将火花字符串转换为日期格式

我有一个日期火花数据框与字符串列的格式为 MM-dd-yyyy，我正在尝试转换成一个日期列。

我试过:

df.select(to_date(df.STRING_COLUMN).alias('new_date')).show()

我得到了一串空值，有人能帮忙吗？

378648 次浏览
小开 
from datetime import datetime
from pyspark.sql.functions import col, udf
from pyspark.sql.types import DateType






# Creation of a dummy dataframe:
df1 = sqlContext.createDataFrame([("11/25/1991","11/24/1991","11/30/1991"),
("11/25/1391","11/24/1992","11/30/1992")], schema=['first', 'second', 'third'])


# Setting an user define function:
# This function converts the string cell into a date:
func =  udf (lambda x: datetime.strptime(x, '%m/%d/%Y'), DateType())


df = df1.withColumn('test', func(col('first')))


df.show()


df.printSchema()

Here is the output:

+----------+----------+----------+----------+
|     first|    second|     third|      test|
+----------+----------+----------+----------+
|11/25/1991|11/24/1991|11/30/1991|1991-01-25|
|11/25/1391|11/24/1992|11/30/1992|1391-01-17|
+----------+----------+----------+----------+


root
|-- first: string (nullable = true)
|-- second: string (nullable = true)
|-- third: string (nullable = true)
|-- test: date (nullable = true)
小开
最佳答案

更新 (1/10/2018) :

对于 Spark 2.2 + ，最好的方法可能是使用 to_date或 to_timestamp函数，它们都支持 format参数。来自文件:

>>> from pyspark.sql.functions import to_timestamp
>>> df = spark.createDataFrame([('1997-02-28 10:30:00',)], ['t'])
>>> df.select(to_timestamp(df.t, 'yyyy-MM-dd HH:mm:ss').alias('dt')).collect()
[Row(dt=datetime.datetime(1997, 2, 28, 10, 30))]

原答 (适用于 Spark < 2.2)

在没有 UDF 的情况下这样做是可能的(更好?) :

from pyspark.sql.functions import unix_timestamp, from_unixtime


df = spark.createDataFrame(
[("11/25/1991",), ("11/24/1991",), ("11/30/1991",)],
['date_str']
)


df2 = df.select(
'date_str',
from_unixtime(unix_timestamp('date_str', 'MM/dd/yyy')).alias('date')
)


print(df2)
#DataFrame[date_str: string, date: timestamp]


df2.show(truncate=False)
#+----------+-------------------+
#|date_str  |date               |
#+----------+-------------------+
#|11/25/1991|1991-11-25 00:00:00|
#|11/24/1991|1991-11-24 00:00:00|
#|11/30/1991|1991-11-30 00:00:00|
#+----------+-------------------+
小开

The strptime() approach does not work for me. I get another cleaner solution, using cast:

from pyspark.sql.types import DateType
spark_df1 = spark_df.withColumn("record_date",spark_df['order_submitted_date'].cast(DateType()))
#below is the result
spark_df1.select('order_submitted_date','record_date').show(10,False)


+---------------------+-----------+
|order_submitted_date |record_date|
+---------------------+-----------+
|2015-08-19 12:54:16.0|2015-08-19 |
|2016-04-14 13:55:50.0|2016-04-14 |
|2013-10-11 18:23:36.0|2013-10-11 |
|2015-08-19 20:18:55.0|2015-08-19 |
|2015-08-20 12:07:40.0|2015-08-20 |
|2013-10-11 21:24:12.0|2013-10-11 |
|2013-10-11 23:29:28.0|2013-10-11 |
|2015-08-20 16:59:35.0|2015-08-20 |
|2015-08-20 17:32:03.0|2015-08-20 |
|2016-04-13 16:56:21.0|2016-04-13 |
小开

Try this:

df = spark.createDataFrame([('2018-07-27 10:30:00',)], ['Date_col'])
df.select(from_unixtime(unix_timestamp(df.Date_col, 'yyyy-MM-dd HH:mm:ss')).alias('dt_col'))
df.show()
+-------------------+
|           Date_col|
+-------------------+
|2018-07-27 10:30:00|
+-------------------+
小开

在已接受的答案的更新中，你没有看到 to_date函数的示例，因此使用它的另一个解决方案是:

from pyspark.sql import functions as F


df = df.withColumn(
'new_date',
F.to_date(
F.unix_timestamp('STRINGCOLUMN', 'MM-dd-yyyy').cast('timestamp')))
小开

可能没有那么多的答案，所以想分享我的代码，可以帮助某人

from pyspark.sql import SparkSession
from pyspark.sql.functions import to_date


spark = SparkSession.builder.appName("Python Spark SQL basic example")\
.config("spark.some.config.option", "some-value").getOrCreate()




df = spark.createDataFrame([('2019-06-22',)], ['t'])
df1 = df.select(to_date(df.t, 'yyyy-MM-dd').alias('dt'))
print df1
print df1.show()

输出

DataFrame[dt: date]
+----------+
|        dt|
+----------+
|2019-06-22|
+----------+

如果您想要转换 datetime，那么使用以上代码转换为 date，然后使用 to _ time 戳。 let me know if you have any doubt.


提交答案