Pyspark: explode json in column to multiple columns

Question

The data looks like this -

+-----------+-----------+-----------------------------+
|         id|      point|                         data|
+-----------------------------------------------------+
|        abc|          6|{"key1":"124", "key2": "345"}|
|        dfl|          7|{"key1":"777", "key2": "888"}|
|        4bd|          6|{"key1":"111", "key2": "788"}|

I am trying to break it into the following format.

+-----------+-----------+-----------+-----------+
|         id|      point|       key1|       key2|
+------------------------------------------------
|        abc|          6|        124|        345|
|        dfl|          7|        777|        888|
|        4bd|          6|        111|        788|

The explode function explodes the dataframe into multiple rows. But that is not the desired solution.

Note: This solution does not answers my questions. PySpark "explode" dict in column

Answer 1

You can use from_json function also to achieve this,

data = [("abc", 6, '{"key1":"124", "key2": "345"}'),
        ("dfl", 7, '{"key1":"777", "key2": "888"}'),
        ("4bd", 6, '{"key1":"111", "key2": "788"}')]

df = spark.createDataFrame(data, ["id", "point", "data"])


schema = "key1 string, key2 string"


df = df.withColumn("data", from_json(col("data"), schema)) \
       .select("id", "point", "data.key1", "data.key2")

df.show()

Answer 2

Simply do this:

df.select("id", "point", "data.*").show()

It will give you following answer:

Explanation:

To expand a struct type data, 'data.*' can be used. Doing this will expand the data column and the 'key' inside data column will become new columns.

Answer 3

In this approach you just need to set the name of column with Json content. No need to set up the schema. It makes everything automatically.

json_col_name = 'data'
keys = df.select(f"{json_col_name}.*").columns
jsonFields= [f"{json_col_name}.{key} {key}" for key in keys]

main_fields = [key for key in df.columns if key != json_col_name]
df_new = df.selectExpr(main_fields + jsonFields)

Answer 4

As mentioned by @jxc, json_tuple should work fine if you were not able to define the schema beforehand and you only needed to deal with a single level of json string. I think it's more straight forward and easier to use. Strangely, I didn't find anyone else mention this function before.

In my use case, original dataframe schema: StructType(List(StructField(a,StringType,true))), json string column shown as:

+---------------------------------------+
|a                                      |
+---------------------------------------+
|{"k1": "v1", "k2": "2", "k3": {"m": 1}}|
|{"k1": "v11", "k3": "v33"}             |
|{"k1": "v13", "k2": "23"}              |
+---------------------------------------+

Expand json fields into new columns with json_tuple:

from pyspark.sql import functions as F

df = df.select(F.col('a'), 
    F.json_tuple(F.col('a'), 'k1', 'k2', 'k3') \
    .alias('k1', 'k2', 'k3'))

df.schema
df.show(truncate=False)

The document doesn't say much about it, but at least in my use case, new columns extracted by json_tuple are StringType, and it only extract single depth of JSON string.

StructType(List(StructField(k1,StringType,true),StructField(k2,StringType,true),StructField(k3,StringType,true)))

+---------------------------------------+---+----+-------+
|a                                      |k1 |k2  |k3     |
+---------------------------------------+---+----+-------+
|{"k1": "v1", "k2": "2", "k3": {"m": 1}}|v1 |2   |{"m":1}|
|{"k1": "v11", "k3": "v33"}             |v11|null|v33    |
|{"k1": "v13", "k2": "23"}              |v13|23  |null   |
+---------------------------------------+---+----+-------+

Answer 5

All credits to Shrikant Prabhu

You can simply use SQL

SELECT id, point, data.*
FROM original_table

Like this the schema of the new table will adapt if the data changes and you won't have to do anything in your pipelin.

Answer 6

This works for my use case

data1 = spark.read.parquet(path)
json_schema = spark.read.json(data1.rdd.map(lambda row: row.json_col)).schema
data2 = data1.withColumn("data", from_json("json_col", json_schema))
col1 = data2.columns
col1.remove("data")
col2 = data2.select("data.*").columns
append_str ="data."
col3 = [append_str + val for val in col2]
col_list = col1 + col3
data3 = data2.select(*col_list).drop("json_col")

Answer 7

As suggested by @pault, the data field is a string field. since the keys are the same (i.e. 'key1', 'key2') in the JSON string over rows, you might also use json_tuple() (this function is New in version 1.6 based on the documentation)

from pyspark.sql import functions as F

df.select('id', 'point', F.json_tuple('data', 'key1', 'key2').alias('key1', 'key2')).show()

Below is My original post: which is most likely WRONG if the original table is from df.show(truncate=False) and thus the data field is NOT a python data structure.

Since you have exploded the data into rows, I supposed the column data is a Python data structure instead of a string:

from pyspark.sql import functions as F

df.select('id', 'point', F.col('data').getItem('key1').alias('key1'), F.col('data')['key2'].alias('key2')).show()

Answer 8

As long as you are using Spark version 2.1 or higher, pyspark.sql.functions.from_json should get you your desired result, but you would need to first define the required schema

from pyspark.sql.functions import from_json, col
from pyspark.sql.types import StructType, StructField, StringType

schema = StructType(
    [
        StructField('key1', StringType(), True),
        StructField('key2', StringType(), True)
    ]
)

df.withColumn("data", from_json("data", schema))\
    .select(col('id'), col('point'), col('data.*'))\
    .show()

which should give you

+---+-----+----+----+
| id|point|key1|key2|
+---+-----+----+----+
|abc|    6| 124| 345|
|df1|    7| 777| 888|
|4bd|    6| 111| 788|
+---+-----+----+----+