I am trying to convert a .csv file to a .parquet file.
The csv file (Temp.csv
) has the following format
1,Jon,Doe,Denver
I am using the following python code to convert it into parquet
from pyspark import SparkContext
from pyspark.sql import SQLContext
from pyspark.sql.types import *
import os
if __name__ == "__main__":
sc = SparkContext(appName="CSV2Parquet")
sqlContext = SQLContext(sc)
schema = StructType([
StructField("col1", IntegerType(), True),
StructField("col2", StringType(), True),
StructField("col3", StringType(), True),
StructField("col4", StringType(), True)])
dirname = os.path.dirname(os.path.abspath(__file__))
csvfilename = os.path.join(dirname,'Temp.csv')
rdd = sc.textFile(csvfilename).map(lambda line: line.split(","))
df = sqlContext.createDataFrame(rdd, schema)
parquetfilename = os.path.join(dirname,'output.parquet')
df.write.mode('overwrite').parquet(parquetfilename)
The result is only a folder named, output.parquet
and not a parquet file that I'm looking for, followed by the following error on the console.
I have also tried running the following code to face a similar issue.
from pyspark.sql import SparkSession
import os
spark = SparkSession \
.builder \
.appName("Protob Conversion to Parquet") \
.config("spark.some.config.option", "some-value") \
.getOrCreate()
# read csv
dirname = os.path.dirname(os.path.abspath(__file__))
csvfilename = os.path.join(dirname,'Temp.csv')
df = spark.read.csv(csvfilename)
# Displays the content of the DataFrame to stdout
df.show()
parquetfilename = os.path.join(dirname,'output.parquet')
df.write.mode('overwrite').parquet(parquetfilename)
How to best do it? Using windows, python 2.7.
Using the packages pyarrow
and pandas
you can convert CSVs to Parquet without using a JVM in the background:
import pandas as pd
df = pd.read_csv('example.csv')
df.to_parquet('output.parquet')
One limitation in which you will run is that pyarrow
is only available for Python 3.5+ on Windows. Either use Linux/OSX to run the code as Python 2 or upgrade your windows setup to Python 3.6.