Convert Python Dictionary List to PySpark DataFrame
articleArticles 579
imageDiagrams 59
commentComments 309
loyaltyKontext Points 6394
more_vert
This article shows how to convert a Python dictionary list to a DataFrame in Spark using Python.
Example dictionary list
data = [{"Category": 'Category A', "ID": 1, "Value": 12.40},
{"Category": 'Category B', "ID": 2, "Value": 30.10},
{"Category": 'Category C', "ID": 3, "Value": 100.01}
]The above dictionary list will be used as the input.
Solution 1 - Infer schema from dict
In Spark 2.x, schema can be directly inferred from dictionary. The following code snippets directly create the data frame using SparkSession.createDataFrame function.
Code snippet
from pyspark.sql import SparkSession
appName = "Python Example - PySpark Parsing Dictionary as DataFrame"
master = "local"
# Create Spark session
spark = SparkSession.builder \
.appName(appName) \
.master(master) \
.getOrCreate()
# List
data = [{"Category": 'Category A', "ID": 1, "Value": 12.40},
{"Category": 'Category B', "ID": 2, "Value": 30.10},
{"Category": 'Category C', "ID": 3, "Value": 100.01}
]
# Create data frame
df = spark.createDataFrame(data)
print(df.schema)
df.show()Output
The following is the output from the above PySpark script.
session.py:340: UserWarning: inferring schema from dict is deprecated,please use pyspark.sql.Row instead
warnings.warn("inferring schema from dict is deprecated,"
StructType(List(StructField(Category,StringType,true),StructField(ID,LongType,true),StructField(Value,DoubleType,true)))
+----------+---+------+
| Category| ID| Value|
+----------+---+------+
|Category A| 1| 12.4|
|Category B| 2| 30.1|
|Category C| 3|100.01|
+----------+---+------+
The script created a DataFrame with inferred schema as:
StructType(List(StructField(Category,StringType,true),StructField(ID,LongType,true),StructField(Value,DoubleType,true)))
However there is one warning:
Warning: inferring schema from dict is deprecated,please use pyspark.sql.Row instead
Solution 2 - Use pyspark.sql.Row
As the warning message suggests in solution 1, we are going to use pyspark.sql.Row in this solution.
Code snippet
from pyspark.sql import SparkSession, Row
appName = "Python Example - PySpark Parsing Dictionary as DataFrame"
master = "local"
# Create Spark session
spark = SparkSession.builder \
.appName(appName) \
.master(master) \
.getOrCreate()
# List
data = [{"Category": 'Category A', "ID": 1, "Value": 12.40},
{"Category": 'Category B', "ID": 2, "Value": 30.10},
{"Category": 'Category C', "ID": 3, "Value": 100.01}
]
# Create data frame
df = spark.createDataFrame([Row(**i) for i in data])
print(df.schema)
df.show()
In this code snippet, we use pyspark.sql.Row to parse dictionary item. It also uses ** to unpack keywords in each dictionary.
The output is the same as solution 1.
Solution 3 - Explicit schema
Of course, we can explicitly define the schema for the DataFrame.
In the following code snippet, we define the schema based on the data types in the dictionary:
schema = StructType([
StructField('Category', StringType(), False),
StructField('ID', IntegerType(), False),
StructField('Value', DecimalType(scale=2), True)
])Code snippet
from pyspark.sql import SparkSession
from pyspark.sql.types import ArrayType, StructField, StructType, StringType, IntegerType, DecimalType
from decimal import Decimal
appName = "Python Example - PySpark Parsing Dictionary as DataFrame"
master = "local"
# Create Spark session
spark = SparkSession.builder \
.appName(appName) \
.master(master) \
.getOrCreate()
# List
data = [{"Category": 'Category A', "ID": 1, "Value": Decimal(12.40)},
{"Category": 'Category B', "ID": 2, "Value": Decimal(30.10)},
{"Category": 'Category C', "ID": 3, "Value": Decimal(100.01)}
]
schema = StructType([
StructField('Category', StringType(), False),
StructField('ID', IntegerType(), False),
StructField('Value', DecimalType(scale=2), True)
])
# Create data frame
df = spark.createDataFrame(data, schema)
print(df.schema)
df.show()There are many different ways to achieve the same goal. Let me know if you have other options.
info Last modified by Raymond 5 years ago
copyright
This page is subject to Site terms.
comment Comments
S
Swapnil Patil
Swapnil
access_time
5 years ago
link
more_vert
Hi Raymond,
wonderful Article ,Was just confused at below line :
df = spark.createDataFrame([Row(**i) for i in data])
I assume Row class needs input like
row=Row(Category= 'Category A', ID= 1,Value=1) so how this is getting translated here..
or is it like when we give input like a key ,val,it understands and creates schema correctly ?