Question

我是PySpark的新手。我用pandas拉了一个csv文件。并使用registerTempTable函数创建了临时表。

from pyspark.sql import SQLContext
from pyspark.sql import Row
import pandas as pd
sqlc = SQLContext(sc)

aa1 = pd.read_csv("D:\mck1.csv")

aa2 = sqlc.createDataFrame(aa1)

aa2.show()

+--------+-------+----------+------------+---------+------------+-------------------+
|    City|     id|First_Name|Phone_Number|new_date|new      code|           New_date|
+--------+-------+----------+------------+---------+------------+-------------------+
|KOLKATTA|9000007|       AAA|  1111119411| 20080714|          13|2016-08-16 00:00:00|
|KOLKATTA|9000007|       BBB|  1111119421| 20080714|          13|2016-08-06 00:00:00|
|KOLKATTA|9000007|       CCC|  1111119461| 20080714|          13|2016-08-13 00:00:00|
|KOLKATTA|9000007|       DDD|  1111119471| 20080714|          13|2016-08-27 00:00:00|
|KOLKATTA|9000007|       EEE|  1111119491| 20080714|          13|2016-08-15 00:00:00|
|KOLKATTA|9111147|       FFF|  1111119401| 20080714|          13|2016-08-24 00:00:00|
|KOLKATTA|9585458|   FORMULA|  1111110112| 19990930|          13|2016-08-16 00:00:00|
|KOLKATTA|9569878|   APPLEII|  1111110132| 19990930|          13|2016-08-06 00:00:00|

aa3 = aa2.registerTempTable("mytable1")

sqlc.sql(""" select right(phone_number,4) from mytable1 """).show()

现在我尝试使用右侧（phone_number，4）使用电话号码右侧拉出最后四个字符并面对跟随错误

---------------------------------------------------------------------------
Py4JJavaError                             Traceback (most recent call last)
<ipython-input-18-07f08e3d0a8f> in <module>()
----> 1 sqlc.sql(""" select right(Phone_number,4) from mytable1 """).show()

C:\spark-1.4.1-bin-hadoop2.6\python\pyspark\sql\context.pyc in sql(self, sqlQuery)
    500         [Row(f1=1, f2=u'row1'), Row(f1=2, f2=u'row2'), Row(f1=3, f2=u'row3')]
    501         """
--> 502         return DataFrame(self._ssql_ctx.sql(sqlQuery), self)
    503 
    504     @since(1.0)

C:\spark-1.4.1-bin-hadoop2.6\python\lib\py4j-0.8.2.1-src.zip\py4j\java_gateway.py in __call__(self, *args)
    536         answer = self.gateway_client.send_command(command)
    537         return_value = get_return_value(answer, self.gateway_client,
--> 538                 self.target_id, self.name)
    539 
    540         for temp_arg in temp_args:

C:\spark-1.4.1-bin-hadoop2.6\python\lib\py4j-0.8.2.1-src.zip\py4j\protocol.py in get_return_value(answer, gateway_client, target_id, name)
    298                 raise Py4JJavaError(
    299                     'An error occurred while calling {0}{1}{2}.\n'.
--> 300                     format(target_id, '.', name), value)
    301             else:
    302                 raise Py4JError(

Py4JJavaError: An error occurred while calling o55.sql.
: java.lang.RuntimeException: [1.9] failure: ``union'' expected but `right' found

 select right(Phone_number,4) from mytable1 
        ^
    at scala.sys.package$.error(package.scala:27)
    at org.apache.spark.sql.catalyst.AbstractSparkSQLParser.parse(AbstractSparkSQLParser.scala:36)
    at org.apache.spark.sql.catalyst.DefaultParserDialect.parse(ParserDialect.scala:67)
    at org.apache.spark.sql.SQLContext$$anonfun$2.apply(SQLContext.scala:145)

为什么pyspark不支持RIGHT和LEFT功能？如何为列选择四个字符？

Answer 1

查看documentation，您是否尝试过子串函数？

pyspark.sql.functions.substring(str, pos, len)[source]

编辑

根据你的评论，你可以得到最后四个：

from pyspark.sql.functions import substring

df = sqlContext.createDataFrame([('abcdefg',)], ['s',])
df.select(substring(df.s, -4, 4).alias('s')).collect()

Answer 2

而不是正确，请尝试使用rpad：

sqlc.sql(""" select rpad(phone_number, 4, phone_number) from mytable1 """).show()

PySpark SQL中的LEFT和RIGHT函数

2 个答案: