How to use PySpark to load a rolling window from daily files? - csv

I have a large number of fairly large daily files stored in a blog storage engine(S3, Azure datalake exc.. exc..) data1900-01-01.csv, data1900-01-02.csv,....,data2017-04-27.csv. My goal is to preform a rolling N-day linear regression but I am having trouble with the data loading aspect. I am not sure how to do this without nested RDD's.
The schema for every .csv file is the same.
In other words for every date d_t, I need data x_t and to join data (x_t-1, x_t-2,... x_t-N).
How can I use PySpark to load an N-day Window of these daily files? All of the PySpark examples I can find seem to load from one very large file or data set.
Here's an example of my current code:
dates = [('1995-01-03', '1995-01-04', '1995-01-05'), ('1995-01-04', '1995-01-05', '1995-01-06')]
p = sc.parallelize(dates)
def test_run(date_range):
dt0 = date_range[-1] #get the latest date
s = '/daily/data{}.csv'
df0 = spark.read.csv(s.format(dt0), header=True, mode='DROPMALFORM')
file_list = [s.format(dt) for dt in date_range[:-1]] # Get a window of trailing dates
df1 = spark.read.csv(file_list, header=True, mode='DROPMALFORM')
return 1
p.filter(test_run)
p.map(test_run) #fails with same error as p.filter
I'm on PySpark version '2.1.0'
I'm running this on an Azure HDInsight cluster jupyter notebook.
spark here is of type <class 'pyspark.sql.session.SparkSession'>
A smaller more reproducible example is as follows:
p = sc.parallelize([1, 2, 3])
def foo(date_range):
df = spark.createDataFrame([(1, 0, 3)], ["a", "b", "c"])
return 1
p.filter(foo).count()

You are better off with using Dataframes rather than RDD. Dataframe's read.csv api accepts list of paths like -
pathList = ['/path/to/data1900-01-01.csv','/path/to/data1900-01-02.csv']
df = spark.read.csv(pathList)
have a look at documentation for read.csv
You can form the list of paths to date files to your data files by doing some date operation over window of N days like "path/to/data"+datetime.today().strftime("%Y-%m-%d"))+.csv" (This will get you file name of today only but its not hard to figure out date calculation for N days)
However keep in mind that schema of all date csvs should be same for above to work.
edit : When you parallelize list of dates i.e. p, each date gets processed individually by different executors, so input to test_run2 wasnt really as list of dates, it was one individual string like 1995-01-01
Try this instead, see if this works.
# Get the list of dates
date_range = window(dates, N)
s = '/daily/data{}.csv'
dt0 = date_range[-1] # most recent file
df0 = spark.read.csv(s.format(dt0), header=True, mode='DROPMALFORM')
# read previous files
file_list = [s.format(dt) for dt in date_range[:-1]]
df1 = spark.read.csv(file_list, header=True, mode='DROPMALFORM')
r, resid = computeLinearRegression(df0,df1)
r.write.save('daily/r{}.csv'.format(dt0))
resid.write.save('/daily/resid{}.csv'.format(dt0))

Related

loop through a list of tables and save to csv with table name

I have a list of 120 tables and i want to save sample size of first 1000 and last 1000 rows from each table into individual csv files for each table.
How can this be done in code repo or code authoring.
The following code allows to save one table to csv, can anyone help with this to loop through list of tables from a project folder and create individual csv files for each table?
#transform(
my_input = Input('/path/to/input/dataset'),
my_output = Output('/path/to/output/dataset')
)
def compute_function(my_input, my_output):
my_output.write_dataframe(
my_input.dataframe(),
output_format = "csv",
options = {
"compression": "gzip"
}
)
psuedo code
list_of_tables = [table1,table2,table3,...table120]
for tables in list_of_tables:
table = table.limit(1000)
table.write_dataframe(table.dataframe(),output_format = "csv",
options = {
"compression": "gzip"
})
i was able to get it working for one table, how can i just loop through a list of tables and generate it ?
The code for one table
# to get the first and last rows
from transforms.api import transform_df, Input, Output
from pyspark.sql.functions import monotonically_increasing_id
from pyspark.sql.functions import col
table_name = 'stock'
#transform_df(
output=Output(f"foundry/sample/{table_name}_sample"),
my_input=Input(f"foundry/input/{table_name}"),
)
def compute_first_last_1000(my_input):
first_stock_df = my_input.withColumn("index", monotonically_increasing_id())
first_stock_df = first_stock_df.orderBy("index").filter(col("index") < 1000).drop("index")
last_stock_df = my_input.withColumn("index", monotonically_increasing_id())
last_stock_df = last_stock_df.orderBy("index").filter(col("index") < 1000).drop("index")
stock_df = first_stock_df.unionByName(last_stock_df)
return stock_df
# code to save as csv file
table_name = 'stock'
#transform(
output=Output(f"foundry/sample/{table_name}_sample_csv"),
my_input=Input(f"foundry/sample/{table_name}_sample"),
)
def my_compute_function(my_input, output):
df = my_input.dataframe()
with output.filesystem().open('stock.csv', 'w') as stream:
csv_writer = csv.writer(stream)
csv_writer.writerow(df.schema.names)
csv_writer.writerows(df.collect())
Your best strategy here would be to programatically generate your transforms, you can also do a multi output transform if you don't fancy creating 1000 transforms. Something like this (written live into the answer box, non tested code some sintax may be wrong):
# you can generate this programatically
my_inputs = [
'/path/to/input/dataset1',
'/path/to/input/dataset2',
'/path/to/input/dataset3',
# ...
]
for table_path in my_inputs:
#transform_df(
Output(table_path + '_out'),
df=Input(table_path))
def transform(df):
# your logic here
return df
If you need to read the table names rather than hard coding them, then you could use the os.listdir or the os.walk method.
I think the previous answer left out the part about exporting only the first and last N rows. If the table is converted to a dataframe df, then
dfoutput = df.head(N).append(df.tail(N)])
or
dfoutput = df[:N].append(df[-N:])

Python: Reading and Writing HUGE Json files

I am new to python. So please excuse me if I am not asking the questions in pythonic way.
My requirements are as follows:
I need to write python code to implement this requirement.
Will be reading 60 json files as input. Each file is approximately 150 GB.
Sample structure for all 60 json files is as shown below. Please note each file will have only ONE json object. And the huge size of each file is because of the number and size of the "array_element" array contained in that one huge json object.
{
"string_1":"abc",
"string_1":"abc",
"string_1":"abc",
"string_1":"abc",
"string_1":"abc",
"string_1":"abc",
"array_element":[]
}
Transformation logic is simple. I need to merge all the array_element from all 60 files and write it into one HUGE json file. That is almost 150GB X 60 will be the size of the output json file.
Questions for which I am requesting your help on:
For reading: Planning on using "ijson" module's ijson.items(file_object, "array_element"). Could you please tell me if ijson.items will "Yield" (that is NOT load the entire file into memory) one item at a time from "array_element" array in the json file? I dont think json.load is an option here because we cannot hold such a huge dictionalry in-memory.
For writing: I am planning to read each item using ijson.item, and do json.dumps to "encode" and then write it to the file using file_object.write and NOT using json.dump since I cannot have such a huge dictionary in memory to use json.dump. Could you please let me know if f.flush() applied in the code shown below is needed? To my understanding, the internal buffer will automatically get flushed by itself when it is full and the size of the internal buffer is constant and wont dynamically grow to an extent that it will overload the memory? please let me know
Are there any better approach to the ones mentioned above for incrementally reading and writing huge json files?
Code snippet showing above described reading and writing logic:
for input_file in input_files:
with open("input_file.json", "r") as f:
objects = ijson.items(f, "array_element")
for item in objects:
str = json.dumps(item, indent=2)
with open("output.json", "a") as f:
f.write(str)
f.write(",\n")
f.flush()
with open("output.json", "a") as f:
f.seek(0,2)
f.truncate(f.tell() - 1)
f.write("]\n}")
Hope I have asked my questions clearly. Thanks in advance!!
The following program assumes that the input files have a format that is predictable enough to skip JSON parsing for the sake of performance.
My assumptions, inferred from your description, are:
All files have the same encoding.
All files have a single position somewhere at the start where "array_element":[ can be found, after which the "interesting portion" of the file begins
All files have a single position somewhere at the end where ]} marks the end of the "interesting portion"
All "interesting portions" can be joined with commas and still be valid JSON
When all of these points are true, concatenating a predefined header fragment, the respective file ranges, and a footer fragment would produce one large, valid JSON file.
import re
import mmap
head_pattern = re.compile(br'"array_element"\s*:\s*\[\s*', re.S)
tail_pattern = re.compile(br'\s*\]\s*\}\s*$', re.S)
input_files = ['sample1.json', 'sample2.json']
with open('result.json', "wb") as result:
head_bytes = 500
tail_bytes = 50
chunk_bytes = 16 * 1024
result.write(b'{"JSON": "fragment", "array_element": [\n')
for input_file in input_files:
print(input_file)
with open(input_file, "r+b") as f:
mm = mmap.mmap(f.fileno(), 0)
start = head_pattern.search(mm[:head_bytes])
end = tail_pattern.search(mm[-tail_bytes:])
if not (start and end):
print('unexpected file format')
break
start_pos = start.span()[1]
end_pos = mm.size() - end.span()[1] + end.span()[0]
if input_files.index(input_file) > 0:
result.write(b',\n')
pos = start_pos
mm.seek(pos)
while True:
if pos + chunk_bytes >= end_pos:
result.write(mm.read(end_pos - pos))
break
else:
result.write(mm.read(chunk_bytes))
pos += chunk_bytes
result.write(b']\n}')
If the file format is 100% predictable, you can throw out the regular expressions and use mm[:head_bytes].index(b'...') etc for the start/end position arithmetic.

Save content of Spark DataFrame as a single CSV file [duplicate]

This question already has answers here:
Write single CSV file using spark-csv
(16 answers)
Closed 4 years ago.
Say I have a Spark DataFrame which I want to save as CSV file. After Spark 2.0.0 , DataFrameWriter class directly supports saving it as a CSV file.
The default behavior is to save the output in multiple part-*.csv files inside the path provided.
How would I save a DF with :
Path mapping to the exact file name instead of folder
Header available in first line
Save as a single file instead of multiple files.
One way to deal with it, is to coalesce the DF and then save the file.
df.coalesce(1).write.option("header", "true").csv("sample_file.csv")
However this has disadvantage in collecting it on Master machine and needs to have a master with enough memory.
Is it possible to write a single CSV file without using coalesce ? If not, is there a efficient way than the above code ?
Just solved this myself using pyspark with dbutils to get the .csv and rename to the wanted filename.
save_location= "s3a://landing-bucket-test/export/"+year
csv_location = save_location+"temp.folder"
file_location = save_location+'export.csv'
df.repartition(1).write.csv(path=csv_location, mode="append", header="true")
file = dbutils.fs.ls(csv_location)[-1].path
dbutils.fs.cp(file, file_location)
dbutils.fs.rm(csv_location, recurse=True)
This answer can be improved by not using [-1], but the .csv seems to always be last in the folder. Simple and fast solution if you only work on smaller files and can use repartition(1) or coalesce(1).
Use:
df.toPandas().to_csv("sample_file.csv", header=True)
See documentation for details:
https://spark.apache.org/docs/latest/api/python/pyspark.sql.html?highlight=dataframe#pyspark.sql.DataFrame.toPandas
df.coalesce(1).write.option("inferSchema","true").csv("/newFolder",header =
'true',dateFormat = "yyyy-MM-dd HH:mm:ss")
The following scala method works in local or client mode, and writes the df to a single csv of the chosen name. It requires that the df fit into memory, otherwise collect() will blow up.
import org.apache.hadoop.fs.{FileSystem, Path}
val SPARK_WRITE_LOCATION = some_directory
val SPARKSESSION = org.apache.spark.sql.SparkSession
def saveResults(results : DataFrame, filename: String) {
var fs = FileSystem.get(this.SPARKSESSION.sparkContext.hadoopConfiguration)
if (SPARKSESSION.conf.get("spark.master").toString.contains("local")) {
fs = FileSystem.getLocal(new conf.Configuration())
}
val tempWritePath = new Path(SPARK_WRITE_LOCATION)
if (fs.exists(tempWritePath)) {
val x = fs.delete(new Path(SPARK_WRITE_LOCATION), true)
assert(x)
}
if (results.count > 0) {
val hadoopFilepath = new Path(SPARK_WRITE_LOCATION, filename)
val writeStream = fs.create(hadoopFilepath, true)
val bw = new BufferedWriter( new OutputStreamWriter( writeStream, "UTF-8" ) )
val x = results.collect()
for (row : Row <- x) {
val rowString = row.mkString(start = "", sep = ",", end="\n")
bw.write(rowString)
}
bw.close()
writeStream.close()
val resultsWritePath = new Path(WRITE_DIRECTORY, filename)
if (fs.exists(resultsWritePath)) {
fs.delete(resultsWritePath, true)
}
fs.copyToLocalFile(false, hadoopFilepath, resultsWritePath, true)
} else {
System.exit(-1)
}
}
This solution is based on a Shell Script and is not parallelized, but is still very fast, especially on SSDs. It uses cat and output redirection on Unix systems. Suppose that the CSV directory containing partitions is located on /my/csv/dir and that the output file is /my/csv/output.csv:
#!/bin/bash
echo "col1,col2,col3" > /my/csv/output.csv
for i in /my/csv/dir/*.csv ; do
echo "Processing $i"
cat $i >> /my/csv/output.csv
rm $i
done
echo "Done"
It will remove each partition after appending it to the final CSV in order to free space.
"col1,col2,col3" is the CSV header (here we have three columns of name col1, col2 and col3). You must tell Spark to don't put the header in each partition (this is accomplished with .option("header", "false") because the Shell Script will do it.
For those still wanting to do this here's how I got it done using spark 2.1 in scala with some java.nio.file help.
Based on https://fullstackml.com/how-to-export-data-frame-from-apache-spark-3215274ee9d6
val df: org.apache.spark.sql.DataFrame = ??? // data frame to write
val file: java.nio.file.Path = ??? // target output file (i.e. 'out.csv')
import scala.collection.JavaConversions._
// write csv into temp directory which contains the additional spark output files
// could use Files.createTempDirectory instead
val tempDir = file.getParent.resolve(file.getFileName + "_tmp")
df.coalesce(1)
.write.format("com.databricks.spark.csv")
.option("header", "true")
.save(tempDir.toAbsolutePath.toString)
// find the actual csv file
val tmpCsvFile = Files.walk(tempDir, 1).iterator().toSeq.find { p =>
val fname = p.getFileName.toString
fname.startsWith("part-00000") && fname.endsWith(".csv") && Files.isRegularFile(p)
}.get
// move to desired final path
Files.move(tmpCsvFile, file)
// delete temp directory
Files.walk(tempDir)
.sorted(java.util.Comparator.reverseOrder())
.iterator().toSeq
.foreach(Files.delete(_))
The FileUtil.copyMerge() from the Hadoop API should solve your problem.
import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.fs._
def merge(srcPath: String, dstPath: String): Unit = {
val hadoopConfig = new Configuration()
val hdfs = FileSystem.get(hadoopConfig)
FileUtil.copyMerge(hdfs, new Path(srcPath), hdfs, new Path(dstPath), true, hadoopConfig, null)
// the "true" setting deletes the source files once they are merged into the new output
}
See Write single CSV file using spark-csv
This is how distributed computing work! Multiple files inside a directory is exactly how distributed computing works, this is not a problem at all since all software can handle it.
Your question should be "how is it possible to download a CSV composed of multiple files?" -> there are already lof of solutions in SO.
Another approach could be to use Spark as a JDBC source (with the awesome Spark Thrift server), write a SQL query and transform the result to CSV.
In order to prevent OOM in the driver (since the driver will get ALL
the data), use incremental collect
(spark.sql.thriftServer.incrementalCollect=true), more info at
http://www.russellspitzer.com/2017/05/19/Spark-Sql-Thriftserver/.
Small recap about Spark "data partition" concept:
INPUT (X PARTITIONs) -> COMPUTING (Y PARTITIONs) -> OUTPUT (Z PARTITIONs)
Between "stages", data can be transferred between partitions, this is the "shuffle". You want "Z" = 1, but with Y > 1, without shuffle? this is impossible.

feeding a dataframe created from a CSV to MLlib Kmeans: IndexError: list index out of range

because i can not use spark csv i have manually created a dataframe from CSV as follow:
raw_data=sc.textFile("data/ALS.csv").cache()
csv_data=raw_data.map(lambda l:l.split(","))
header=csv_data.first()
csv_data=csv_data.filter(lambda line:line !=header)
row_data=csv_data.map(lambda p :Row (
location_history_id=p[0],
user_id=p[1],
latitude=p[2],
longitude=p[3],
address=p[4],
created_at=p[5],
valid_until=p[6],
timezone_offset_secs=p[7],
opening_times_id=p[8],
timezone_id=p[9]))
location_df = sqlContext.createDataFrame(row_data)
location_df.registerTempTable("locations")
i need only two columns :
lati_longi_df=sqlContext.sql("""SELECT latitude, longitude FROM locations""")
rdd_lati_longi = lati_longi_df.map(lambda data: Vectors.dense([float(c) for c in data]))
rdd_lati_longi.take(2):
[DenseVector([-6.2416, 106.7949]),
DenseVector([-6.2443, 106.7956])]
now it seems that every thing is ready for KMeans training:
clusters = KMeans.train(rdd_lati_longi, 10, maxIterations=30,
runs=10, initializationMode="random")
but i get the following error:
IndexError: list index out of range
First three lines of ALS.csv:
location_history_id,user_id,latitude,longitude,address,created_at,valid_until,timezone_offset_secs,opening_times_id,timezone_id
Why don't you allow spark to parse csv instead? You can enable csv support with something like this:
pyspark --packages com.databricks:spark-csv_2.10:1.4.0

Parallel programming in R

I have a file that consists of multiple JSON objects. I need to read through these files and extract certain fields from the JSON objects. To complicate things, some of the objects do not contain all the fields. I am dealing with a large file of over 200,000 JSON objects. I would like to split job across multiple cores. I have tried to experiment with doSNOW, foreach, and parallel and really do not understand how to do this. The following is my code that I would like to make more efficient.
foreach (i in 2:length(linn)) %dopar% {
json_data <- fromJSON(linn[i])
if(names(json_data)[1]=="info")
next
mLocation <- ifelse('location' %!in% names(json_data$actor),'NULL',json_data$actor$location$displayName)
mRetweetCount <- ifelse('retweetCount' %!in% names(json_data),0,json_data$retweetCount)
mGeo <- ifelse('geo' %!in% names(json_data),c(-0,-0),json_data$geo$coordinates)
tweet <- rbind(tweet,
data.frame(
record.no = i,
id = json_data$id,
objecttype = json_data$actor$objectType,
postedtime = json_data$actor$postedTime,
location = mLocation,
displayname = json_data$generator$displayName,
link = json_data$generator$link,
body = json_data$body,
retweetcount = mRetweetCount,
geo = mGeo)
)
}
Rather than trying to parallelize an iteration, I think you're better off trying to vectorize (hmm, actually most of the below is still iterating...). For instance here we get all our records (no speed gain yet, though see below...)
json_data <- lapply(linn, fromJSON)
For location we pre-allocate a vector of NAs to represent records for which there is no location, then find records that do have a location (maybe there's a better way of doing this...) and update them
mLocation <- rep(NA, length(json_data))
idx <- sapply(json_data, function(x) "location" %in% names(x$actor))
mLocation[idx] <- sapply(json_data[idx], function(x) x$location$displayName)
Finally, create a 200,000 row data frame in a single call (rather than your 'copy and append' pattern, which makes a copy of the first row, then the first and second row, then the first, second, third row, then ... so N-squared rows, in addition to recreating factors and other data.frame specific expenses; this is likely where you spend most of your time)
data.frame(i=seq_along(json_data), location=mLocation)
The idea would be to accumulate all the columns, and then do just one call to data.frame(). I think you could cheat on parsing line-at-a-time, by pasting everything into a single string repersenting a JSON array, and parsing in one call
json_data <- fromJSON(sprintf("[%s]", paste(linn, collapse=",")))