Scala Read File. Loading it from input file Web 1 i have a csv file and i want to read that file and store it in case class.
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Val f = new file (getclass.getclassloader.getresource (movies.txt).getpath) import scala.io.source source.fromfile (f).getlines.foreach { line => count+=1 // custom code } share improve this answer follow answered jul 16, 2015. 8 move your movies.txt to resources dir, then you can do the following: Web this worked for me (in the spark shell): But in case of my csv file there are some data which have already comma itself. Assume you have a file path and you want to read some text data from it: Web 3 i think relying on java libraries is what (almost?) everyone would do, the scala library included. Web what's a simple and canonical way to read an entire file into memory in scala? Web let us see some methods how to read files over scala: Reading scala file from console we can read file from console and check for the data and do certain operations over. See for instance the source code of scala.io.source.
So the problem how to split data from that. Val f = new file (getclass.getclassloader.getresource (movies.txt).getpath) import scala.io.source source.fromfile (f).getlines.foreach { line => count+=1 // custom code } share improve this answer follow answered jul 16, 2015. Web this worked for me (in the spark shell): Web what's a simple and canonical way to read an entire file into memory in scala? See for instance the source code of scala.io.source. Use a slightly longer approach that properly closes the file. Web 3 answers sorted by: Web how to read gzip'd file in scala ask question asked 10 years, 4 months ago modified 5 years, 1 month ago viewed 15k times 30 in java, i'd wrap a gzipinputstream over a fileinputstream and be done. So the problem how to split data from that. How is the equivalent done in scala? Web 3 i think relying on java libraries is what (almost?) everyone would do, the scala library included.
