I need to perform update insert (Upsert) on old data with new data.
Pseudo Code:
old_data = spark.read.parquet('s3://bucket/old_data/')
new_data= spark.read.parquet('s3://bucket/new_data/')
common_records = old_data.join(new_data,on=opk,how="inner")
non_match_records = old_data.join(new_data,on=opk,how="left_anti")
new_records = new_data.join(old_data,on=opk,how="left_anti")
dfs = [common_records , non_match_records , new_records ]
final_data = reduce(DataFrame.unionAll, dfs)
final_data .cache()
final_data.write.parquet('s3://bucket/old_data/')
Error :
Even I cached data, Still It's Looking for old_data path, Is there anyway to directly write to old data s3 path.
I have tried it to write to some temp path and read from it and write to main path Like below It worked but It's taking time process when I have data in Billons.
final_data.write.parquet('s3://bucket/temp/')
df = spark.read.parquet('s3://bucket/temp/')
df.write.parquet('s3://bucket/old_data/')
I want to reduce this Temp writing & reading part.
Thanks in advance :)
You need to perform an action to trigger dataframe caching. Thus you should modify the last lines of your code snippet as follow:
...
final_data = final_data.cache()
final_data.count()
final_data.write.parquet('s3://bucket/old_data/')
By performing a count action on your dataframe, you trigger caching process and then be able to write on same directory from where your read.
However, I don't know if this will improve performance of your application as cache fallback to disk write when dataframe is too huge for memory. If your use case is updating parquet files, I advise you to look at DeltaLake that was created to solve this.
I am using spark streaming and I want to save each batch of spark streaming on my local in Avro format. I have used saveAsNewAPIHadoopFile to save data in Avro format. This works well. But it overwrites the existing file. Next batch data will overwrite the old data. Is there any way to save Avro file in common directory? I tried by adding some properties of Hadoop job conf for adding a prefix in the file name. But not working any properties.
dstream.foreachRDD {
rdd.saveAsNewAPIHadoopFile(
path,
classOf[AvroKey[T]],
classOf[NullWritable],
classOf[AvroKeyOutputFormat[T]],
job.getConfiguration()
)
}
Try this -
You can make your process split into 2 steps :
Step-01 :- Write Avro file using saveAsNewAPIHadoopFile to <temp-path>
Step-02 :- Move file from <temp-path> to <actual-target-path>
This will definitely solve your problem for now. I will share my thoughts if I get to fulfill this scenario in one step instead of two.
Hope this is helpful.
I have a directory in an azure data lake that has the following path:
'adl://home/../psgdata/clusters/iptiqadata-prod-cluster-eus2-01/psgdata/mib'
Within this directory there are a number of other directories (50) that have the format 20190404.
The directory 'adl://home/../psgdata/clusters/iptiqadata-prod-cluster-eus2-01/psgdata/mib/20180404' contains 100 or so xml files which I am working with.
I can create an rdd for each of the sub-folders which works fine, but ideally I want to pass only the top path, and have spark recursively find the files. I have read other SO posts and tried using a wildcard thus:
pathWild = 'adl://home/../psgdata/clusters/iptiqadata-prod-cluster-eus2-01/psgdata/mib/*'
rdd = sc.wholeTextFiles(pathWild)
rdd.count()
But it just freezes and does nothing at all, seems to completely destroy the kernel. I am working in Jupyter on Spark 2.x. New to spark. Thanks!
Try this:
pathWild = 'adl://home/../psgdata/clusters/iptiqadata-prod-cluster-eus2-01/psgdata/mib/*/*'
response = "mi_or_chd_5"
outcome = sqlc.sql("""select eid,{response} as response
from outcomes
where {response} IS NOT NULL""".format(response=response))
outcome.write.parquet(response, mode="overwrite") # Success
print outcome.schema
StructType(List(StructField(eid,IntegerType,true),StructField(response,ShortType,true)))
But then:
outcome2 = sqlc.read.parquet(response) # fail
fails with:
AnalysisException: u'Unable to infer schema for Parquet. It must be specified manually.;'
in
/usr/local/lib/python2.7/dist-packages/pyspark-2.1.0+hadoop2.7-py2.7.egg/pyspark/sql/utils.pyc in deco(*a, **kw)
The documentation for parquet says the format is self describing, and the full schema was available when the parquet file was saved. What gives?
Using Spark 2.1.1. Also fails in 2.2.0.
Found this bug report, but was fixed in
2.0.1, 2.1.0.
UPDATE: This work when on connected with master="local", and fails when connected to master="mysparkcluster".
This error usually occurs when you try to read an empty directory as parquet.
Probably your outcome Dataframe is empty.
You could check if the DataFrame is empty with outcome.rdd.isEmpty() before writing it.
In my case, the error occured because I was trying to read a parquet file which started with an underscore (e.g. _lots_of_data.parquet). Not sure why this was an issue, but removing the leading underscore solved the problem.
See also:
Re: Spark-2.0.0 fails reading a parquet dataset generated by Spark-1.6.2
I'm using AWS Glue and I received this error while reading data from a data catalog table (location: s3 bucket).
After bit of analysis I realised that, this is due to file not available in file location(in my case s3 bucket path).
Glue was trying to apply data catalog table schema on a file which doesn't exist.
After copying file into s3 bucket file location, issue got resolved.
Hope this helps someone who encounters/encountered an error in AWS Glue.
This case occurs when you try to read a table that is empty. If the table had correctly inserted data, there should be no problem.
Besides that with parquet, the same thing happens with ORC.
I see there are already so many Answers. But the issue I faced was my Spark job was trying to read a file which is being overwritten by another Spark job that was previously started. It sounds bad, but I did that mistake.
Just to emphasize #Davos answer in a comment, you will encounter this exact exception error, if your file name has a dot . or an underscore _ at start of the filename
val df = spark.read.format("csv").option("delimiter", "|").option("header", "false")
.load("/Users/myuser/_HEADER_0")
org.apache.spark.sql.AnalysisException:
Unable to infer schema for CSV. It must be specified manually.;
Solution is to rename the file and try again (e.g. _HEADER rename to HEADER)
Happened to me for a parquet file that was in the process of being written to.
Just need to wait for it to be completely written.
In my case, the error occurred because the filename contained underscores. Rewriting / reading the file without underscores (hyphens were OK) solved the problem...
For me this happened when I thought loading the correct file path but instead pointed a incorrect folder
I ran into a similar problem with reading a csv
spark.read.csv("s3a://bucket/spark/csv_dir/.")
gave an error of:
org.apache.spark.sql.AnalysisException: Unable to infer schema for CSV. It must be specified manually.;
I found if I removed the trailing . and then it works. ie:
spark.read.csv("s3a://bucket/spark/csv_dir/")
I tested this for parquet adding a trailing . and you get an error of:
org.apache.spark.sql.AnalysisException: Unable to infer schema for Parquet. It must be specified manually.;
I just encountered the same problem but none of the solutions here work for me. I try to merge the row groups of my parquet files on hdfs by first reading them and write it to another place using:
df = spark.read.parquet('somewhere')
df.write.parquet('somewhere else')
But later when I query it with
spark.sql('SELECT sth FROM parquet.`hdfs://host:port/parquetfolder/` WHERE .. ')
It shows the same problem.
I finally solve this by using pyarrow:
df = spark.read.parquet('somewhere')
pdf = df.toPandas()
adf = pa.Table.from_pandas(pdf) # import pyarrow as pa
fs = pa.hdfs.connect()
fw = fs.open(path, 'wb')
pq.write_table(adf, fw) # import pyarrow.parquet as pq
fw.close()
Check if .parquet files available at the response path. I am assuming, either files don't exist or it may be exist in some internal(partitioned) folders. If files are available under multiple hierarchy folders then append /* for each folder.
As in my case .parquet files were under 3 folders from base_path, so I have given path as base_path/*/*/*
As others mentioned, in my case this error appeared when I was reading S3 keys that did not exist.
A solution is filter-in keys that do exist:
import com.amazonaws.services.s3.AmazonS3URI
import org.apache.hadoop.fs.{FileSystem, Path}
import org.apache.spark.sql.SparkSession
import java.net.URI
def addEndpointToUrl(url: String, domain: String = "s3.amazonaws.com"): String = {
val uri = new URI(url)
val hostWithEndpoint = uri.getHost + "." + domain
new URI(uri.getScheme, uri.getUserInfo, hostWithEndpoint, uri.getPort, uri.getPath, uri.getQuery, uri.getFragment).toString
}
def createS3URI(url: String): AmazonS3URI = {
try {
// try to instantiate AmazonS3URI with url
new AmazonS3URI(url)
} catch {
case e: IllegalArgumentException if e.getMessage.
startsWith("Invalid S3 URI: hostname does not appear to be a valid S3 endpoint") => {
new AmazonS3URI(addEndpointToUrl(url))
}
}
}
def s3FileExists(spark: SparkSession, url: String): Boolean = {
val amazonS3Uri: AmazonS3URI = createS3URI(url)
val s3BucketUri = new URI(s"${amazonS3Uri.getURI().getScheme}://${amazonS3Uri.getBucket}")
FileSystem
.get(s3BucketUri, spark.sparkContext.hadoopConfiguration)
.exists(new Path(url))
}
and you can use it as:
val partitions = List(yesterday, today, tomorrow)
.map(f => somepath + "/date=" + f)
.filter(f => s3FileExists(spark, f))
val df = spark.read.parquet(partitions: _*)
For that solution I took some code out of spark-redshift project, here.
You are just loading a parquet file , Of course parquet had valid
schema. Otherwise it would not been saved as parquet. This error means
-
Either parquet file does not exist . (99.99% cases this is the issue. Spark error messages are often less obvious)
Somehow the parquet file got corrupted or Or It's not a parquet file at all
I ran into this issue because of folder in folder issue.
for example folderA.parquet was supposed to have partion.... but instead it have folderB.parquet which inside have partition.
Resolution,
transfer the file to parent folder and delete the subfolder.
you can read with /*
outcome2 = sqlc.read.parquet(f"{response}/*") # work for me
My problem was moving the files to another location and not changing the path in the files in the folder _spark_metadata.
To fix it do something like:
sed -i "s|/old/path|/new/path|g" ** .*
Seems this issue can be caused by a lot of reasons, I am providing another scenario:
By default the spark parquet source is using "partition inferring" which means it requires the file path to be partition in Key=Value pairs and the loads happens at the root.
To avoid this, if we assure all the leaf files have identical schema, then we can use
df = spark.read.format("parquet")\
.option("recursiveFileLookup", "true")
to disable the "partition inferring" manually. It basically load files one by one using the parquet's logical_type.
I tried the below spark scala code and got the output as mentioned below.
I have tried to pass the inputs to script, but it didn't receive and when i used collect the print statement i used in the script appeared twice.
My simple and very basic perl script first:
#!/usr/bin/perl
print("arguments $ARGV[0] \n"); // Just print the arguments.
My Spark code:
object PipesExample {
def main(args:Array[String]){
val conf = new SparkConf();
val sc = new SparkContext(conf);
val distScript = "/home/srinivas/test.pl"
sc.addFile(distScript)
val rdd = sc.parallelize(Array("srini"))
val piped = rdd.pipe(Seq(SparkFiles.get("test.pl")))
println(" output " + piped.collect().mkString(" "));
}
}
Output looked like this..
output arguments arguments
1) What mistake i have done to make it fail receiving the arguments.?
2) Why it executed twice.?
If it looks too basic, please apologize me. I was trying to understand to the best and want to clear my doubts.
From my experience, it is executed twice because spark divides your RDD into two partitions and each partition is passed to your external script.
The reason your application couldn't pick test.pl file is, the file is in some node's location. But the application master is created in one of the nodes in cluster. So prolly if the file isn't in that node, it can't pick the file.
You should always save the file in HDFS or S3 to access the external files. Or else pass the HDFS file location through spark command options.