I am looking at any library (in java) that can help me generate a dummy JSON file to test my code for e.g The JSON file can contain random user profile data-name, address, zipcode
I searched StackOverflow and found this link, found the following link : How to generate JSON string in Java?
I think the suggested library https://github.com/DiUS/java-faker, seems to be useful, however because of security constraints I cannot use this particular library. Are there any more recommendations?
Use for instance Faker, like that:
#!/usr/bin/env python3
from json import dumps
from faker import Faker
fake = Faker()
def user():
return dict(
name=fake.name(),
address=fake.address(),
bio=fake.text()
)
print('[')
try:
while True:
print(dumps(user()))
print(',')
except KeyboardInterrupt:
# XXX: json array can not end with a comma
print(dumps(user()))
print(']')
You can use it like that:
python3 fake_user.py > users.json
Use Ctrl+C to stop it when the file is big enough
Related
I have a pipeline in NiFi that pulls down some invalid JSON that I need to clean up. The best solution I've concocted is to run a Python script via ExecuteStreamCommand and simultaneously clean/split it up in one fell swoop. However, even though I use sys.stdout.write() in my for loop, only the original JSON comes out in the output stream in NiFi.
Am I misusing sys.stdout.write() or is this possible, but I've just done something wrong? My end goal is for each line of the json to be a new flow file, i.e. file 1 is {"fruit":"apple",..., file 2 is {"fruit":"cherry",..., and so on.
example JSON
{"fruit":"apple", "vegetable":"celery", "location":{"country":"nor\\way", "city":"oslo", }, "color":"blue"}
{"fruit":"cherry", "vegetable":"kale", "location":{"country":"france", "city":"calais", }, "color":"green"}
{"fruit":"peach", "vegetable":"peas", "location":{"country":"united\\kingdom", "city":"london", }, "color":"yellow"}
script
import json
import re
import sys
flow_file = sys.stdin.read()
try:
load = json.loads(flow_file)
sys.stdout.write(flow_file)
except:
flow_file_esc = re.sub(r"[(\\)]", "", flow_file)
for f in flow_file_esc.splitlines():
sys.stdout.write(str(f))
Can you clean the file first with ReplaceText and then split it with SplitJson, SplitRecord, or ForkRecord?
If you need to combine the two operations and want to script it, you could try ExecuteScript with Jython (since it doesn't look like you're using native CPython libraries), I have some simple examples in my cookbook and my blog.
I am parsing 350 txt files having json data using python. I am able to retrieve 62 of those object and store them on mysql database, but after that I am getting an error saying JSONDecodeError: ExtraData
Python:
import os
import ast
import json
import mysql.connector as mariadb
from mysql.connector.constants import ClientFlag
mariadb_connection = mariadb.connect(user='root', password='137800000', database='shaproject',client_flags=[ClientFlag.LOCAL_FILES])
cursor = mariadb_connection.cursor()
sql3 = """INSERT INTO shaproject.alttwo (alttwo_id,responses) VALUES """
os.chdir('F:/Code Blocks/SEM 2/DM/Project/350/For Merge Disqus')
current_list_dir=os.listdir()
print(current_list_dir)
cur_cwd=os.getcwd()
cur_cwd=cur_cwd.replace('\\','/')
twoid=1
for every_file in current_list_dir:
file=open(cur_cwd + "/" + every_file)
utffile=file.read()
data=json.loads(utffile)
for i in range(0,len(data['response'])):
data123 = json.dumps(data['response'][i])
tup=(twoid,data123)
print(sql3+str(tup))
twoid+=1
cursor.execute(sql3+str(tup)+";")
print(tup)
mariadb_connection.commit()
I have searched online and found that multiple dump statements are resulting in this error. But I am unable to resolve it.
You want to use glob.
Rather than os.listdir(), which is too permissive,
use glob to focus on just the *.json files.
Print out the name of the file before asking .loads() to parse it.
Rename any badly formatted files to .txt rather than .json, in order to skip them.
Note that you can pass the open file directly to .load(), if you wish.
Closing open files would be a good thing.
Rather than a direct assignment (with no close()!)
you would be better off with with:
with open(cur_cwd + "/" + every_file) as file:
data = json.load(file)
Talking about current current working directory seems
both repetitive and redundant.
It would suffice to call it cwd.
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
I have a csv data file containing commas within a column value. For example,
value_1,value_2,value_3
AAA_A,BBB,B,CCC_C
Here, the values are "AAA_A","BBB,B","CCC_C". But, when trying to split the line by comma, it is giving me 4 values, i.e. "AAA_A","BBB","B","CCC_C".
How to get the right values after splitting the line by commas in PySpark?
Use spark-csv class from databriks.
Delimiters between quotes, by default ("), are ignored.
Example:
val df = sqlContext.read
.format("com.databricks.spark.csv")
.option("header", "true") // Use first line of all files as header
.option("inferSchema", "true") // Automatically infer data types
.load("cars.csv")
For more info, review https://github.com/databricks/spark-csv
If your quote is (') instance of ("), you could configure with this class.
EDIT:
For python API:
df = sqlContext.read.format('com.databricks.spark.csv').options(header='true', inferschema='true').load('cars.csv')
Best regards.
If you do not mind the extra package dependency, you could use Pandas to parse the CSV file. It handles internal commas just fine.
Dependencies:
from pyspark import SparkContext
from pyspark.sql import SQLContext
import pandas as pd
Read the whole file at once into a Spark DataFrame:
sc = SparkContext('local','example') # if using locally
sql_sc = SQLContext(sc)
pandas_df = pd.read_csv('file.csv') # assuming the file contains a header
# If no header:
# pandas_df = pd.read_csv('file.csv', names = ['column 1','column 2'])
s_df = sql_sc.createDataFrame(pandas_df)
Or, even more data-consciously, you can chunk the data into a Spark RDD then DF:
chunk_100k = pd.read_csv('file.csv', chunksize=100000)
for chunky in chunk_100k:
Spark_temp_rdd = sc.parallelize(chunky.values.tolist())
try:
Spark_full_rdd += Spark_temp_rdd
except NameError:
Spark_full_rdd = Spark_temp_rdd
del Spark_temp_rdd
Spark_DF = Spark_full_rdd.toDF(['column 1','column 2'])
I'm (really) new to Pyspark, but have been using Pandas for the past years. What I'm going to put here might not be ultimately the best solution, but it works for me so I think it's worth posting here.
I'm encountering the same issue loading in a CSV file with extra comma embedded in one special field, which triggered an error if using Pyspark, but had no problem if using Pandas. So I looked around for a solution to deal with this extra delimiter, and the following piece of code solved my issue:
df = sqlContext.read.format('csv').option('header','true').option('maxColumns','3').option('escape','"').load('cars.csv')
I personally like to force the 'maxColumns' parameter to allow only a specific number of columns. So if the "BBB,B" somehow got parsed into two strings, spark is going to give an error message and print the whole line for you. And the 'escape' option is the one that really fixed my issue. I don't know if this helps, but hopefully that's something to run experiments with.
I have a working code for parsing a JSON output using KornShell by treating it as a string of characters. The issue I have is that the vendor keeps changing the position of the field that I am intersted in. I understand in JSON, we can parse it by key-value pairs.
Is there something out there that can do this? I am intersted in a specific field and I would like to use it to run the checks on the status of another RESTAPI call.
My sample json output is like this:
JSONDATA value :
{
"status": "success",
"job-execution-id": 396805,
"job-execution-user": "flexapp",
"job-execution-trigger": "RESTAPI"
}
I would need the job-execution-id value to monitor this job through the rest of the script.
I am using the following command to parse it:
RUNJOB=$(print ${DATA} |cut -f3 -d':'|cut -f1 -d','| tr -d [:blank:]) >> ${LOGDIR}/${LOGFILE}
The problem with this is, it is field delimited by :. The field position has been known to be changed by the vendors during releases.
So I am trying to see if I can use a utility out there that would always give me the key-value pair of "job-execution-id": 396805, no matter where it is in the json output.
I started looking at jsawk, and it requires the js interpreter to be installed on our machines which I don't want. Any hint on how to go about finding which RPM that I need to solve it?
I am using RHEL5.5.
Any help is greatly appreciated.
The ast-open project has libdss (and a dss wrapper) which supposedly could be used with ksh. Documentation is sparse and is limited to a few messages on the ast-user mailing list.
The regression tests for libdss contain some json and xml examples.
I'll try to find more info.
Python is included by default with CentOS so one thing you could do is pass your JSON string to a Python script and use Python's JSON parser. You can then grab the value written out by the script. An example you could modify to meet your needs is below.
Note that by specifying other dictionary keys in the Python script you can get any of the values you need without having to worry about the order changing.
Python script:
#get_job_execution_id.py
# The try/except is because you'll probably have Python 2.4 on CentOS 5.5,
# and the straight "import json" statement won't work unless you have Python 2.6+.
try:
import json
except:
import simplejson as json
import sys
json_data = sys.argv[1]
data = json.loads(json_data)
job_execution_id = data['job-execution-id']
sys.stdout.write(str(job_execution_id))
Kornshell script that executes it:
#get_job_execution_id.sh
#!/bin/ksh
JSON_DATA='{"status":"success","job-execution-id":396805,"job-execution-user":"flexapp","job-execution-trigger":"RESTAPI"}'
EXECUTION_ID=`python get_execution_id.py "$JSON_DATA"`
echo $EXECUTION_ID