如何在pyspark中拆除CLOB?

本文介绍了如何在pyspark中拆除CLOB?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我从Oracle压缩了数据，并且该表的列包含CLOB DataType，我将其设置为String以获取HDFS中的数据.现在，我必须拆除CLOB数据，并在Hive中为此创建一个单独的表.

I sqooped data from Oracle and the table had a column with CLOB DataType, I made it String to get the data in HDFS. Now I have to dismantle the CLOB Data and create a separate table for that in Hive.

我有txt格式的HDFS文件.我可以隔离CLOB数据，并希望为CLOB制作DataFrame

I have the HDFS file in txt format. I can segregate the CLOB data and be hoping to make DataFrame for CLOB

CLOB的格式如下:

[name] Bob [Age] 21 [City] London [work] No,
[name] Steve [Age] 51 [City] London [work] Yes,
.....
around a million rows like this

sc.setLogLevel("WARN")
log_txt=sc.textFile("/path/to/data/sample_data.txt")
header = log_txt.first()

log_txt = log_txt.filter(lambda line: line != header)
log_txt.take(10)
  [u'0\\tdog\\t20160906182001\\tgoogle.com', u'1\\tcat\\t20151231120504\\tgoogle.com']

temp_var = log_txt.map(lambda k: k.split("\\t"))

log_df=temp_var.toDF(header.split("\\t"))

log_df = log_df.withColumn("field1Int", log_df["field1"].cast(IntegerType()))
log_df = log_df.withColumn("field3TimeStamp", log_df["field1"].cast(TimestampType()))

log_df.schema
StructType(List(StructField(field1,StringType,true),StructField(field2,StringType,true),StructField(field3,StringType,true),StructField(field4,StringType,true),StructField(field1Int,IntegerType,true),StructField(field3TimeStamp,TimestampType,true)))

这就是我创建DataFrame的方式.

This is how I have created DataFrame.

我需要您的帮助来确定如何拆卸CLOB(采用字符串数据类型).并在其上创建一个表格.

拆卸后，我希望表格具有以下列:

After dismantling, I expect the Table to have following Columns like:

+---------+---------------+----------+-----+
|Name     |Age            | City     | Work|
+---------+---------------+----------+-----+
|      Bob|           21  |London    | No  |
|    Steve|           51  |London    |Yes  |
+---------+---------------+----------+-----+

任何帮助将不胜感激.

推荐答案

在这里:

import re
from pyspark.sql import Row

rdd = sc.parallelize(["[name] Bob [Age] 21 [City] London [work] No",
                      "[name] Steve [Age] 51 [City] London [work] Yes",
                      "[name] Steve [Age] [City] London [work] Yes"])

def clob_to_table(line):
    m = re.search(r"\[name\](.*)?\[Age\](.*)?\[City\](.*)?\[work\](.*)?", line)
    return Row(name=m.group(1), age=m.group(2), city=m.group(3), work=m.group(4))

rdd = rdd.map(clob_to_table)

df = spark.createDataFrame(rdd)
df.show()

+----+--------+-------+----+
| age|    city|   name|work|
+----+--------+-------+----+
| 21 | London |   Bob |  No|
| 51 | London | Steve | Yes|
|    | London | Steve | Yes
+----+--------+-------+----+

这篇关于如何在pyspark中拆除CLOB?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

Clob