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Creating a DataFrame from RDD while specifying DateType() in schema



2019 Community Moderator ElectionHow to remove items from a list while iterating?Delete column from pandas DataFrame by column nameSelect rows from a DataFrame based on values in a column in pandasGet list from pandas DataFrame column headersDifference between DataFrame, Dataset, and RDD in SparkConvert json string to modified RDDSpark SQL(v2.0) UDAF in Scala returns empty stringEnforcing a schema on RDD while converting them to DataFrameHow do I apply schema with nullable = false to json readingpyspark load csv file into dataframe using a schema










1















I am creating a DataFrame from RDD and one of the value is a date. I don't know how to specify DateType() in schema.



Let me illustrate the problem at hand -



One way we can load the date into the DataFrame is by first specifying it as string and converting it to proper date using to_date() function.



from pyspark.sql.types import Row, StructType, StructField, StringType, IntegerType, DateType
from pyspark.sql.functions import col, to_date
values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
rdd= values.map(lambda t: Row(A=t[0],date=t[1]))

# Importing date as String in Schema
schema = StructType([StructField('A', IntegerType(), True), StructField('date', StringType(), True)])
df = sqlContext.createDataFrame(rdd, schema)

# Finally converting the string into date using to_date() function.
df = df.withColumn('date',to_date(col('date'), 'yyyy-MM-dd'))
df.show()
+---+----------+
| A| date|
+---+----------+
| 3|2012-02-02|
| 5|2018-08-08|
+---+----------+

df.printSchema()
root
|-- A: integer (nullable = true)
|-- date: date (nullable = true)


Is there a way, where we could use DateType() in the schema and avoid having to convert string to date explicitly?



Something like this -



values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
rdd= values.map(lambda t: Row(A=t[0],date=t[1]))
# Somewhere we would need to specify date format 'yyyy-MM-dd' too, don't know where though.
schema = StructType([StructField('A', DateType(), True), StructField('date', DateType(), True)])


UPDATE: As suggested by @user10465355, following code works -



import datetime
schema = StructType([
StructField('A', IntegerType(), True),
StructField('date', DateType(), True)
])
rdd= values.map(lambda t: Row(A=t[0],date=datetime.datetime.strptime(t[1], "%Y-%m-%d")))
df = sqlContext.createDataFrame(rdd, schema)
df.show()
+---+----------+
| A| date|
+---+----------+
| 3|2012-02-02|
| 5|2018-08-08|
+---+----------+
df.printSchema()
root
|-- A: integer (nullable = true)
|-- date: date (nullable = true)









share|improve this question




























    1















    I am creating a DataFrame from RDD and one of the value is a date. I don't know how to specify DateType() in schema.



    Let me illustrate the problem at hand -



    One way we can load the date into the DataFrame is by first specifying it as string and converting it to proper date using to_date() function.



    from pyspark.sql.types import Row, StructType, StructField, StringType, IntegerType, DateType
    from pyspark.sql.functions import col, to_date
    values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
    rdd= values.map(lambda t: Row(A=t[0],date=t[1]))

    # Importing date as String in Schema
    schema = StructType([StructField('A', IntegerType(), True), StructField('date', StringType(), True)])
    df = sqlContext.createDataFrame(rdd, schema)

    # Finally converting the string into date using to_date() function.
    df = df.withColumn('date',to_date(col('date'), 'yyyy-MM-dd'))
    df.show()
    +---+----------+
    | A| date|
    +---+----------+
    | 3|2012-02-02|
    | 5|2018-08-08|
    +---+----------+

    df.printSchema()
    root
    |-- A: integer (nullable = true)
    |-- date: date (nullable = true)


    Is there a way, where we could use DateType() in the schema and avoid having to convert string to date explicitly?



    Something like this -



    values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
    rdd= values.map(lambda t: Row(A=t[0],date=t[1]))
    # Somewhere we would need to specify date format 'yyyy-MM-dd' too, don't know where though.
    schema = StructType([StructField('A', DateType(), True), StructField('date', DateType(), True)])


    UPDATE: As suggested by @user10465355, following code works -



    import datetime
    schema = StructType([
    StructField('A', IntegerType(), True),
    StructField('date', DateType(), True)
    ])
    rdd= values.map(lambda t: Row(A=t[0],date=datetime.datetime.strptime(t[1], "%Y-%m-%d")))
    df = sqlContext.createDataFrame(rdd, schema)
    df.show()
    +---+----------+
    | A| date|
    +---+----------+
    | 3|2012-02-02|
    | 5|2018-08-08|
    +---+----------+
    df.printSchema()
    root
    |-- A: integer (nullable = true)
    |-- date: date (nullable = true)









    share|improve this question


























      1












      1








      1








      I am creating a DataFrame from RDD and one of the value is a date. I don't know how to specify DateType() in schema.



      Let me illustrate the problem at hand -



      One way we can load the date into the DataFrame is by first specifying it as string and converting it to proper date using to_date() function.



      from pyspark.sql.types import Row, StructType, StructField, StringType, IntegerType, DateType
      from pyspark.sql.functions import col, to_date
      values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
      rdd= values.map(lambda t: Row(A=t[0],date=t[1]))

      # Importing date as String in Schema
      schema = StructType([StructField('A', IntegerType(), True), StructField('date', StringType(), True)])
      df = sqlContext.createDataFrame(rdd, schema)

      # Finally converting the string into date using to_date() function.
      df = df.withColumn('date',to_date(col('date'), 'yyyy-MM-dd'))
      df.show()
      +---+----------+
      | A| date|
      +---+----------+
      | 3|2012-02-02|
      | 5|2018-08-08|
      +---+----------+

      df.printSchema()
      root
      |-- A: integer (nullable = true)
      |-- date: date (nullable = true)


      Is there a way, where we could use DateType() in the schema and avoid having to convert string to date explicitly?



      Something like this -



      values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
      rdd= values.map(lambda t: Row(A=t[0],date=t[1]))
      # Somewhere we would need to specify date format 'yyyy-MM-dd' too, don't know where though.
      schema = StructType([StructField('A', DateType(), True), StructField('date', DateType(), True)])


      UPDATE: As suggested by @user10465355, following code works -



      import datetime
      schema = StructType([
      StructField('A', IntegerType(), True),
      StructField('date', DateType(), True)
      ])
      rdd= values.map(lambda t: Row(A=t[0],date=datetime.datetime.strptime(t[1], "%Y-%m-%d")))
      df = sqlContext.createDataFrame(rdd, schema)
      df.show()
      +---+----------+
      | A| date|
      +---+----------+
      | 3|2012-02-02|
      | 5|2018-08-08|
      +---+----------+
      df.printSchema()
      root
      |-- A: integer (nullable = true)
      |-- date: date (nullable = true)









      share|improve this question
















      I am creating a DataFrame from RDD and one of the value is a date. I don't know how to specify DateType() in schema.



      Let me illustrate the problem at hand -



      One way we can load the date into the DataFrame is by first specifying it as string and converting it to proper date using to_date() function.



      from pyspark.sql.types import Row, StructType, StructField, StringType, IntegerType, DateType
      from pyspark.sql.functions import col, to_date
      values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
      rdd= values.map(lambda t: Row(A=t[0],date=t[1]))

      # Importing date as String in Schema
      schema = StructType([StructField('A', IntegerType(), True), StructField('date', StringType(), True)])
      df = sqlContext.createDataFrame(rdd, schema)

      # Finally converting the string into date using to_date() function.
      df = df.withColumn('date',to_date(col('date'), 'yyyy-MM-dd'))
      df.show()
      +---+----------+
      | A| date|
      +---+----------+
      | 3|2012-02-02|
      | 5|2018-08-08|
      +---+----------+

      df.printSchema()
      root
      |-- A: integer (nullable = true)
      |-- date: date (nullable = true)


      Is there a way, where we could use DateType() in the schema and avoid having to convert string to date explicitly?



      Something like this -



      values=sc.parallelize([(3,'2012-02-02'),(5,'2018-08-08')])
      rdd= values.map(lambda t: Row(A=t[0],date=t[1]))
      # Somewhere we would need to specify date format 'yyyy-MM-dd' too, don't know where though.
      schema = StructType([StructField('A', DateType(), True), StructField('date', DateType(), True)])


      UPDATE: As suggested by @user10465355, following code works -



      import datetime
      schema = StructType([
      StructField('A', IntegerType(), True),
      StructField('date', DateType(), True)
      ])
      rdd= values.map(lambda t: Row(A=t[0],date=datetime.datetime.strptime(t[1], "%Y-%m-%d")))
      df = sqlContext.createDataFrame(rdd, schema)
      df.show()
      +---+----------+
      | A| date|
      +---+----------+
      | 3|2012-02-02|
      | 5|2018-08-08|
      +---+----------+
      df.printSchema()
      root
      |-- A: integer (nullable = true)
      |-- date: date (nullable = true)






      python apache-spark pyspark






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 7 at 13:05







      cph_sto

















      asked Mar 7 at 7:51









      cph_stocph_sto

      2,3282422




      2,3282422






















          1 Answer
          1






          active

          oldest

          votes


















          3














          Long story short, schema used with RDD of external object is not intended to be used that way - declared types should reflect the actual state of the data, not the desired one.



          In other words to allow:



          schema = StructType([
          StructField('A', IntegerType(), True),
          StructField('date', DateType(), True)
          ])


          the data corresponding to date field should use datetime.date. So for example with your RDD[Tuple[int, str]]:



          import datetime

          spark.createDataFrame(
          # Since values from the question are just two element tuples
          # we can use mapValues to transform the "value"
          # but in general case you'll need map
          values.mapValues(datetime.date.fromisoformat),
          schema
          )


          The closest you can get to desired behavior is to convert data (RDD[Row]) with JSON reader, using dicts



          from pyspark.sql import Row

          spark.read.schema(schema).json(rdd.map(Row.asDict))


          or better explicit JSON dumps:



          import json
          spark.read.schema(schema).json(rdd.map(Row.asDict).map(json.dumps))


          but that's of course much more expensive than explicit casting, which BTW, is easy to automate in simple cases like the one you describe:



          from pyspark.sql.functions import col

          (spark
          .createDataFrame(values, ("a", "date"))
          .select([col(f.name).cast(f.dataType) for f in schema]))





          share|improve this answer

























          • Thank you so much for your efforts. Appreciated!! Your code, with slight modification, works :) I just made a post-update.

            – cph_sto
            Mar 7 at 13:06











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          Long story short, schema used with RDD of external object is not intended to be used that way - declared types should reflect the actual state of the data, not the desired one.



          In other words to allow:



          schema = StructType([
          StructField('A', IntegerType(), True),
          StructField('date', DateType(), True)
          ])


          the data corresponding to date field should use datetime.date. So for example with your RDD[Tuple[int, str]]:



          import datetime

          spark.createDataFrame(
          # Since values from the question are just two element tuples
          # we can use mapValues to transform the "value"
          # but in general case you'll need map
          values.mapValues(datetime.date.fromisoformat),
          schema
          )


          The closest you can get to desired behavior is to convert data (RDD[Row]) with JSON reader, using dicts



          from pyspark.sql import Row

          spark.read.schema(schema).json(rdd.map(Row.asDict))


          or better explicit JSON dumps:



          import json
          spark.read.schema(schema).json(rdd.map(Row.asDict).map(json.dumps))


          but that's of course much more expensive than explicit casting, which BTW, is easy to automate in simple cases like the one you describe:



          from pyspark.sql.functions import col

          (spark
          .createDataFrame(values, ("a", "date"))
          .select([col(f.name).cast(f.dataType) for f in schema]))





          share|improve this answer

























          • Thank you so much for your efforts. Appreciated!! Your code, with slight modification, works :) I just made a post-update.

            – cph_sto
            Mar 7 at 13:06
















          3














          Long story short, schema used with RDD of external object is not intended to be used that way - declared types should reflect the actual state of the data, not the desired one.



          In other words to allow:



          schema = StructType([
          StructField('A', IntegerType(), True),
          StructField('date', DateType(), True)
          ])


          the data corresponding to date field should use datetime.date. So for example with your RDD[Tuple[int, str]]:



          import datetime

          spark.createDataFrame(
          # Since values from the question are just two element tuples
          # we can use mapValues to transform the "value"
          # but in general case you'll need map
          values.mapValues(datetime.date.fromisoformat),
          schema
          )


          The closest you can get to desired behavior is to convert data (RDD[Row]) with JSON reader, using dicts



          from pyspark.sql import Row

          spark.read.schema(schema).json(rdd.map(Row.asDict))


          or better explicit JSON dumps:



          import json
          spark.read.schema(schema).json(rdd.map(Row.asDict).map(json.dumps))


          but that's of course much more expensive than explicit casting, which BTW, is easy to automate in simple cases like the one you describe:



          from pyspark.sql.functions import col

          (spark
          .createDataFrame(values, ("a", "date"))
          .select([col(f.name).cast(f.dataType) for f in schema]))





          share|improve this answer

























          • Thank you so much for your efforts. Appreciated!! Your code, with slight modification, works :) I just made a post-update.

            – cph_sto
            Mar 7 at 13:06














          3












          3








          3







          Long story short, schema used with RDD of external object is not intended to be used that way - declared types should reflect the actual state of the data, not the desired one.



          In other words to allow:



          schema = StructType([
          StructField('A', IntegerType(), True),
          StructField('date', DateType(), True)
          ])


          the data corresponding to date field should use datetime.date. So for example with your RDD[Tuple[int, str]]:



          import datetime

          spark.createDataFrame(
          # Since values from the question are just two element tuples
          # we can use mapValues to transform the "value"
          # but in general case you'll need map
          values.mapValues(datetime.date.fromisoformat),
          schema
          )


          The closest you can get to desired behavior is to convert data (RDD[Row]) with JSON reader, using dicts



          from pyspark.sql import Row

          spark.read.schema(schema).json(rdd.map(Row.asDict))


          or better explicit JSON dumps:



          import json
          spark.read.schema(schema).json(rdd.map(Row.asDict).map(json.dumps))


          but that's of course much more expensive than explicit casting, which BTW, is easy to automate in simple cases like the one you describe:



          from pyspark.sql.functions import col

          (spark
          .createDataFrame(values, ("a", "date"))
          .select([col(f.name).cast(f.dataType) for f in schema]))





          share|improve this answer















          Long story short, schema used with RDD of external object is not intended to be used that way - declared types should reflect the actual state of the data, not the desired one.



          In other words to allow:



          schema = StructType([
          StructField('A', IntegerType(), True),
          StructField('date', DateType(), True)
          ])


          the data corresponding to date field should use datetime.date. So for example with your RDD[Tuple[int, str]]:



          import datetime

          spark.createDataFrame(
          # Since values from the question are just two element tuples
          # we can use mapValues to transform the "value"
          # but in general case you'll need map
          values.mapValues(datetime.date.fromisoformat),
          schema
          )


          The closest you can get to desired behavior is to convert data (RDD[Row]) with JSON reader, using dicts



          from pyspark.sql import Row

          spark.read.schema(schema).json(rdd.map(Row.asDict))


          or better explicit JSON dumps:



          import json
          spark.read.schema(schema).json(rdd.map(Row.asDict).map(json.dumps))


          but that's of course much more expensive than explicit casting, which BTW, is easy to automate in simple cases like the one you describe:



          from pyspark.sql.functions import col

          (spark
          .createDataFrame(values, ("a", "date"))
          .select([col(f.name).cast(f.dataType) for f in schema]))






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Mar 7 at 13:13

























          answered Mar 7 at 10:48









          user10465355user10465355

          2,1172419




          2,1172419












          • Thank you so much for your efforts. Appreciated!! Your code, with slight modification, works :) I just made a post-update.

            – cph_sto
            Mar 7 at 13:06


















          • Thank you so much for your efforts. Appreciated!! Your code, with slight modification, works :) I just made a post-update.

            – cph_sto
            Mar 7 at 13:06

















          Thank you so much for your efforts. Appreciated!! Your code, with slight modification, works :) I just made a post-update.

          – cph_sto
          Mar 7 at 13:06






          Thank you so much for your efforts. Appreciated!! Your code, with slight modification, works :) I just made a post-update.

          – cph_sto
          Mar 7 at 13:06




















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