Suppose I have a txt file named "input.txt" and I want to use scala to read it in. The dimension of the file is not available in the beginning.
So, how to construct such an Array[Array[Float]]? What I want is a simple and neat way rather than write some code like in Java to iterates over lines and parse each number. I think functional programming should be quite good at it.. but cannot think of one up to now.
Best Regards
If your input is correct, you can do it in such way:
val source = io.Source.fromFile("input.txt")
val data = source.getLines().map(line => line.split(" ").map(_.toFloat)).toArray
source.close()
Update: for additional information about using Source check this thread
Related
I am new to functional-style programming and scala, so my question might seem to be bit primitive.
Is there a specific way to read csv file in scala using functional style? Also, how the inner joins are performed for combining 2 csv files in scala using functional style?
I know spark and generally use data frame but don't have any idea in scala and finding it tough to search on google as well, since don't have much knowledge about it. Also, if anyone knows good links for the functional style programming for scala it would be great help.
The question is indeed too broad.
Is there a specific way to read csv file in scala using functional
style?
So far I don't know of a king's road to parse CSVs completely hassle-free.
CSV parsing includes
going through the input line-by-line
understanding, what to do with the (optional) header
accurately parsing each line each line according to the CSV specification
turning line parts into a business object
I recommend to
turn your input into Iterator[String]
split each line into parts using a library of your choice (e.g. opencsv)
manually create a desired domain object from the line parts
Here is an simple example which (ignores error handling and potential header)
case class Person(name: String, street: String)
val lineParser = new CSVParserBuilder().withSeparator(',').build()
val lines: Iterator[String] = Source.fromInputStream(new FileInputStream("file.csv")).getLines()
val parsedObjects: Iterator[Person] = lines.map(line => {
val parts: Array[String] = lineParser.parseLine(line)
Person(parts(0), parts(1))
})
I'm new to Spark and Scala so I might have misunderstood some basic things here. I'm trying to train Sparks word2vec model on my own data. According to their documentation, one way to do this is
val input = sc.textFile("text8").map(line => line.split(" ").toSeq)
val word2vec = new Word2Vec()
val model = word2vec.fit(input)
The text8 dataset contains one line of many words, meaning that input will become an RDD[Seq[String]].
After massaging my own dataset, which has one word per line, using different maps etc. I'm left with an RDD[String], but I can't seem to be able to train the word2vec model on it. I tried doing input.map(v => Seq(v)) which does actually give an RDD[Seq[String]], but that will give one sequence for each word, which I guess is totally wrong.
How can I wrap a sequence around my strings, or is there something else I have missed?
EDIT
So I kind of figured it out. From my clean being an RDD[String] I do val input = sc.parallelize(Seq(clean.collect().toSeq)). This gives me the correct data structure (RDD[Seq[String]]) to fit the word2vec model. However, running collect on a large dataset gives me out of memory error. I'm not quite sure how they intend the fitting to be done? Maybe it is not really parallelizable. Or maybe I'm supposed to have several semi-long sequences of strings inside and RDD, instead of one long sequence like I have now?
It seems that the documentation is updated in an other location (even though I was looking at the "latest" docs). New docs are at: https://spark.apache.org/docs/latest/ml-features.html
The new example drops the text8 example file alltogether. I'm doubting whether the original example ever worked as intended. The RDD input to word2vec should be a set of lists of strings, typically sentences or otherwise constructed n-grams.
Example included for other lost souls:
val documentDF = sqlContext.createDataFrame(Seq(
"Hi I heard about Spark".split(" "),
"I wish Java could use case classes".split(" "),
"Logistic regression models are neat".split(" ")
).map(Tuple1.apply)).toDF("text")
// Learn a mapping from words to Vectors.
val word2Vec = new Word2Vec()
.setInputCol("text")
.setOutputCol("result")
.setVectorSize(3)
.setMinCount(0)
val model = word2Vec.fit(documentDF)
Why not
input.map(v => v.split(" "))
or whatever would be an appropriate delimiter to split your words on. This will give you the desired sequence of strings - but with valid words.
As I can recall, word2vec in ml take dataframe as argument and word2vec in mllib can take rdd as argument. The example you posted is for word2vec in ml. Here is the official guide: https://spark.apache.org/docs/latest/mllib-feature-extraction.html#word2vec
it is easy for Hadoop to use .replace() for example
String[] valArray = value.toString().replace("\N", "")
But it dosen't work in Spark,I write Scala in Spark-shell like below
val outFile=inFile.map(x=>x.replace("\N",""))
So,how to deal with it?
For some reason your x is an Array[String]. How did you get it like that? You can .toString.replace it if you like, but that will probably not get you what you want (and would give the wrong output in java anyway); you probably want to do another layer of map, inFile.map(x => x.map(_.replace("\N","")))
I'm using the native parser combinator library in scala, and I'd like to use it to parse a number of large files. I have my combinators set up, but the file that I'm trying to parse is too large to be read into memory all at once. I'd like to be able to stream from an input file through my parser and read it back to disk so that I don't need to store it all in memory at once.My current system looks something like this:
val f = Source.fromFile("myfile")
parser.parse(parser.document.+, f.reader).get.map{_.writeToFile}
f.close
This reads the whole file in as it parses, which I'd like to avoid.
There is no easy or built-in way to accomplish this using scala's parser combinators, which provide a facility for implementing parsing expression grammars.
Operators such as ||| (longest match) are largely incompatible with a stream parsing model, as they require extensive backtracking capabilities. In order to accomplish what you are trying to do, you would need to re-formulate your grammar such that no backtracking is required, ever. This is generally much harder than it sounds.
As mentioned by others, your best bet would be to look into a preliminary phase where you chunk your input (e.g. by line) so that you can handle a portion of the stream at a time.
One easy way of doing it is to grab an Iterator from the Source object and then walk through the lines like so:
val source = Source.fromFile("myFile")
val lines = source.getLines
for (line <- lines) {
// Do magic with the line-value
}
source.close // Close the file
But you will need to be able to use the lines one by one in your parser of course.
Source: https://groups.google.com/forum/#!topic/scala-user/LPzpXo3sUVE
You might try the StreamReader class that is part of the parsing package.
You would use it something like:
val f = StreamReader( fromFile("myfile","UTF-8").reader() )
parseAll( parser, f )
The longest match as one poster above mentioned combined with regex's using source.subSequence(0, source.length) means even StreamReader doesn't help.
The best kludgy answer I have is use getLines as others have mentioned, and chunk as the accepted answer mentions. My particular input required me to chunk 2 lines at a time. You could build an iterator out of the chunks you build to make it slightly less ugly.
Is there an idiomatic Scala solution to obtain only the file names from File.listFiles?
Perhaps something like:
val names = new File(dir).listFiles.somemagicTrait(_getName)
and have names become a List[String]?
I know I can just loop and add them to a mutable list.
how about?
new File(dir).listFiles.map(_.getName).toList
I'm always wary of answering the wrong part of the question, but as Jean-Phillipe commented, you can get an array of the names from
new File(dir).list
and if you really need a list call toList on that.