I'm trying to write a dataframe in spark to an HDFS location and I expect that if I'm adding the partitionBy notation Spark will create partition
(similar to writing in Parquet format)
folder in form of
partition_column_name=partition_value
( i.e partition_date=2016-05-03). To do so, I ran the following command :
(df.write
.partitionBy('partition_date')
.mode('overwrite')
.format("com.databricks.spark.csv")
.save('/tmp/af_organic'))
but partition folders had not been created
any idea what sould I do in order for spark DF automatically create those folders?
Thanks,
Spark 2.0.0+:
Built-in csv format supports partitioning out of the box so you should be able to simply use:
df.write.partitionBy('partition_date').mode(mode).format("csv").save(path)
without including any additional packages.
Spark < 2.0.0:
At this moment (v1.4.0) spark-csv doesn't support partitionBy (see databricks/spark-csv#123) but you can adjust built-in sources to achieve what you want.
You can try two different approaches. Assuming your data is relatively simple (no complex strings and need for character escaping) and looks more or less like this:
df = sc.parallelize([
("foo", 1, 2.0, 4.0), ("bar", -1, 3.5, -0.1)
]).toDF(["k", "x1", "x2", "x3"])
You can manually prepare values for writing:
from pyspark.sql.functions import col, concat_ws
key = col("k")
values = concat_ws(",", *[col(x) for x in df.columns[1:]])
kvs = df.select(key, values)
and write using text source
kvs.write.partitionBy("k").text("/tmp/foo")
df_foo = (sqlContext.read.format("com.databricks.spark.csv")
.options(inferSchema="true")
.load("/tmp/foo/k=foo"))
df_foo.printSchema()
## root
## |-- C0: integer (nullable = true)
## |-- C1: double (nullable = true)
## |-- C2: double (nullable = true)
In more complex cases you can try to use proper CSV parser to preprocess values in a similar way, either by using UDF or mapping over RDD, but it will be significantly more expensive.
If CSV format is not a hard requirement you can also use JSON writer which supports partitionBy out-of-the-box:
df.write.partitionBy("k").json("/tmp/bar")
as well as partition discovery on read.
Related
I read a .csv file to create a data frame and I want to write the data to a kafka topic. The code is the following
df = spark.read.format("csv").option("header", "true").load(f'{file_location}')
kafka_df = df.selectExpr("to_json(struct(*)) AS value").selectExpr("CAST(value AS STRING)")
kafka_df.show(truncate=False)
And the data frame looks like this:
value
"{""id"":""d215e9f1-4d0c-42da-8f65-1f4ae72077b3"",""latitude"":""-63.571457254062715"",""longitude"":""-155.7055842710919""}"
"{""id"":""ca3d75b3-86e3-438f-b74f-c690e875ba52"",""latitude"":""-53.36506636464281"",""longitude"":""30.069167069917597""}"
"{""id"":""29e66862-9248-4af7-9126-6880ceb3b45f"",""latitude"":""-23.767505281795835"",""longitude"":""174.593140405442""}"
"{""id"":""451a7e21-6d5e-42c3-85a8-13c740a058a9"",""latitude"":""13.02054867061598"",""longitude"":""20.328402498420786""}"
"{""id"":""09d6c11d-7aae-4d17-8cd8-183157794893"",""latitude"":""-81.48976715040848"",""longitude"":""1.1995769642056189""}"
"{""id"":""393e8760-ef40-482a-a039-d263af3379ba"",""latitude"":""-71.73949722379649"",""longitude"":""112.59922770487054""}"
"{""id"":""d6db8fcf-ee83-41cf-9ec2-5c2909c18534"",""latitude"":""-4.034680969008576"",""longitude"":""60.59645511854336""}"
After I wrote it to Kafka I want to read it and transform the binary data from column "value" back to json string but the result is that the value contains only the id, not the whole string. Any ideea why?
from pyspark.sql import functions as F
df = consume_from_event_hub(topic, bootstrap_servers, config, consumer_group)
string_df = df.select(F.col("value").cast("string"))
string_df.display()
value
794541bc-30e6-4c16-9cd0-3c5c8995a3a4
20ea5b50-0baa-47e3-b921-f9a3ac8873e2
598d2fc1-c919-4498-9226-dd5749d92fc5
86cd5b2b-1c57-466a-a3c8-721811ab6959
807de968-c070-4b8b-86f6-00a865474c35
e708789c-e877-44b8-9504-86fd9a20ef91
9133a888-2e8d-4a5a-87ce-4a53e63b67fc
cd5e3e0d-8b02-45ee-8634-7e056d49bf3b
the CSV the format is this
id,latitude,longitude
bd6d98e1-d1da-4f41-94ba-8dbd8c8fce42,-86.06318155350924,-108.14300138138589
c39e84c6-8d7b-4cc5-b925-68a5ea406d52,74.20752175171859,-129.9453606091319
011e5fb8-6ab7-4ee9-97bb-acafc2c71e15,19.302250885973592,-103.2154291337162
You need to remove selectExpr("CAST(value AS STRING)") since to_json already returns a string column
from pyspark.sql.functions import col, to_json, struct
df = spark.read.format("csv").option("header", "true").option("inferSchema", "true").load(f'{file_location}')
kafka_df = df.select(to_json(struct(col("*"))).alias("value"))
kafka_df.show(truncate=False)
I'm not sure what's wrong with the consumer. That should have worked unless consume_from_event_hub does something specifically to extract the ID column
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.
When I create a dataframe from json file, the fields from the json file are sorted by default in the dataframe. How to avoid this sorting?
Jsonfile having one json message per line:
{"name":"john","age":10,"class":2}
{"name":"rambo","age":11,"class":3}
When I create Data frame from this file as:
val jDF = sqlContext.read.json("/user/inputfiles/sample.json")
a DF is created as jDF: org.apache.spark.sql.DataFrame = [age: bigint, class: bigint, name: string]
. In the DF the fields are sorted by default.
How do we avoid this from happening?
Im unable to understand what is going wrong here.
Appreciate any help in sorting out the problem.
For Question 1:
A simple way is to do select on the DataFrame:
val newDF = jDF.select("name","age","class")
The order of parameters is the order of the columns you want.
But this could be verbose if there are many columns and you have to define the order yourself.
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
I have a file on s3 in json format(filename=a). I read it and create a dataframe (df) using sqlContext.read.json. On checking df.printSchema; the schema is not what I want. So I specify my own schema with double and string type.
Then I reload the json data in a dataframe (df3) specifying the above schema but when I do df3.head(1) I see "None" values for some of my variables.
See code below -
df = sqlContext.read.json(os.path.join('file:///data','a'))
print df.count()
df.printSchema()
df.na.fill(0)
After specifying my own schema (sch). Since the schema code is long I haven't included it here.
sch=StructType(List(StructField(x,DoubleType,true),StructField(y,DoubleType,true)))
f = sc.textFile(os.path.join('file:///data','a'))
f_json = f.map(lambda x: json.loads(x))
df3 = sqlContext.createDataFrame(f_json, sch)
df3.head(1)
[Row(x=85.7, y=None)]
I obtain 'None' values for all my columns with DoubleType (datatype) when I do df3.head(1).Am I doing something wrong when I reload the df3 dataframe?
I was able to take care of "None" by doing df.na.fill(0)!