Pyspark: Parse a column of json strings

I have a pyspark dataframe consisting of one column, called json, where each row is a unicode string of json. I'd like to parse each row and return a new dataframe where each row is the parsed json.

# Sample Data Frame
jstr1 = u'{"header":{"id":12345,"foo":"bar"},"body":{"id":111000,"name":"foobar","sub_json":{"id":54321,"sub_sub_json":{"col1":20,"col2":"somethong"}}}}'
jstr2 = u'{"header":{"id":12346,"foo":"baz"},"body":{"id":111002,"name":"barfoo","sub_json":{"id":23456,"sub_sub_json":{"col1":30,"col2":"something else"}}}}'
jstr3 = u'{"header":{"id":43256,"foo":"foobaz"},"body":{"id":20192,"name":"bazbar","sub_json":{"id":39283,"sub_sub_json":{"col1":50,"col2":"another thing"}}}}'
df = sql_context.createDataFrame([Row(json=jstr1),Row(json=jstr2),Row(json=jstr3)])

I've tried mapping over each row with json.loads:

(df
  .select('json')
  .rdd
  .map(lambda x: json.loads(x))
  .toDF()
).show()

But this returns a TypeError: expected string or buffer

I suspect that part of the problem is that when converting from a dataframe to an rdd, the schema information is lost, so I've also tried manually entering in the schema info:

schema = StructType([StructField('json', StringType(), True)])
rdd = (df
  .select('json')
  .rdd
  .map(lambda x: json.loads(x))
)
new_df = sql_context.createDataFrame(rdd, schema)
new_df.show()

But I get the same TypeError.

Looking at this answer, it looks like flattening out the rows with flatMap might be useful here, but I'm not having success with that either:

schema = StructType([StructField('json', StringType(), True)])
rdd = (df
  .select('json')
  .rdd
  .flatMap(lambda x: x)
  .flatMap(lambda x: json.loads(x))
  .map(lambda x: x.get('body'))
)
new_df = sql_context.createDataFrame(rdd, schema)
new_df.show()

I get this error: AttributeError: 'unicode' object has no attribute 'get'.

For Spark 2.1+, you can use from_json which allows the preservation of the other non-json columns within the dataframe as follows:

from pyspark.sql.functions import from_json, col
json_schema = spark.read.json(df.rdd.map(lambda row: row.json)).schema
df.withColumn('json', from_json(col('json'), json_schema))

You let Spark derive the schema of the json string column. Then the df.json column is no longer a StringType, but the correctly decoded json structure, i.e., nested StrucType and all the other columns of df are preserved as-is.

You can access the json content as follows:

df.select(col('json.header').alias('header'))

Converting a dataframe with json strings to structured dataframe is'a actually quite simple in spark if you convert the dataframe to RDD of strings before (see: http://spark.apache.org/docs/latest/sql-programming-guide.html#json-datasets)

For example:

>>> new_df = sql_context.read.json(df.rdd.map(lambda r: r.json))
>>> new_df.printSchema()
root
 |-- body: struct (nullable = true)
 |    |-- id: long (nullable = true)
 |    |-- name: string (nullable = true)
 |    |-- sub_json: struct (nullable = true)
 |    |    |-- id: long (nullable = true)
 |    |    |-- sub_sub_json: struct (nullable = true)
 |    |    |    |-- col1: long (nullable = true)
 |    |    |    |-- col2: string (nullable = true)
 |-- header: struct (nullable = true)
 |    |-- foo: string (nullable = true)
 |    |-- id: long (nullable = true)

Existing answers do not work if your JSON is anything but perfectly/traditionally formatted. For example, the RDD-based schema inference expects JSON in curly-braces {} and will provide an incorrect schema (resulting in null values) if, for example, your data looks like:

[  {    "a": 1.0,    "b": 1  },  {    "a": 0.0,    "b": 2  }]

I wrote a function to work around this issue by sanitizing JSON such that it lives in another JSON object:

def parseJSONCols(df, *cols, sanitize=True):
    """Auto infer the schema of a json column and parse into a struct.

    rdd-based schema inference works if you have well-formatted JSON,
    like ``{"key": "value", ...}``, but breaks if your 'JSON' is just a
    string (``"data"``) or is an array (``[1, 2, 3]``). In those cases you
    can fix everything by wrapping the data in another JSON object
    (``{"key": [1, 2, 3]}``). The ``sanitize`` option (default True)
    automatically performs the wrapping and unwrapping.

    The schema inference is based on this
    `SO Post <https://stackoverflow.com/a/45880574)/>`_.

    Parameters
    ----------
    df : pyspark dataframe
        Dataframe containing the JSON cols.
    *cols : string(s)
        Names of the columns containing JSON.
    sanitize : boolean
        Flag indicating whether you'd like to sanitize your records
        by wrapping and unwrapping them in another JSON object layer.

    Returns
    -------
    pyspark dataframe
        A dataframe with the decoded columns.
    """
    res = df
    for i in cols:

        # sanitize if requested.
        if sanitize:
            res = (
                res.withColumn(
                    i,
                    psf.concat(psf.lit('{"data": '), i, psf.lit('}'))
                )
            )
        # infer schema and apply it
        schema = spark.read.json(res.rdd.map(lambda x: x[i])).schema
        res = res.withColumn(i, psf.from_json(psf.col(i), schema))

        # unpack the wrapped object if needed
        if sanitize:
            res = res.withColumn(i, psf.col(i).data)
    return res

Note: psf = pyspark.sql.functions.

Here's a concise (spark SQL) version of @nolan-conaway's parseJSONCols function.

SELECT 
explode(
    from_json(
        concat('{"data":', 
               '[{"a": 1.0,"b": 1},{"a": 0.0,"b": 2}]', 
               '}'), 
        'data array<struct<a:DOUBLE, b:INT>>'
    ).data) as data;

> PS. I've added the explode function as well :P

You'll need to know some HIVE SQL types

from pyspark.sql import SparkSession
spark = SparkSession.builder.getOrCreate()

def map2json(dict):
    import json
    return json.dumps(dict)
from pyspark.sql.types import StringType
spark.udf.register("map2json", lambda dict: map2json(dict), StringType())

spark.sql("select map2json(map('a', '1'))").show()

If you don't know the schema of each JSON (and it can be different) you can use :

from pyspark.sql import SparkSession
spark = SparkSession.builder.getOrCreate()
 
# ... here you get your DF

# Assuming the first column of your DF is the JSON to parse
my_df = spark.read.json(my_df.rdd.map(lambda x: x[0]))

Note that it won't keep any other column present in your dataset. From : https://github.com/apache/spark/pull/22775