PySpark Dataframe from Python Dictionary without Pandas - pyspark

Question

I am trying to convert the following Python dict into PySpark DataFrame but I am not getting expected output.
dict_lst = {'letters': ['a', 'b', 'c'],
'numbers': [10, 20, 30]}
df_dict = sc.parallelize([dict_lst]).toDF() # Result not as expected
df_dict.show()
Is there a way to do this without using Pandas?

Answer 1

Quoting myself:
I find it's useful to think of the argument to createDataFrame() as a list of tuples where each entry in the list corresponds to a row in the DataFrame and each element of the tuple corresponds to a column.
So the easiest thing is to convert your dictionary into this format. You can easily do this using zip():
column_names, data = zip(*dict_lst.items())
spark.createDataFrame(zip(*data), column_names).show()
#+-------+-------+
#|letters|numbers|
#+-------+-------+
#| a| 10|
#| b| 20|
#| c| 30|
#+-------+-------+
The above assumes that all of the lists are the same length. If this is not the case, you would have to use itertools.izip_longest (python2) or itertools.zip_longest (python3).
from itertools import izip_longest as zip_longest # use this for python2
#from itertools import zip_longest # use this for python3
dict_lst = {'letters': ['a', 'b', 'c'],
'numbers': [10, 20, 30, 40]}
column_names, data = zip(*dict_lst.items())
spark.createDataFrame(zip_longest(*data), column_names).show()
#+-------+-------+
#|letters|numbers|
#+-------+-------+
#| a| 10|
#| b| 20|
#| c| 30|
#| null| 40|
#+-------+-------+

Answer 2

Your dict_lst is not really the format you want to adopt to create a dataframe. It would be better if you had a list of dict instead of a dict of list.
This code creates a DataFrame from you dict of list :
from pyspark.sql import SQLContext, Row
sqlContext = SQLContext(sc)
dict_lst = {'letters': ['a', 'b', 'c'],
'numbers': [10, 20, 30]}
values_lst = dict_lst.values()
nb_rows = [len(lst) for lst in values_lst]
assert min(nb_rows)==max(nb_rows) #We must have the same nb of elem for each key
row_lst = []
columns = dict_lst.keys()
for i in range(nb_rows[0]):
row_values = [lst[i] for lst in values_lst]
row_dict = {column: value for column, value in zip(columns, row_values)}
row = Row(**row_dict)
row_lst.append(row)
df = sqlContext.createDataFrame(row_lst)

Answer 3

Using pault's answer above I imposed a specific schema on my dataframe as follows:
import pyspark
from pyspark.sql import SparkSession, functions
spark = SparkSession.builder.appName('dictToDF').getOrCreate()
get data:
dict_lst = {'letters': ['a', 'b', 'c'],'numbers': [10, 20, 30]}
data = dict_lst.values()
create schema:
from pyspark.sql.types import *
myschema= StructType([ StructField("letters", StringType(), True)\
,StructField("numbers", IntegerType(), True)\
])
create df from dictionary - with schema:
df=spark.createDataFrame(zip(*data), schema = myschema)
df.show()
+-------+-------+
|letters|numbers|
+-------+-------+
| a| 10|
| b| 20|
| c| 30|
+-------+-------+
show df schema:
df.printSchema()
root
|-- letters: string (nullable = true)
|-- numbers: integer (nullable = true)

Answer 4

You can also use a Python List to quickly prototype a DataFrame. The idea is based from Databricks's tutorial.
df = spark.createDataFrame(
[(1, "a"),
(1, "a"),
(1, "b")],
("id", "value"))
df.show()
+---+-----+
| id|value|
+---+-----+
| 1| a|
| 1| a|
| 1| b|
+---+-----+

Answer 5

Try this out :
dict_lst = [{'letters': 'a', 'numbers': 10},
{'letters': 'b', 'numbers': 20},
{'letters': 'c', 'numbers': 30}]
df_dict = sc.parallelize(dict_lst).toDF() # Result as expected
Output:
>>> df_dict.show()
+-------+-------+
|letters|numbers|
+-------+-------+
| a| 10|
| b| 20|
| c| 30|
+-------+-------+

Answer 6

The most efficient approach is to use Pandas
import pandas as pd
spark.createDataFrame(pd.DataFrame(dict_lst))