I have about 200 nested JSON files with varying levels of nesting from one to three. Each JSON file consist of more than thousand data points. The keys of the values are same in all the files. My objective is to combine the data in all the files in a tabular format in a single CSV file so that I can read all the data and analyze it. I am looking for a simpler python code with brief explanation of each steps of the code to help in understanding the whole sequence of the code.
You can use this code snippet.
First of all install pandas using
pip install pandas
After that, you can use this code to convert JSON files to CSV.
# code to save all data to a single file
import pandas as pd
import glob
path = './path to directory/*.json'
files = glob.glob(path)
data_frames = []
for file in files:
f = open(file, 'r')
data_frames.append(pd.read_json(f))
f.close()
pd.concat(data_frames).to_csv("data.csv")
# code to save CSV data to individual files
import pandas as pd
import glob
path = './path to directory/*.json'
files = glob.glob(path)
for file in files:
f = open(file, 'r')
jsonData = pd.read_json(f.read())
jsonData.to_csv(f.name+".csv")
f.close()
I have got a parquet file / folder (about 1GB) that I would like to load into my local Cassandra DB. Unfortunately I could not find any way (except via SPARK (in Scala)) to directly load this file into CDB. If I blow out the parquet file into CSV it'll just get way too huge for my laptop.
I am setting up a Cassandra DB for a big data analytics case (I've got about 25TB in raw data that we need to get searchable fast). Right now I am running some local tests on how to optimally design the keyspaces, indices and tables before move to Cassandra as a Service on a Hyperscaler. Converting the data to CSV is not an option as this blows up too much.
COPY firmographics.company (col1,col2,col3.....) FROM 'C:\Users\Public\Downloads\companies.csv' WITH DELIMITER='\t' AND HEADER=TRUE;
Turns out, like Alex Ott said, it's easy enough to just write this up in SPARK. Below my code:
import findspark
from pyspark.sql import SparkSession
findspark.init()
spark = SparkSession\
.builder\
.appName("Spark Exploration App")\
.config('spark.jars.packages', 'com.datastax.spark:spark-cassandra-connector_2.11:2.3.2')\
.getOrCreate()
import pandas as pd
df = spark.read.parquet("/PATH/TO/FILE/")
import time
start = time.time()
df2.drop('filename').write\
.format("org.apache.spark.sql.cassandra")\
.mode('append')\
.options(table="few_com", keyspace="bmbr")\
.save()
end = time.time()
print(end - start)
I have made a connection to my HDFS using the following command
import pyarrow as pa
import pyarrow.parquet as pq
fs = pa.hdfs.connect(self.namenode, self.port, user=self.username, kerb_ticket = self.cert)
I'm using the following command to read a parquet file
fs.read_parquet()
but there is not read method for regular text files (e.g. a csv file). How can I read a csv file using pyarrow.
You need to create a file-like object and use the CSV module directly. See pyarrow.csv.read_csv
You can set up a spark session to connect to hdfs, then read it from there.
ss = SparkSession.builder.appName(...)
csv_file = ss.read.csv('/user/file.csv')
Another way is to open the file first, then read it using csv.csv_read
Here is what I used at the end.
from pyarrow import csv
file = 'hdfs://user/file.csv'
with fs.open(file, 'rb') as f:
csv_file = csv.read_csv(f)
Suppose that df is a dataframe in Spark. The way to write df into a single CSV file is
df.coalesce(1).write.option("header", "true").csv("name.csv")
This will write the dataframe into a CSV file contained in a folder called name.csv but the actual CSV file will be called something like part-00000-af091215-57c0-45c4-a521-cd7d9afb5e54.csv.
I would like to know if it is possible to avoid the folder name.csv and to have the actual CSV file called name.csv and not part-00000-af091215-57c0-45c4-a521-cd7d9afb5e54.csv. The reason is that I need to write several CSV files which later on I will read together in Python, but my Python code makes use of the actual CSV names and also needs to have all the single CSV files in a folder (and not a folder of folders).
Any help is appreciated.
A possible solution could be convert the Spark dataframe to a pandas dataframe and save it as csv:
df.toPandas().to_csv("<path>/<filename>")
EDIT: As caujka or snark suggest, this works for small dataframes that fits into driver. It works for real cases that you want to save aggregated data or a sample of the dataframe. Don't use this method for big datasets.
If you want to use only the python standard library this is an easy function that will write to a single file. You don't have to mess with tempfiles or going through another dir.
import csv
def spark_to_csv(df, file_path):
""" Converts spark dataframe to CSV file """
with open(file_path, "w") as f:
writer = csv.DictWriter(f, fieldnames=df.columns)
writer.writerow(dict(zip(fieldnames, fieldnames)))
for row in df.toLocalIterator():
writer.writerow(row.asDict())
If the result size is comparable to spark driver node's free memory, you may have problems with converting the dataframe to pandas.
I would tell spark to save to some temporary location, and then copy the individual csv files into desired folder. Something like this:
import os
import shutil
TEMPORARY_TARGET="big/storage/name"
DESIRED_TARGET="/export/report.csv"
df.coalesce(1).write.option("header", "true").csv(TEMPORARY_TARGET)
part_filename = next(entry for entry in os.listdir(TEMPORARY_TARGET) if entry.startswith('part-'))
temporary_csv = os.path.join(TEMPORARY_TARGET, part_filename)
shutil.copyfile(temporary_csv, DESIRED_TARGET)
If you work with databricks, spark operates with files like dbfs:/mnt/..., and to use python's file operations on them, you need to change the path into /dbfs/mnt/... or (more native to databricks) replace shutil.copyfile with dbutils.fs.cp.
A more databricks'y' solution is here:
TEMPORARY_TARGET="dbfs:/my_folder/filename"
DESIRED_TARGET="dbfs:/my_folder/filename.csv"
spark_df.coalesce(1).write.option("header", "true").csv(TEMPORARY_TARGET)
temporary_csv = os.path.join(TEMPORARY_TARGET, dbutils.fs.ls(TEMPORARY_TARGET)[3][1])
dbutils.fs.cp(temporary_csv, DESIRED_TARGET)
Note if you are working from Koalas data frame you can replace spark_df with koalas_df.to_spark()
For pyspark, you can convert to pandas dataframe and then save it.
df.toPandas().to_csv("<path>/<filename.csv>", header=True, index=False)
There is no dataframe spark API which writes/creates a single file instead of directory as a result of write operation.
Below both options will create one single file inside directory along with standard files (_SUCCESS , _committed , _started).
1. df.coalesce(1).write.mode("overwrite").format("com.databricks.spark.csv").option("header",
"true").csv("PATH/FOLDER_NAME/x.csv")
2. df.repartition(1).write.mode("overwrite").format("com.databricks.spark.csv").option("header",
"true").csv("PATH/FOLDER_NAME/x.csv")
If you don't use coalesce(1) or repartition(1) and take advantage of sparks parallelism for writing files then it will create multiple data files inside directory.
You need to write function in driver which will combine all data file parts to single file(cat part-00000* singlefilename ) once write operation is done.
I had the same problem and used python's NamedTemporaryFile library to solve this.
from tempfile import NamedTemporaryFile
s3 = boto3.resource('s3')
with NamedTemporaryFile() as tmp:
df.coalesce(1).write.format('csv').options(header=True).save(tmp.name)
s3.meta.client.upload_file(tmp.name, S3_BUCKET, S3_FOLDER + 'name.csv')
https://boto3.amazonaws.com/v1/documentation/api/latest/guide/s3-uploading-files.html for more info on upload_file()
Create temp folder inside output folder. Copy file part-00000* with the file name to output folder. Delete the temp folder. Python code snippet to do the same in Databricks.
fpath=output+'/'+'temp'
def file_exists(path):
try:
dbutils.fs.ls(path)
return True
except Exception as e:
if 'java.io.FileNotFoundException' in str(e):
return False
else:
raise
if file_exists(fpath):
dbutils.fs.rm(fpath)
df.coalesce(1).write.option("header", "true").csv(fpath)
else:
df.coalesce(1).write.option("header", "true").csv(fpath)
fname=([x.name for x in dbutils.fs.ls(fpath) if x.name.startswith('part-00000')])
dbutils.fs.cp(fpath+"/"+fname[0], output+"/"+"name.csv")
dbutils.fs.rm(fpath, True)
You can go with pyarrow, as it provides file pointer for hdfs file system. You can write your content to file pointer as a usual file writing. Code example:
import pyarrow.fs as fs
HDFS_HOST: str = 'hdfs://<your_hdfs_name_service>'
FILENAME_PATH: str = '/user/your/hdfs/file/path/<file_name>'
hadoop_file_system = fs.HadoopFileSystem(host=HDFS_HOST)
with hadoop_file_system.open_output_stream(path=FILENAME_PATH) as f:
f.write("Hello from pyarrow!".encode())
This will create a single file with the specified name.
To initiate pyarrow you should define environment CLASSPATH properly, set the output of hadoop classpath --glob to it
df.write.mode("overwrite").format("com.databricks.spark.csv").option("header", "true").csv("PATH/FOLDER_NAME/x.csv")
you can use this and if you don't want to give the name of CSV everytime you can write UDF or create an array of the CSV file name and give it to this it will work
How to convert json to parquet in streaming with Spark?
Acutually i have to ssh from a server, recieve a big json file, convert it to parquet, and upload it on hadoop.
I there a way to do this in a pipelined way?
They are backup files so I have a directory with a predefined amount of files that don't change in size in time
Something like:
scp host /dev/stdout | spark-submit myprogram.py | hadoop /dir/
edit:
Actually I'm working on this:
sc = SparkContext(appName="Test")
sqlContext = SQLContext(sc)
sqlContext.setConf("spark.sql.parquet.compression.codec.", "gzip")
#Since i couldn't get the stdio, went for a pipe:
with open("mypipe", "r") as o:
while True:
line = o.readline()
print "Processing: " + line
lineRDD = sc.parallelize([line])
df = sqlContext.jsonRDD(lineRDD)
#Create and append
df.write.parquet("file:///home/user/spark/test", mode="append")
print "Done."
This is working fine, but the resulting parquet is very large (280kb for 4 lines 2 columns json). Any improvements?
If anyone is interested, I managed to resolve this using the .pipe() method.
https://spark.apache.org/docs/latest/api/python/pyspark.html?highlight=pipe#pyspark.RDD.pipe