Spark DataFrame TimestampType - 如何从字段中获取年,月,日值？

Question

我有Spark DataFrame,带有(5)个顶行,如下所示:

[Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=1, value=638.55),
 Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=2, value=638.55),
 Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=3, value=638.55),
 Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=4, value=638.55),
 Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=5, value=638.55)]

它的架构定义为:

elevDF.printSchema()

root
 |-- date: timestamp (nullable = true)
 |-- hour: long (nullable = true)
 |-- value: double (nullable = true)

如何从"日期"字段中获取年,月,日值？

Answer 1

您可以pyspark.sql.functions.year像使用任何其他RDD一样使用simple :

import datetime
from pyspark.sql.functions import year, month, dayofmonth

elevDF = sc.parallelize([
    (datetime.datetime(1984, 1, 1, 0, 0), 1, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 2, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 3, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 4, 638.55),
    (datetime.datetime(1984, 1, 1, 0, 0), 5, 638.55)
]).toDF(["date", "hour", "value"])

elevDF.select(
    year("date").alias('year'), 
    month("date").alias('month'), 
    dayofmonth("date").alias('day')
).show()
# +----+-----+---+
# |year|month|day|
# +----+-----+---+
# |1984|    1|  1|
# |1984|    1|  1|
# |1984|    1|  1|
# |1984|    1|  1|
# |1984|    1|  1|
# +----+-----+---+

结果是:

elevDF = sqlContext.createDataFrame(sc.parallelize([
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=1, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=2, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=3, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=4, value=638.55),
        Row(date=datetime.datetime(1984, 1, 1, 0, 0), hour=5, value=638.55)]))

(elevDF
 .map(lambda (date, hour, value): (date.year, date.month, date.day))
 .collect())

顺便说一下:pyspark.sql.functions.month无论如何要存储一个小时,所以单独保存它似乎是浪费记忆.

从Spark 1.5开始,您可以使用许多日期处理功能

[(1984, 1, 1), (1984, 1, 1), (1984, 1, 1), (1984, 1, 1), (1984, 1, 1)]

Answer 2

您可以使用的功能pyspark.sql.functions:像功能year,month,etc

请参考此处:https://spark.apache.org/docs/latest/api/python/pyspark.sql.html#pyspark.sql.DataFrame

from pyspark.sql.functions import *

newdf = elevDF.select(year(elevDF.date).alias('dt_year'), month(elevDF.date).alias('dt_month'), dayofmonth(elevDF.date).alias('dt_day'), dayofyear(elevDF.date).alias('dt_dayofy'), hour(elevDF.date).alias('dt_hour'), minute(elevDF.date).alias('dt_min'), weekofyear(elevDF.date).alias('dt_week_no'), unix_timestamp(elevDF.date).alias('dt_int'))

newdf.show()


+-------+--------+------+---------+-------+------+----------+----------+
|dt_year|dt_month|dt_day|dt_dayofy|dt_hour|dt_min|dt_week_no|    dt_int|
+-------+--------+------+---------+-------+------+----------+----------+
|   2015|       9|     6|      249|      0|     0|        36|1441497601|
|   2015|       9|     6|      249|      0|     0|        36|1441497601|
|   2015|       9|     6|      249|      0|     0|        36|1441497603|
|   2015|       9|     6|      249|      0|     1|        36|1441497694|
|   2015|       9|     6|      249|      0|    20|        36|1441498808|
|   2015|       9|     6|      249|      0|    20|        36|1441498811|
|   2015|       9|     6|      249|      0|    20|        36|1441498815|