Question

我正在创建 time_interval 列，并将其添加到Pyspark中现有的 Data-frame 中。理想情况下，time_interval的格式应为“ HHmm ”，分钟应四舍五入到最接近的15分钟标记（815、830、845、900等）。

我有Spark sql代码，可以为我执行逻辑操作，但是如何获取串联为字符串列的值并将其插入现有的Data-frame中呢？

time_interval = sqlContext.sql("select extract(hour from current_timestamp())||floor(extract(minute from current_timestamp())/15)*15")

time_interval.show()

+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
|concat(CAST(hour(current_timestamp()) AS STRING), CAST((FLOOR((CAST(minute(current_timestamp()) AS DOUBLE) / CAST(15 AS DOUBLE))) * CAST(15 AS BIGINT)) AS STRING))|
+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
|                                                                                                                                                               1045|
+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+

baseDF = sqlContext.sql("select * from test_table")
newBase = baseDF.withColumn("time_interval", lit(str(time_interval)))

newBase.select("time_interval").show()

+--------------------+
|       time_interval|
+--------------------+
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
|DataFrame[concat(...|
+--------------------+
only showing top 20 rows

因此，实际的预期结果应该只是在我正在创建的新列中显示实际的字符串值，而不是数据框中的此串联值。如下所示：

newBase.select("time_interval").show(1)
+-------------+
|time_interval|
+-------------+
|    1045     |                                                                                                                                           
+-------------+

Answer 1

由于time_interval是一种数据帧类型，因此在这种情况下需要 collect 和extract the required value out from dataframe。

尝试这种方式：

newBase = baseDF.withColumn("time_interval", lit(str(time_interval.collect()[0][0])))
newBase.show()

（或）

通过使用 select(expr()) 功能：

newBase = baseDF.select("*",expr("string(extract(hour from current_timestamp())||floor(extract(minute from current_timestamp())/15)*15) AS time_interval"))

如pault所述，使用 selectExpr() 函数：

newBase = baseDF.selectExpr("*","string(extract(hour from current_timestamp())||floor(extract(minute from current_timestamp())/15)*15) AS time_interval")

示例：

>>> from pyspark.sql.functions import *
>>> from pyspark.sql.types import IntegerType
>>> time_interval = spark.sql("select extract(hour from current_timestamp())||floor(extract(minute from current_timestamp())/15)*15")
>>> baseDF=spark.createDataFrame([1,2,3,4],IntegerType())
>>> newBase = baseDF.withColumn("time_interval", lit(str(time_interval.collect()[0][0])))
>>> newBase.show()
+-----+-------------+
|value|time_interval|
+-----+-------------+
|    1|         1245|
|    2|         1245|
|    3|         1245|
|    4|         1245|
+-----+-------------+

如何将一个数据帧的连接值插入Pyspark中的另一个数据帧？

1 个答案: