How to create a Spark dataframe with timestamp?-CodePudding

How can I create this Spark dataframe with timestamp data type in one step using python? Here is how I do it in two steps. Using spark 3.1.2

from pyspark.sql.functions import *
from pyspark.sql.types import *

schema_sdf = StructType([ 
    StructField("ts", TimestampType(), True),
    StructField("myColumn", LongType(), True),
    ])

sdf = spark.createDataFrame( ( [ ( to_timestamp(lit("2022-06-29 12:01:19.000")), 0 ) ] ), schema=schema_sdf )

CodePudding user response：

PySpark does not automatically interpret timestamp values from strings. I mostly use the following syntax to create the df and then to cast column type to timestamp:

from pyspark.sql import functions as F

sdf = spark.createDataFrame([("2022-06-29 12:01:19.000", 0 )], ["ts", "myColumn"])
sdf = sdf.withColumn("ts", F.col("ts").cast("timestamp"))

sdf.printSchema()
# root
#  |-- ts: timestamp (nullable = true)
#  |-- myColumn: long (nullable = true)

Long format was automatically inferred, but for timestamp we needed a cast.

On the other hand, even without casting, you are able to use functions which need timestamp as input:

sdf = spark.createDataFrame([("2022-06-29 12:01:19.000", 0 )], ["ts", "myColumn"])
sdf.printSchema()
# root
#  |-- ts: string (nullable = true)
#  |-- myColumn: long (nullable = true)

sdf.selectExpr("extract(year from ts)").show()
#  --------------------- 
# |extract(year FROM ts)|
#  --------------------- 
# |                 2022|
#  ---------------------