Spark DStream has method saveAsTextFiles(prefix, [suffix]) which can be used to save data on HDFS but this function does not accept any path parameter.
myDStream.saveAsTextFiles("prefix_","_suffix")
By default , it is saving data into current logged in user directory on HDFS i.e. if you are running application with root user then data is stored in
/user/root/prefix_TIMESTAMP_suffx
How do I change output directory?
Thanks
Give it a path to the desired HDFS directory as the prefix argument:
myDStream.saveAsTextFiles("hdfs://my/custom/path/prefix_","_suffix")
Related
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 some results from a Spark application saved in the HDFS as files called part-r-0000X (X= 0, 1, etc.). And, because I want to join the whole content in a file, I'm using the following command:
hdfs dfs -getmerge srcDir destLocalFile
The previous command is used in a bash script which makes empty the output directory (where the part-r-... files are saved) and, inside a loop, executes the above getmerge command.
The thing is I need to use the resultant file in another Spark program which need that merged file as input in the HDFS. So I'm saving it as local and then I upload it to the HDFS.
I've thought another option which is write the file from the Spark program in this way:
outputData.coalesce(1, false).saveAsTextFile(outPathHDFS)
But I've read coalesce() doesn't help with the performance.
Any other ideas? suggestions? Thanks!
You wish to merge all the files into a single one so that you can load all the files at once into a Spark rdd, is my guess.
Let the files be in Parts(0,1,....) in HDFS.
Why not load it with wholetextFiles, which actually does what you need.
wholeTextFiles(path, minPartitions=None, use_unicode=True)[source]
Read a directory of text files from HDFS, a local file system (available on all nodes), or any Hadoop-supported file system URI. Each file is read as a single record and returned in a key-value pair, where the key is the path of each file, the value is the content of each file.
If use_unicode is False, the strings will be kept as str (encoding as utf-8), which is faster and smaller than unicode. (Added in Spark 1.2)
For example, if you have the following files:
hdfs://a-hdfs-path/part-00000 hdfs://a-hdfs-path/part-00001 ... hdfs://a-hdfs-path/part-nnnnn
Do rdd = sparkContext.wholeTextFiles(“hdfs://a-hdfs-path”), then rdd contains:
(a-hdfs-path/part-00000, its content) (a-hdfs-path/part-00001, its content) ... (a-hdfs-path/part-nnnnn, its content)
Try SPARK BucketBy.
This is a nice feature via df.write.saveAsTable(), but this format can only be read by SPARK. Data shows up in Hive metastore but cannot be read by Hive, IMPALA.
The best solution that I've found so far was:
outputData.saveAsTextFile(outPath, classOf[org.apache.hadoop.io.compress.GzipCodec])
Which saves the outputData in compressed part-0000X.gz files under the outPath directory.
And, from the other Spark app, it reads those files using this:
val inputData = sc.textFile(inDir + "part-00*", numPartition)
Where inDir corresponds to the outPath.
Scala 2.12 and Spark 2.2.1 here. I used the following code to write the contents of a DataFrame to S3:
myDF.write.mode(SaveMode.Overwrite)
.parquet("s3n://com.example.mybucket/mydata.parquet")
When I go to com.example.mybucket on S3 I actually see a directory called "mydata.parquet", as well as file called "mydata.parquet_$folder$"!!! If I go into the mydata.parquet directory I see two files under it:
_SUCCESS; and
part-<big-UUID>.snappy.parquet
Whereas I was just expecting to see a single file called mydata.parquet living in the root of the bucket.
Is something wrong here (if so, what?!?) or is this expected with the Parquet file format? If its expected, which is the actual Parquet file that I should read from:
mydata.parquet directory?; or
mydata.parquet_$folder$ file?; or
mydata.parquet/part-<big-UUID>.snappy.parquet?
Thanks!
The mydata.parquet/part-<big-UUID>.snappy.parquet is the actual parquet data file. However, often tools like Spark break data sets into multiple part files, and expect to be pointed to a directory that contains multiple files. The _SUCCESS file is a simple flag indicating that the write operation has completed.
According to the api to save the parqueat file it saves inside the folder you provide. Sucess is incidation that the process is completed scuesffuly.
S3 create those $folder if you write directly commit to s3. What happens is it writes to temporory folders and copies to the final destination inside the s3. The reason is there no concept of rename.
Look at the s3-distcp and also DirectCommiter for performance issue.
The $folder$ marker is used by s3n/amazon's emrfs to indicate "empty directory". ignore.
The _SUCCESS file is, as the others note, a 0-byte file. ignore
all other .parquet files in the directory are the output; the number you end up with depends on the number of tasks executed on the input
When spark uses a directory (tree) as a source of data, all files beginning with _ or . are ignored; s3n will strip out those $folder$ things too. So if you use the path for a new query, it will only pick up that parquet file.
I had this code running before
df = sc.wholeTextFiles('./dbs-*.json,./uob-*.json').flatMap(lambda x: flattenTransactionFile(json.loads(x[1]))).toDF()
But it appears that now, I get
Py4JJavaError: An error occurred while calling o24.partitions.
: org.apache.hadoop.mapreduce.lib.input.InvalidInputException: Input Pattern hdfs://localhost:9000/user/jiewmeng/dbs-*.json matches 0 files
Input Pattern hdfs://localhost:9000/user/jiewmeng/uob-*.json matches 0 files
at org.apache.hadoop.mapreduce.lib.input.FileInputFormat.singleThreadedListStatus(FileInputFormat.java:330)
at org.apache.hadoop.mapreduce.lib.input.FileInputFormat.listStatus(FileInputFormat.java:272)
at org.apache.spark.input.WholeTextFileInputFormat.setMinPartitions(WholeTextFileInputFormat.scala:55)
Looks like spark is trying to use Hadoop? How can I use a local file? Also why the sudden failure? Since I managed to use ./dbs-*.json before?
By default, the location of the file is relative to your directory in HDFS. In order to refer local file system, you need to use file:///your_local_path
For e.g. in cloudera VM, if I say
sc.textFile('myfile')
it will assume the HDFS path
/user/cloudera/myfile
where as to mention my local home directory I would say
sc.textFile('file:///home/cloudera/myfile')
I wrote a spark application in which i want to save dataframe in local filesystem.Spark needs to write a file in local filesystem. Then I use java.io.FileReader and FileWriter to read the local file written by spark , do some modification and then write it back in again in local filesystem. So the filepath I need to use is constant . for ex: file:////name.txt , This will be used for both dataframa.save and java fileReader and fileWriter
i used api like this:
dataframe.save(/abc/name.txt)
But spark is saving this file into HDFS. Do we have any env variable which needs to be set to make spark save file into local fs??
Thanks
try dataframe.save("file:///<LOCAL_PATH>/name.txt)