Is it possible to add extra meta data to DataFrames?
Reason
I have Spark DataFrames for which I need to keep extra information. Example: A DataFrame, for which I want to "remember" the highest used index in an Integer id column.
Current solution
I use a separate DataFrame to store this information. Of course, keeping this information separately is tedious and error-prone.
Is there a better solution to store such extra information on DataFrames?
To expand and Scala-fy nealmcb's answer (the question was tagged scala, not python, so I don't think this answer will be off-topic or redundant), suppose you have a DataFrame:
import org.apache.spark.sql
val df = sc.parallelize(Seq.fill(100) { scala.util.Random.nextInt() }).toDF("randInt")
And some way to get the max or whatever you want to memoize on the DataFrame:
val randIntMax = df.rdd.map { case sql.Row(randInt: Int) => randInt }.reduce(math.max)
sql.types.Metadata can only hold strings, booleans, some types of numbers, and other metadata structures. So we have to use a Long:
val metadata = new sql.types.MetadataBuilder().putLong("columnMax", randIntMax).build()
DataFrame.withColumn() actually has an overload that permits supplying a metadata argument at the end, but it's inexplicably marked [private], so we just do what it does — use Column.as(alias, metadata):
val newColumn = df.col("randInt").as("randInt_withMax", metadata)
val dfWithMax = df.withColumn("randInt_withMax", newColumn)
dfWithMax now has (a column with) the metadata you want!
dfWithMax.schema.foreach(field => println(s"${field.name}: metadata=${field.metadata}"))
> randInt: metadata={}
> randInt_withMax: metadata={"columnMax":2094414111}
Or programmatically and type-safely (sort of; Metadata.getLong() and others do not return Option and may throw a "key not found" exception):
dfWithMax.schema("randInt_withMax").metadata.getLong("columnMax")
> res29: Long = 209341992
Attaching the max to a column makes sense in your case, but in the general case of attaching metadata to a DataFrame and not a column in particular, it appears you'd have to take the wrapper route described by the other answers.
As of Spark 1.2, StructType schemas have a metadata attribute which can hold an arbitrary mapping / dictionary of information for each Column in a Dataframe. E.g. (when used with the separate spark-csv library):
customSchema = StructType([
StructField("cat_id", IntegerType(), True,
{'description': "Unique id, primary key"}),
StructField("cat_title", StringType(), True,
{'description': "Name of the category, with underscores"}) ])
categoryDumpDF = (sqlContext.read.format('com.databricks.spark.csv')
.options(header='false')
.load(csvFilename, schema = customSchema) )
f = categoryDumpDF.schema.fields
["%s (%s): %s" % (t.name, t.dataType, t.metadata) for t in f]
["cat_id (IntegerType): {u'description': u'Unique id, primary key'}",
"cat_title (StringType): {u'description': u'Name of the category, with underscores.'}"]
This was added in [SPARK-3569] Add metadata field to StructField - ASF JIRA, and designed for use in Machine Learning pipelines to track information about the features stored in columns, like categorical/continuous, number categories, category-to-index map. See the SPARK-3569: Add metadata field to StructField design document.
I'd like to see this used more widely, e.g. for descriptions and documentation of columns, the unit of measurement used in the column, coordinate axis information, etc.
Issues include how to appropriately preserve or manipulate the metadata information when the column is transformed, how to handle multiple sorts of metadata, how to make it all extensible, etc.
For the benefit of those thinking of expanding this functionality in Spark dataframes, I reference some analogous discussions around Pandas.
For example, see xray - bring the labeled data power of pandas to the physical sciences which supports metadata for labeled arrays.
And see the discussion of metadata for Pandas at Allow custom metadata to be attached to panel/df/series? · Issue #2485 · pydata/pandas.
See also discussion related to units: ENH: unit of measurement / physical quantities · Issue #10349 · pydata/pandas
If you want to have less tedious work, I think you can add an implicit conversion between DataFrame and your custom wrapper (haven't tested it yet though).
implicit class WrappedDataFrame(val df: DataFrame) {
var metadata = scala.collection.mutable.Map[String, Long]()
def addToMetaData(key: String, value: Long) {
metadata += key -> value
}
...[other methods you consider useful, getters, setters, whatever]...
}
If the implicit wrapper is in DataFrame's scope, you can just use normal DataFrame as if it was your wrapper, ie.:
df.addtoMetaData("size", 100)
This way also makes your metadata mutable, so you should not be forced to compute it only once and carry it around.
I would store a wrapper around your dataframe. For example:
case class MyDFWrapper(dataFrame: DataFrame, metadata: Map[String, Long])
val maxIndex = df1.agg("index" ->"MAX").head.getLong(0)
MyDFWrapper(df1, Map("maxIndex" -> maxIndex))
A lot of people saw the word "metadata" and went straight to "column metadata". This does not seem to be what you wanted, and was not what I wanted when I had a similar problem. Ultimately, the problem here is that a DataFrame is an immutable data structure that, whenever an operation is performed on it, the data passes on but the rest of the DataFrame does not. This means that you can't simply put a wrapper on it, because as soon as you perform an operation you've got a whole new DataFrame (potentially of a completely new type, especially with Scala/Spark's tendencies toward implicit conversions). Finally, if the DataFrame ever escapes its wrapper, there's no way to reconstruct the metadata from the DataFrame.
I had this problem in Spark Streaming, which focuses on RDDs (the underlying datastructure of the DataFrame as well) and came to one simple conclusion: the only place to store the metadata is in the name of the RDD. An RDD name is never used by the core Spark system except for reporting, so it's safe to repurpose it. Then, you can create your wrapper based on the RDD name, with an explicit conversion between any DataFrame and your wrapper, complete with metadata.
Unfortunately, this does still leave you with the problem of immutability and new RDDs being created with every operation. The RDD name (our metadata field) is lost with each new RDD. That means you need a way to re-add the name to your new RDD. This can be solved by providing a method that takes a function as an argument. It can extract the metadata before the function, call the function and get the new RDD/DataFrame, then name it with the metadata:
def withMetadata(fn: (df: DataFrame) => DataFrame): MetaDataFrame = {
val meta = df.rdd.name
val result = fn(wrappedFrame)
result.rdd.setName(meta)
MetaDataFrame(result)
}
Your wrapping class (MetaDataFrame) can provide convenience methods for parsing and setting metadata values, as well as implicit conversions back and forth between Spark DataFrame and MetaDataFrame. As long as you run all your mutations through the withMetadata method, your metadata will carry along though your entire transformation pipeline. Using this method for every call is a bit of a hassle, yes, but the simple reality is that there is not a first-class metadata concept in Spark.
Related
I am using Spark to read multiple parquet files into a single RDD, using standard wildcard path conventions. In other words, I'm doing something like this:
val myRdd = spark.read.parquet("s3://my-bucket/my-folder/**/*.parquet")
However, sometimes these Parquet files will have different schemas. When I'm doing my transforms on the RDD, I can try and differentiate between them in the map functions, by looking for the existence (or absence) of certain columns. However a surefire way to know which schema a given row in the RDD uses - and the way I'm asking about specifically here - is to know which file path I'm looking at.
Is there any way, on an RDD level, to tell which specific parquet file the current row came from? So imagine my code looks something like this, currently (this is a simplified example):
val mapFunction = new MapFunction[Row, (String, Row)] {
override def call(row: Row): (String, Row) = myJob.transform(row)
}
val pairRdd = myRdd.map(mapFunction, encoder=kryo[(String, Row)]
Within the myJob.transform( ) code, I'm decorating the result with other values, converting it to a pair RDD, and do some other transforms as well.
I make use of the row.getAs( ... ) method to look up particular column values, and that's a really useful method. I'm wondering if there are any similar methods (e.g. row.getInputFile( ) or something like that) to get the name of the specific file that I'm currently operating on?
Since I'm passing in wildcards to read multiple parquet files into a single RDD, I don't have any insight into which file I'm operating on. If nothing else, I'd love a way to decorate the RDD rows with the input file name. Is this possible?
You can add a new column for the file name as shown below
import org.apache.spark.sql.functions._
val myDF = spark.read.parquet("s3://my-bucket/my-folder/**/*.parquet").withColumn("inputFile", input_file_name())
I work with Spark 1.6.1 in Scala.
I have one dataframe, and I want to create different dataframe and only want to read 1 time.
For example one dataframe have two columns ID and TYPE, and I want to create two dataframe one with the value of type = A and other with type value = B.
I've checked another posts on stackoverflow, but found only the option to read the dataframe 2 times.
However, I would like another solution with the best performance possible.
Kinds regards.
Spark will read from the data source multiple times if you perform multiple actions on the data. The way to aviod this is to use cache(). In this way, the data will be saved to memory after the first action, which will make subsequent actions using the data faster.
Your two dataframes can be created in this way, requiring only one read of the data source.
val df = spark.read.csv(path).cache()
val dfA = df.filter($"TYPE" === "A").drop("TYPE")
val dfB = df.filter($"TYPE" === "B").drop("TYPE")
The "TYPE" column is dropped as it should be unnecessary after the separation.
I am new to spark. I have some json data that comes as an HttpResponse. I'll need to store this data in hive tables. Every HttpGet request returns a json which will be a single row in the table. Due to this, I am having to write single rows as files in the hive table directory.
But I feel having too many small files will reduce the speed and efficiency. So is there a way I can recursively add new rows to the Dataframe and write it to the hive table directory all at once. I feel this will also reduce the runtime of my spark code.
Example:
for(i <- 1 to 10){
newDF = hiveContext.read.json("path")
df = df.union(newDF)
}
df.write()
I understand that the dataframes are immutable. Is there a way to achieve this?
Any help would be appreciated. Thank you.
You are mostly on the right track, what you want to do is to obtain multiple single records as a Seq[DataFrame], and then reduce the Seq[DataFrame] to a single DataFrame by unioning them.
Going from the code you provided:
val BatchSize = 100
val HiveTableName = "table"
(0 until BatchSize).
map(_ => hiveContext.read.json("path")).
reduce(_ union _).
write.insertInto(HiveTableName)
Alternatively, if you want to perform the HTTP requests as you go, we can do that too. Let's assume you have a function that does the HTTP request and converts it into a DataFrame:
def obtainRecord(...): DataFrame = ???
You can do something along the lines of:
val HiveTableName = "table"
val OtherHiveTableName = "other_table"
val jsonArray = ???
val batched: DataFrame =
jsonArray.
map { parameter =>
obtainRecord(parameter)
}.
reduce(_ union _)
batched.write.insertInto(HiveTableName)
batched.select($"...").write.insertInto(OtherHiveTableName)
You are clearly misusing Spark. Apache Spark is analytical system, not a database API. There is no benefit of using Spark to modify Hive database like this. It will only bring a severe performance penalty without benefiting from any of the Spark features, including distributed processing.
Instead you should use Hive client directly to perform transactional operations.
If you can batch-download all of the data (for example with a script using curl or some other program) and store it in a file first (or many files, spark can load an entire directory at once) you can then load that file(or files) all at once into spark to do your processing. I would also check to see it the webapi as any endpoints to fetch all the data you need instead of just one record at a time.
In order to run a few ML algorithms, I need to create extra columns of data. Each of these columns involves some fairly intense calculations that involves keeping moving averages and recording information as you go through each row (and updating it meanwhile). I've done a mock through with a simple Python script and it works, and I am currently looking to translate it to a Scala Spark script that could be run on a larger data set.
The issue is it seems that for these to be highly efficient, using Spark SQL, it is preferred to use the built in syntax and operations (which are SQL-like). Encoding the logic in a SQL expression seems to be a very thought-intensive process, so I'm wondering what the downsides will be if I just manually create the new column values by iterating through each row, keeping track of variables and inserting the column value at the end.
You can convert an rdd into dataframe. Then use map on the data frame and process each row as you wish. If you need to add new column, then you can use, withColumn. However this will only allow one column to be added and it happens for the entire dataframe. If you want more columns to be added, then inside map method,
a. you can gather new values based on the calculations
b. Add these new column values to main rdd as below
val newColumns: Seq[Any] = Seq(newcol1,newcol2)
Row.fromSeq(row.toSeq.init ++ newColumns)
Here row, is the reference of row in map method
c. Create new schema as below
val newColumnsStructType = StructType{Seq(new StructField("newcolName1",IntegerType),new StructField("newColName2", IntegerType))
d. Add to the old schema
val newSchema = StructType(mainDataFrame.schema.init ++ newColumnsStructType)
e. Create new dataframe with new columns
val newDataFrame = sqlContext.createDataFrame(newRDD, newSchema)
Is it possible and what would be the most efficient neat method to add a column to Data Frame?
More specifically, column may serve as Row IDs for the existing Data Frame.
In a simplified case, reading from file and not tokenizing it, I can think of something as below (in Scala), but it completes with errors (at line 3), and anyways doesn't look like the best route possible:
var dataDF = sc.textFile("path/file").toDF()
val rowDF = sc.parallelize(1 to DataDF.count().toInt).toDF("ID")
dataDF = dataDF.withColumn("ID", rowDF("ID"))
It's been a while since I posted the question and it seems that some other people would like to get an answer as well. Below is what I found.
So the original task was to append a column with row identificators (basically, a sequence 1 to numRows) to any given data frame, so the rows order/presence can be tracked (e.g. when you sample). This can be achieved by something along these lines:
sqlContext.textFile(file).
zipWithIndex().
map(case(d, i)=>i.toString + delimiter + d).
map(_.split(delimiter)).
map(s=>Row.fromSeq(s.toSeq))
Regarding the general case of appending any column to any data frame:
The "closest" to this functionality in Spark API are withColumn and withColumnRenamed. According to Scala docs, the former Returns a new DataFrame by adding a column. In my opinion, this is a bit confusing and incomplete definition. Both of these functions can operate on this data frame only, i.e. given two data frames df1 and df2 with column col:
val df = df1.withColumn("newCol", df1("col") + 1) // -- OK
val df = df1.withColumn("newCol", df2("col") + 1) // -- FAIL
So unless you can manage to transform a column in an existing dataframe to the shape you need, you can't use withColumn or withColumnRenamed for appending arbitrary columns (standalone or other data frames).
As it was commented above, the workaround solution may be to use a join - this would be pretty messy, although possible - attaching the unique keys like above with zipWithIndex to both data frames or columns might work. Although efficiency is ...
It's clear that appending a column to the data frame is not an easy functionality for distributed environment and there may not be very efficient, neat method for that at all. But I think that it's still very important to have this core functionality available, even with performance warnings.
not sure if it works in spark 1.3 but in spark 1.5 I use withColumn:
import sqlContext.implicits._
import org.apache.spark.sql.functions._
df.withColumn("newName",lit("newValue"))
I use this when I need to use a value that is not related to existing columns of the dataframe
This is similar to #NehaM's answer but simpler
I took help from above answer. However, I find it incomplete if we want to change a DataFrame and current APIs are little different in Spark 1.6.
zipWithIndex() returns a Tuple of (Row, Long) which contains each row and corresponding index. We can use it to create new Row according to our need.
val rdd = df.rdd.zipWithIndex()
.map(indexedRow => Row.fromSeq(indexedRow._2.toString +: indexedRow._1.toSeq))
val newstructure = StructType(Seq(StructField("Row number", StringType, true)).++(df.schema.fields))
sqlContext.createDataFrame(rdd, newstructure ).show
I hope this will be helpful.
You can use row_number with Window function as below to get the distinct id for each rows in a dataframe.
df.withColumn("ID", row_number() over Window.orderBy("any column name in the dataframe"))
You can also use monotonically_increasing_id for the same as
df.withColumn("ID", monotonically_increasing_id())
And there are some other ways too.