在 pyspark 中解码字符串 URL 列？

Question

我正在使用 python 2.7 和 spark 2.2.0 版。我在 pyspark 中创建了一个数据框，它具有 String 列类型并包含 URLs。

df = spark.createDataFrame([('example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0',)], ['url'])
df.show(1, False)

+-------------------------------------------------------------------------------------------------------+
|url                                                                                                    |
+-------------------------------------------------------------------------------------------------------+
|example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0|
+-------------------------------------------------------------------------------------------------------+

因此，为了解码列中的所有 URL，我尝试使用 urllib 并创建一个像这样使用它的 udf

from pyspark.sql.types import StringType
from pyspark.sql.functions import udf
decode_url = udf(lambda val: (urllib.unquote(val).decode('utf8'), StringType()))

然后在我的专栏上应用 udf 之后，我期待这样的数据

+---------------------------------+
|url                              |
+---------------------------------+
|example.com?title=правовая+защита|
+---------------------------------+

但是我得到这样的错误

UnicodeEncodeError: 'ascii' codec can't encode characters in position 18-33: ordinal not in range(128)

    at org.apache.spark.api.python.PythonRunner$$anon.read(PythonRDD.scala:193)
    at org.apache.spark.api.python.PythonRunner$$anon.<init>(PythonRDD.scala:234)
    at org.apache.spark.api.python.PythonRunner.compute(PythonRDD.scala:152)
    at org.apache.spark.sql.execution.python.BatchEvalPythonExec$$anonfun$doExecute.apply(BatchEvalPythonExec.scala:144)
    at org.apache.spark.sql.execution.python.BatchEvalPythonExec$$anonfun$doExecute.apply(BatchEvalPythonExec.scala:87)
    at org.apache.spark.rdd.RDD$$anonfun$mapPartitions$$anonfun$apply.apply(RDD.scala:797)
    at org.apache.spark.rdd.RDD$$anonfun$mapPartitions$$anonfun$apply.apply(RDD.scala:797)
    at org.apache.spark.rdd.MapPartitionsRDD.compute(MapPartitionsRDD.scala:38)
    at org.apache.spark.rdd.RDD.computeOrReadCheckpoint(RDD.scala:323)
    at org.apache.spark.rdd.RDD.iterator(RDD.scala:287)
    at org.apache.spark.rdd.MapPartitionsRDD.compute(MapPartitionsRDD.scala:38)
    at org.apache.spark.rdd.RDD.computeOrReadCheckpoint(RDD.scala:323)
    at org.apache.spark.rdd.RDD.iterator(RDD.scala:287)
    at org.apache.spark.rdd.MapPartitionsRDD.compute(MapPartitionsRDD.scala:38)
    at org.apache.spark.rdd.RDD.computeOrReadCheckpoint(RDD.scala:323)
    at org.apache.spark.rdd.RDD.iterator(RDD.scala:287)
    at org.apache.spark.scheduler.ResultTask.runTask(ResultTask.scala:87)
    at org.apache.spark.scheduler.Task.run(Task.scala:108)
    at org.apache.spark.executor.Executor$TaskRunner.run(Executor.scala:338)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1149)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:624)
    at java.lang.Thread.run(Thread.java:748)

但是如果我从列中手动取出一些 url 并尝试对其进行解码，它工作正常

import urllib 
url='example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0'
print urllib.unquote(url).decode('utf8')
example.com?title=правовая+защита

Answer 1

似乎在后台进行了一些奇怪的编码。你为什么不自己明确编码？

>>> decode_udf= udf(lambda val: urllib.unquote(val.encode('utf-8')).decode('utf-8'), StringType())
>>> df.withColumn('decoded_url', decode_udf('url')).show(truncate=False)
+-------------------------------------------------------------------------------------------------------+---------------------------------+
|url                                                                                                    |decoded_url                      |
+-------------------------------------------------------------------------------------------------------+---------------------------------+
|example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0|example.com?title=правовая+защита|
+-------------------------------------------------------------------------------------------------------+---------------------------------+
``

Answer 2

如果你想在 spark sql 中这样做，你可以这样做：

注意：编码字段名称为 dctr

输入： im_segments%3Debejz4gv%2Ck1GmZLwg%2C8zY92P4g%2Cka6ee4eb%2CgPKlZXXb%2CqkVvpGk9%2Cky1ee4Dk%2CgvqKoW0b%2CgO5l6Zrk%2CgO5lGpdk%2CxkD6AYgm%2CgO5rENWk%2Cg7VrxvDb

预期输出： im_segments=ebejz4gv,k1GmZLwg,8zY92P4g,ka6ee4eb,gPKlZXXb,qkVvpGk9,ky1ee4Dk,gvqKoW0b,gO5l6Zrk,gO5lGpdk,xkD6AYgm,gO5rENWk,g7VrxvDb

答案：

select distinct reflect('java.net.URLDecoder','decode', dctr , 'utf-8') as dctr from table

在 pyspark 中解码字符串 URL 列？

Decoding a String URL column in pyspark?

python

urllib

python-2.7

pyspark

pyspark-sql