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将pyspark字符串转换为日期格式

[英]Convert pyspark string to date format

I have a date pyspark dataframe with a string column in the format of MM-dd-yyyy and I am attempting to convert this into a date column.我有一个带有MM-dd-yyyy格式的字符串列的日期 pyspark 数据框,我正在尝试将其转换为日期列。

I tried:我试过:

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

and I get a string of nulls.我得到一串空值。 Can anyone help?任何人都可以帮忙吗?

Update (1/10/2018):更新(1/10/2018):

For Spark 2.2+ the best way to do this is probably using the to_date or to_timestamp functions, which both support the format argument.对于 Spark 2.2+,最好的方法可能是使用to_dateto_timestamp函数,它们都支持format参数。 From the docs:从文档:

>>> 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))]

Original Answer (for Spark < 2.2)原始答案(对于 Spark < 2.2)

It is possible (preferrable?) to do this without a udf:在没有 udf 的情况下可以(最好?)做到这一点:

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|
#+----------+-------------------+
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:这是输出:

+----------+----------+----------+----------+
|     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)

The strptime() approach does not work for me. strptime() 方法对我不起作用。 I get another cleaner solution, using cast:我得到了另一个更清洁的解决方案,使用演员:

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 |

In the accepted answer's update you don't see the example for the to_date function, so another solution using it would be:在接受的答案的更新中,您没有看到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')))

possibly not so many answers so thinking to share my code which can help someone可能没有那么多答案,所以想分享我可以帮助某人的代码

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()

output输出

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

the above code to convert to date if you want to convert datetime then use to_timestamp.上面的代码转换为日期如果你想转换日期时间然后使用to_timestamp。 let me know if you have any doubt.如果您有任何疑问,请告诉我。

Try this:试试这个:

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|  
+-------------------+  

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