如何从同时列出数据和架构的 JSON 文件创建 Spark-SQL dataframe

Question

conf = SparkConf().setAppName("PySpark").setMaster("local")
sc = SparkContext(conf=conf)
sqlContext = SQLContext(sc)

file = sqlContext.read.json(json_file_path)
file.show()

Outputs:

+--------------------+--------------------+
|                data|              schema|
+--------------------+--------------------+
|[[The battery is ...|[[[index, integer...|
+--------------------+--------------------+

How do I extract the data using my own created schema. My schema code is:

from pyspark.sql.types import ArrayType, StructField, StructType, StringType, IntegerType
schema = StructType([
    StructField('index', IntegerType(), True),
    StructField('content', StringType(), True),
    StructField('label', IntegerType(), True),
    StructField('label_1', StringType(), True ),
    StructField('label_2', StringType(), True ),
    StructField('label_3', IntegerType(), True ),
    StructField('label_4', IntegerType(), True )])

I have tried:

file.withColumn("data", from_json("data", schema))\
    .show()

But I receive the following error:

 cannot resolve 'from_json(`data`)' due to data type mismatch: argument 1 requires string type, however, '`data`' is of array<struct<content:string,index:bigint,label:bigint,label_1:string,label_2:string,label_3:double,label_4:timestamp>> type.;;

Answer 1

The read method already recognized the schema in the back.

Try running file.printSchema() and it should show more-less the schema that you want.

The way unpack the data is to run:

file = file.select(explode("data").as("exploded_data"))

If you want, you can take it to next level with:

file.select(file.col("exploded_data.*"))

This will flatten out the schema.

Disclaimer: This is scala code, python might need tiny adjustments