本文介绍了Spark DataFrame TimestampType - 如何从字段中获取年、月、日值?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个带有 take(5) 顶行的 Spark DataFrame,如下所示:
I have Spark DataFrame with take(5) top rows as follows:
[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?
How do I get the Year, Month, Day values from the 'date' field?
推荐答案
从 Spark 1.5 开始,您可以使用许多日期处理函数:
Since Spark 1.5 you can use a number of date processing functions:
pyspark.sql.functions.year
pyspark.sql.functions.month
pyspark.sql.functions.dayofmonth
pyspark.sql.functions.dayofweek()
pyspark.sql.functions.dayofyear
pyspark.sql.functions.weekofyear()
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|
# +----+-----+---+
你可以像使用任何其他 RDD 一样使用简单的 map
:
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())
结果是:
[(1984, 1, 1), (1984, 1, 1), (1984, 1, 1), (1984, 1, 1), (1984, 1, 1)]
顺便说一句:datetime.datetime
无论如何都会存储一个小时,所以单独保存它似乎是在浪费内存.
Btw: datetime.datetime
stores an hour anyway so keeping it separately seems to be a waste of memory.
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