Pyspark: Parse a column of json strings

Question

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 dataf

Answer 1

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)

Answer 2

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

Answer 3

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.

Answer 4

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'))