alias("t_in"),
          getFirst(C_text)(hiveDF("texst")).alias("t_c"),
          getFirst(ADD_text)(hiveDF("texst")).alias("t_add"),
                            .
                            .
                            .
                            .

        getFirst(R_text)(hiveDF("texst")).alias("t_r") 

        )

最后，我将这个数据框插入到Hive表中：

tbl_separate_fields.registerTempTable("tbl_separate_fields") 
hiveContext.sql("INSERT INTO TABLE TABLE_INSERT PARTITION (date_data)  SELECT * FROM tbl_separate_fields")

这个解决方案对整个数据帧的持续时间为1小时，因此我希望优化并减少执行时间。有什么解决方案吗？

我们正在使用Hadoop 2.7.1和Apache-Spark 1.5.1。Spark的配置如下：

val conf = new SparkConf().set("spark.storage.memoryFraction", "0.1")
val sc = new SparkContext(conf)
val hiveContext = new org.apache.spark.sql.hive.HiveContext(sc)

提前感谢。

编辑数据：

+---------+--------------+------------------------------------------------------------------------------------------------------+
| 国家    | 日期         |                                                 文本                                                 |
+---------+--------------+------------------------------------------------------------------------------------------------------+
| "美国"  | "2016-10-03" | "T_D: QQWE\nT_NAME: name_1\nT_IN: ind_1\nT_C: c1ws12\nT_ADD: Sec_1_P\n ...........\nT_R: 45ee"       |
| "美国"  | "2016-10-03" | "T_NAME: name_2\nT_D: QQAA\nT_IN: ind_2\nT_C: c1ws12 ...........\nT_R: 46ee"                         |
| .       | .            | .                                                                                                    |
| .       | .            | .                                                                                                    |
| "美国"  | "2016-10-03" | "T_NAME: name_300000\nT_ADD: Sec_1_P\nT_IN: ind_65\nT_C: c1ws12\n ...........\nT_R: 47aa"            |
+---------+--------------+------------------------------------------------------------------------------------------------------+

- E.Aarón

1个回答

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- Tzach Zohar · Accepted Answer

在这种情况下使用正则表达式既慢又不稳定。如果您知道所有记录具有相同的结构，即所有“文本”值具有相同的“部分”数量和顺序，则以下代码将适用（对于任何列数），主要利用org.apache.spark.sql.functions中的split函数：

import org.apache.spark.sql.functions._

// first - split "text" column values into Arrays
val textAsArray: DataFrame = inputDF
  .withColumn("as_array", split(col("text"), "\n"))
  .drop("text")
  .cache()

// get a sample (first row) to get column names, can be skipped if you want to hard-code them:
val sampleText = textAsArray.first().getAs[mutable.WrappedArray[String]]("as_array").toArray
val columnNames: Array[(String, Int)] = sampleText.map(_.split(": ")(0)).zipWithIndex

// add Column per columnName with the right value and drop the no-longer-needed as_array column
val withValueColumns: DataFrame = columnNames.foldLeft(textAsArray) {
  case (df, (colName, index)) => df.withColumn(colName, split(col("as_array").getItem(index), ": ").getItem(1))
}.drop("as_array")

withValueColumns.show()
// for the sample data I created, 
// with just 4 "parts" in "text" column, this prints:
// +-------+----------+----+------+-----+------+
// |country| date_data| T_D|T_NAME| T_IN|   T_C|
// +-------+----------+----+------+-----+------+
// |   EEUU|2016-10-03|QQWE|name_1|ind_1|c1ws12|
// |   EEUU|2016-10-03|QQAA|name_2|ind_2|c1ws12|
// +-------+----------+----+------+-----+------+

或者，如果上述假设不成立，你可以使用一个UDF将文本列转换为Map，然后对所需列的硬编码列表执行类似的reduceLeft操作:

import sqlContext.implicits._

// sample data: not the same order, not all records have all columns:
val inputDF: DataFrame = sc.parallelize(Seq(
  ("EEUU", "2016-10-03", "T_D: QQWE\nT_NAME: name_1\nT_IN: ind_1\nT_C: c1ws12"),
  ("EEUU", "2016-10-03", "T_D: QQAA\nT_IN: ind_2\nT_NAME: name_2")
)).toDF("country", "date_data", "text")

// hard-coded list of expected column names:
val columnNames: Seq[String] = Seq("T_D", "T_NAME", "T_IN", "T_C")

// UDF to convert text into key-value map
val asMap = udf[Map[String, String], String] { s =>
  s.split("\n").map(_.split(": ")).map { case Array(k, v) => k -> v }.toMap
}


val textAsMap = inputDF.withColumn("textAsMap", asMap(col("text"))).drop("text")

// for each column name - lookup the value in the map
val withValueColumns: DataFrame = columnNames.foldLeft(textAsMap) {
  case (df, colName) => df.withColumn(colName, col("textAsMap").getItem(colName))
}.drop("textAsMap")

withValueColumns.show()
// prints:
// +-------+----------+----+------+-----+------+
// |country| date_data| T_D|T_NAME| T_IN|   T_C|
// +-------+----------+----+------+-----+------+
// |   EEUU|2016-10-03|QQWE|name_1|ind_1|c1ws12|
// |   EEUU|2016-10-03|QQAA|name_2|ind_2|  null|
// +-------+----------+----+------+-----+------+