Question

我在pandas中有一个pyspark数据框。我想将此数据框创建/加载到hive表中。

pd_df = pandas data frame

id                    int64
TEST_TIME             datetime64[ns]
status_time           object
GROUP                 object
test_type             object
dtype: object

    id TEST_TIME            status_time                 GROUP       test_type

0   1 2017-03-12 02:19:51                                       Driver started
1   2 2017-03-12 02:19:53  2017-03-11 18:13:43.577   ALARM      AL_PT2334_L
2   3 2017-03-12 02:19:53  2017-03-11 18:13:43.577   ALARM      AL_Turb_CNet_Ch_A_Fault
3   4 2017-03-12 02:19:53  2017-03-11 18:13:43.577   ALARM      AL_Encl_Fire_Sys_Trouble
4   5 2017-03-12 02:19:54  2017-03-11 18:13:44.611  STATUS      ST_Engine_Turning_Mode

现在我将pandas数据框转换为火花数据框，如下所示。

spark_df = sqlContext.createDataFrame(pd_df)


+---+-------------------+--------------------+------+--------------------+
| id|          TEST_TIME|         status_time| GROUP|           test_type|
+---+-------------------+--------------------+------+--------------------+
|  1|1489285191000000000|                    |      |      Driver started|
|  2|1489285193000000000|2017-03-11 18:13:...| ALARM|         AL_PT2334_L|
|  3|1489285193000000000|2017-03-11 18:13:...| ALARM|AL_Turb_CNet_Ch_A...|
|  4|1489285193000000000|2017-03-11 18:13:...| ALARM|AL_Encl_Fire_Sys_...|
|  5|1489285194000000000|2017-03-11 18:13:...|STATUS|ST_Engine_Turning...|
+---+-------------------+--------------------+------+--------------------+

DataFrame[id: bigint, TEST_TIME: bigint, status_time: string, GROUP: string, test_type: string]

我希望TEST_TIME列成为时间戳列，但我得到的是bigint。

我希望timestamp与pd_df中的spark_df完全相同。

我在将pandas数据框转换为spark数据框

时做了如下操作

spark_df = sqlContext.createDataFrame(pd_df).withColumn("TEST_TIME", (F.unix_timestamp("TEST_TIME") + 28800).cast('timestamp'))

我得到以下错误

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/cloudera/parcels/CDH-5.8.0-1.cdh5.8.0.p0.42/lib/spark/python/pyspark/sql/dataframe.py", line 1314, in withColumn
    return DataFrame(self._jdf.withColumn(colName, col._jc), self.sql_ctx)
  File "/opt/cloudera/parcels/CDH-5.8.0-1.cdh5.8.0.p0.42/lib/spark/python/lib/py4j-0.9-src.zip/py4j/java_gateway.py", line 813, in __call__
  File "/opt/cloudera/parcels/CDH-5.8.0-1.cdh5.8.0.p0.42/lib/spark/python/pyspark/sql/utils.py", line 51, in deco
    raise AnalysisException(s.split(': ', 1)[1], stackTrace)
pyspark.sql.utils.AnalysisException: u"cannot resolve 'unixtimestamp(TEST_TIME,yyyy-MM-dd HH:mm:ss)' due to data type mismatch: argument 1 requires (string or date or timestamp) type, however, 'TEST_TIME' is of bigint type.;"

我如何实现我想要的目标

Answer 1

将datetime64类型的pandas dataframe列转换为python datetime对象，如下所示： pd_df['TEST_TIME'] = pandas.Series(pd_df['TEST_TIME'].dt.to_pydatetime(), dtype=object)

然后像你一样创建火花数据框。

Answer 2

只需将其转换为正确的范围（从纳秒到秒）并投射

df.withColumn(
  "TEST_TIME",
  (F.col("TEST_TIME") / F.pow(F.lit(1000), F.lit(3))).cast('timestamp'))

用于激活数据帧的Pandas将datetime数据类型转换为bigint

2 个答案: