PySpark 数据帧将异常字符串格式转换为时间戳

Question

I am using PySpark through Spark 1.5.0. I have an unusual String format in rows of a column for datetime values. It looks like this:

Row[(datetime='2016_08_21 11_31_08')]

Is there a way to convert this unorthodox yyyy_mm_dd hh_mm_dd format into a Timestamp? Something that can eventually come along the lines of

df = df.withColumn("date_time",df.datetime.astype('Timestamp'))

I had thought that Spark SQL functions like regexp_replace could work, but of course I need to replace _ with - in the date half and _ with : in the time part.

I was thinking I could split the column in 2 using substring and count backward from the end of time. Then do the 'regexp_replace' separately, then concatenate. But this seems to many operations? Is there an easier way?

Answer 1

Spark >= 2.2

from pyspark.sql.functions import to_timestamp

(sc
    .parallelize([Row(dt='2016_08_21 11_31_08')])
    .toDF()
    .withColumn("parsed", to_timestamp("dt", "yyyy_MM_dd HH_mm_ss"))
    .show(1, False))

## +-------------------+-------------------+
## |dt                 |parsed             |
## +-------------------+-------------------+
## |2016_08_21 11_31_08|2016-08-21 11:31:08|
## +-------------------+-------------------+

Spark < 2.2

It is nothing that unix_timestamp cannot handle:

from pyspark.sql import Row
from pyspark.sql.functions import unix_timestamp

(sc
    .parallelize([Row(dt='2016_08_21 11_31_08')])
    .toDF()
    .withColumn("parsed", unix_timestamp("dt", "yyyy_MM_dd HH_mm_ss")
    # For Spark <= 1.5
    # See issues.apache.org/jira/browse/SPARK-11724 
    .cast("double")
    .cast("timestamp"))
    .show(1, False))

## +-------------------+---------------------+
## |dt                 |parsed               |
## +-------------------+---------------------+
## |2016_08_21 11_31_08|2016-08-21 11:31:08.0|
## +-------------------+---------------------+

In both cases the format string should be compatible with Java SimpleDateFormat .

Answer 2

zero323's answer answers the question, but I wanted to add that if your datetime string has a standard format, you should be able to cast it directly into timestamp type:

df.withColumn('datetime', col('datetime_str').cast('timestamp'))

It has the advantage of handling milliseconds , while unix_timestamp only has only second-precision ( to_timestamp works with milliseconds too but requires Spark >= 2.2 as zero323 stated). I tested it on Spark 2.3.0, using the following format: '2016-07-13 14:33:53.979' (with milliseconds, but it also works without them).

Answer 3

我完全同意所选的答案，但是我想将格式设置为 'yyyy_MM_dd HH_mm_ss' 以避免出现诸如 '2019_01_27 16_00_00' -> Note hours > 12 这样的时间戳问题

Answer 4

I add some more code lines from Florent F's answer for better understanding and running the snippet in local machine:

import os, pdb, sys
import pyspark
from pyspark.sql import SparkSession
from pyspark.sql import Row
from pyspark.sql.types import StructType, ArrayType  
from pyspark.sql.types import StringType
from pyspark.sql.functions import col

sc = pyspark.SparkContext('local[*]')
spark = SparkSession.builder.getOrCreate()

# preparing some example data - df1 with String type and df2 with Timestamp type
df1 = sc.parallelize([{"key":"a", "date":"2016-02-01"}, 
    {"key":"b", "date":"2016-02-02"}]).toDF()
df1.show()

df2 = df1.withColumn('datetime', col('date').cast("timestamp"))
df2.show()