I'd like to print rdd data using scala such as below
res1.foreach{case(userid,tags)=>println(s"${userid}${"\t"}${tags.topicInterests.map(_.id).mkString(",")}")}
And now ,i want to save the detail to local file instead of println,how can i implement it?
Use saveAsTextFile() method of the RDD as shown below:
val strRdd = res1.map{case(userid,tags)=>(s"${userid}${"\t"}${tags.topicInterests.map(_.id).mkString(",")}")}
strRdd.saveAsTextFile("/home/test_user/result")
Note that, saveAsTextFile method takes a path(absolute or relative) to a folder/directory and not a file. The RDD data will be written as part files inside the given directory. In this case, a directory called result will be created with part files inside it.
There will be as many part files as the number of partitions in the strRdd. If the path /home/test_user/result already exists, your code will fail. So you will have to use a non-existing directory only.
Bonus info: The same saveAsTextFile method also works on other file systems like HDFS, S3 etc by taking the URL to the target directories instead of just paths.
Related
I want to update a CSV file depending on some condition, for that I read the file, made all the needed update, however when I tried to write it I'm getting a FileNotFoundException.
I think that it is due to the writing process, because when I access the path (where the input/output file were located) I find it empty.
Is there a better way to update a file? And if not, how can I resolve the FileNotFoundException error?
you can do it either by writing a temporary table/csv or using checkpointing :
This works :
sparkSession.sparkContext.setCheckpointDir("tmp")
ss.read.csv("test.csv") // read existing csv
.withColumn("test",lit(1)) // modify
.checkpoint(eager = true) // checkpoint, write to disk
.write.mode("overwrite")
.csv("test.csv") // write to same location
I’ve been looking for a while now for a way to get all filenames in a directory and its sub-directories in Hadoop file system (hdfs).
I found out I can use these commands to get it :
sc.hadoopConfiguration.set("mapreduce.input.fileinputformat.input.dir.recursive", "true")
sc.wholeTextFiles(path).map(_._1)
Here is "wholeTextFiles" documentation:
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.
Parameters:
path - Directory to the input data files, the path can be
comma separated paths as the list of inputs.
minPartitions - A
suggestion value of the minimal splitting number for input data.
Returns:
RDD representing tuples of file path and the corresponding
file content
Note: Small files are preferred, large file is also
allowable, but may cause bad performance., On some filesystems,
.../path/* can be a more efficient way to read all files in a
directory rather than .../path/ or .../path, Partitioning is
determined by data locality. This may result in too few partitions by
default.
As you can see "wholeTextFiles" returns a pair RDD with both the filenames and their content. So I tried mapping it and taking only the file names, but I suspect it still reads the files.
The reason I suspect so: if I try to count (for example) and I get the spark equivalent of "out of memory" (losing executors and not being able to complete the tasks).
I would rather use Spark to achieve this goal the fastest way possible, however, if there are other ways with a reasonable performance I would be happy to give them a try.
EDIT:
To clear it - I want to do it using Spark, I know I can do it using HDFS commands and such thing - I would like to know how to do such thing with the existing tools provided with Spark and maybe an explanation on how I can make "wholeTextFiles" not reading the text itself (kind of like how transformations only happen after an action and some of the "commands" never really happen).
Thank you very much!
This is the way to list out all the files till the depth of last subdirectory....and is with out using wholetextfiles
and is recursive call till the depth of subdirectories...
val lb = new scala.collection.mutable[String] // variable to hold final list of files
def getAllFiles(path:String, sc: SparkContext):scala.collection.mutable.ListBuffer[String] = {
val conf = sc.hadoopConfiguration
val fs = FileSystem.get(conf)
val files: RemoteIterator[LocatedFileStatus] = fs.listLocatedStatus(new Path(path))
while(files.hasNext) {// if subdirectories exist then has next is true
var filepath = files.next.getPath.toString
//println(filepath)
lb += (filepath)
getAllFiles(filepath, sc) // recursive call
}
println(lb)
lb
}
Thats it. it was tested with success. you can use as is..
I have a directory with some subfolders which content different parquet files. Something like this:
2017-09-05
10-00
part00000.parquet
part00001.parquet
11-00
part00000.parquet
part00001.parquet
12-00
part00000.parquet
part00001.parquet
What I want is by passing the path to the directory 05-09 to get a list of names of all parquet files.
I was able to achieve it, but in a very inefficient way:
val allParquetFiles = sc.wholeTextFiles("C:/MyDocs/2017-09-05/*/*.parquet")
allParquetFiles.keys.foreach((k) => println("The path to the file is: "+k))
So each key is the name I am looking for, but this process requires me to load all files as well, which then I can't use, since I get them in binary (and I don't know how to convert them into a dataframe).
Once I have the keys (so the list of filePaths) I am planning to invoke:
val myParquetDF = sqlContext.read.parquet(filePath);
As you may have already understood I am quite new in Spark. So please if there is a faster or easier approach to read a list of parquet files located in different folders, please let me know.
My Partial Solution: I wasn't able to get all paths for all filenames in a folder, but I was able to get the content of all files of that type into the same dataframe. Which was my ultimate goal. In case someone may need it in the future, I used the following line:
val df = sqlContext.read.parquet("C:/MyDocs/2017-05-09/*/*.parquet")
Thanks for your time
You can do it using the hdfs api like this
import org.apache.hadoop.fs._
import org.apache.hadoop.conf._
val fs = FileSystem.get(new Configuration())
val files = ( fs.listStatus(new Path("C:/MyDocs/2017-09-05/*/*.parquet")) ).map(_.getPath.toString)
First, it is better to avoid using wholeTextFiles. This method reads the whole file at once. Try to use textFile method. read more
Second, if you need to get all files recursively in one directory, you can achieve it by textFile method:
sc.hadoopConfiguration.set("mapreduce.input.fileinputformat.input.dir.recursive", "true")
This configuration will enable recursive search (works for spark jobs as for mapreduce jobs). And then just invoke sc.textFile(path).
I need to save DataFrame in CSV or parquet format (as a single file) and then open it again. The amount of data will not exceed 60Mb, so a single file is reasonable solution. This simple task provides me a lot of headache... This is what I tried:
To read the file if it exists:
df = sqlContext
.read.parquet("s3n://bucket/myTest.parquet")
.toDF("key", "value", "date", "qty")
To write the file:
df.write.parquet("s3n://bucket/myTest.parquet")
This does not work because:
1) write creates the folder myTest.parquet with hadoopish files that later I cannot read with .read.parquet("s3n://bucket/myTest.parquet"). In fact I don't care about multiple hadoopish files, unless I can later read them easily into DataFrame. Is it possible?
2) I am always working with the same file myTest.parquet that I am updating and overwriting in S3. It tells me that the file cannot be saved because it already exists.
So, can someone indicate me a right way to do the read/write loop? The file format doesn't matter for me (csv,parquet,csv,hadoopish files) unleass I can make the read and write loop.
You can save your DataFrame with saveAsTable("TableName") and read it with table("TableName"). And the location can be set by spark.sql.warehouse.dir. And you can overwrite a file with mode(SaveMode.Ignore). You can read here more from the official documentation.
In Java it would look like this:
SparkSession spark = ...
spark.conf().set("spark.sql.warehouse.dir", "hdfs://localhost:9000/tables");
Dataset<Row> data = ...
data.write().mode(SaveMode.Overwrite).saveAsTable("TableName");
Now you can read from the Data with:
spark.read().table("TableName");
I'm trying to read/monitor txt files from a Hadoop file system directory. But I've noticed all txt files inside this directory are directories themselves as showed in this example bellow:
/crawlerOutput/b6b95b75148cdac44cd55d93fe2bbaa76aa5cccecf3d723c5e47d361b28663be-1427922269.txt/_SUCCESS
/crawlerOutput/b6b95b75148cdac44cd55d93fe2bbaa76aa5cccecf3d723c5e47d361b28663be-1427922269.txt/part-00000
/crawlerOutput/b6b95b75148cdac44cd55d93fe2bbaa76aa5cccecf3d723c5e47d361b28663be-1427922269.txt/part-00001
I'd want read all the data inside the part's files. I'm trying to use the following code as showed in this snippet:
val testData = ssc.textFileStream("/crawlerOutput/*/*")
But, unfortunately it said it doesn't exist /crawlerOutput/*/*. Doesn't textFileStream accept wildcards? What should I do to solve this problem?
The textFileStream() is just a wrapper for fileStream() and does not support subdirectories (see https://spark.apache.org/docs/1.3.0/streaming-programming-guide.html).
You would need to list the specific directories to monitor. If you need to detect new directories a StreamingListener could be used to check then stop streaming context and restart with new values.
Just thinking out loud.. If you intend to process each subdirectory once and just want to detect these new directories then potentially key off another location that may contain job info or a file token that once present could be consumed in the streaming context and call the appropriate textFile() to ingest the new path.