Scala source getlines

fromFile("example. val in = scala. Source类来打开和读取文件。 Aug 9, 2022 · As a quick note about reading files, if you ignore possible exceptions, you can use code like this read a text file into an Array, List, or Seq using Scala: val bufferedSource = io. toList方法从输入流中读取文件的内容，并将其转换为包含每行数据的列表。 最后，我们使用 inputStream. Nov 7, 2016 · using(Source. fromFile which returns a BufferedSource. 7. /**. val lines = Source. File. Feb 22, 2020 · Solution. object FileIO {. Sourceobject no7 { def main (args: Array [String Feb 4, 2018 · Thanks. 我们执行以下步骤在 Scala 中读取文件： 首先，我们指定文件名及其完整路径。 使用 Source. Type members. Source: BufferedSource. During creation of the list li2 the iterator is read, so becomes empty. 3 participants. Source Object: scala. r 是一个Regex对象 Jul 28, 2020 · A reasonable algorithm would be: for each line, split it into words; parse each word into an Int; An implementation of that algorithm: io. answered Jun 1, 2012 at 8:34. scala and looks as follows ( line 350 ): val contents = scala. Dec 1, 2020 · I’m trying to update some old code to Scala 3. txt" val source = Source. length, line) f. next return the next element in the iteration and moves the iterator one step ahead. toArray. Returns a scala. Supply the empty argument list ` ()` explicitly to invoke method getLines, …. mkString("\n") Exercise 2 Write a Scala program that reads a file with tabs, replaces each tab with spaces so that tab stops are at n-column boundaries, and writes the result to the same file. Source类，然后使用fromFile方法读取文件的内容，并通过getLines方法将文件的每一行读取为一个字符串。 最后，我们使用 toList 方法转换成列表，并通过 foreach 方法打印每一行。 Jan 5, 2011 · val file: FileOps = Path ("file") // write bytes. mkString } Jan 30, 2023 · 在 Scala 中方法 2：逐行读取文件. This time, this is a Scala collection, which you can apply map, flatMap, filter, toList, mkString and all the niceties we’re used to. getLines() // will split at every line break, regardless if quoted or not The getLines gets something like: Jan 10, 2013 · +1 for the comment that you can use IOUtils in Scala: a proven, reliable, optimized library. fromFile. Source. This recipe demonstrates three approaches: A simple use of the scala. fromFile(fname) for (line <- f. yet, if I debug this code, stop on the next line, go and physically change the file on the HDD, the values returned by this iterator correspond to the file from before the Nov 10, 2012 · Assuming we have a source s that we're opening and closing elsewhere, this is all you need: val lines: IndexedSeq[IndexedSeq[Char]] = s. fromFile creates a source from the input file. txt"). fromFile(filename). While this is likely not a problem in the productive sources (see output below), this has in the past led Mar 14, 2019 · an issue if Source. toList Oct 8, 2012 · scala. * Scalaでファイルの入出力. fromFile method. getLines() for (line <- lines Dec 3, 2014 · I'm doing it this way right now: //println(line) lst = line. parsing - Parser combinators (scala-parser-combinators. toArray . fromFile . get the encoding right. Source val filename = "fileopen. You want a Scala HTTP client you can use to make GET request calls. Stream. 8, but people are now actively discussing such a library. def fromInputStream(is: InputStream)(implicit codec: Codec): BufferedSource =. Using(Source. It can be used in this case with scala. Source Aug 31, 2015 · It is mostly a UTF-8 file but some of the field values are in different encodings, so there is no encoding in which the whole file is valid, but I need to parse it anyway. csv"). isFile. As far as I understand, it is supposed to return null so I wrap it into Option. Nov 17, 2015 · } } import scala. parseInt(s) } thirdLargest(intStream) The file 10m. splitAt (1) Side note: scala. collection. scala> val holmes = Source. Jul 3, 2016 · Because Source. Here is a hint: Source. fromURL method FileIO. In Scala, for reading a text file and uploading it into an array, a common approach is. However, not quite where I was looking for help. zip库来解压缩文件。 以下是使用scala. nio. 所以说，scala，其实主要就是针对某些特定领域的一些复杂系统的，比较适用的 Jan 12, 2022 · The . I'd hope that Source would release the file handle once the contents of the file had been read but it doesn't seem to do that. hasnext checks if there is next element available to iterate. Close Source. fromFile method to read a csv file. Stepper for the elements of this collection. g. fromFile opens the file, and getLines returns an iterator over the lines in the file. getLines() #3 . Solution. mkString 以字符串形式输出文件内容; 将字符串转换为数字，可以用toInt或toDouble方法; 使用Java的PrintWriter来写入文本文件- “正则”. import scala. Russ January 7, 2020, 12:34am 1. Prince John Wesley. Source和java. 在开始迭代文件的行数之前，我们首先需要读取文件。Scala提供了多种读取文件的方法，我们将使用最简单和常用的方法，即使用scala. The offending code is located in file ReplDriver. txt") val lines = Source. Predef to support treating strings as sequences of characters. In every example on internet they are doing: import scala. getLines() This now returns an Iterator of the lines in the file. toList. getLines() // Get the iterator object. val string = io. Input is buffered in a buffer of size bufferSize. For collections holding primitive values, the Stepper can be used as an iterator which doesn't box the elements. Source读取文件，首先需要创建一个Source对象，然后调用其getLines方法来获取文件的每一行内容。. csv") while yields a URL which can then convert into a Stream, or whatever. I am not even able to read the file usin scala. There is another operator for reference equality (and you don't need it). fromFile and create a string using mkString it appends a newLine but if I do . toIterator if file. collection and its sub-packages contain Scala's collections framework Jan 31, 2017 · from the Scala API stated here fromFile is a method defined on the Source companion object. Source in Scala 2. It is better to use the "\\W+" regexp rather than ("[. It'd be nice if Scala had an IO library and not something that is exists only to support their compiler. fromFile(file, "UTF-8" ). bufferedSource. The flip side of this presents itself is the programmer needs to be Jan 14, 2022 · You should remove the . toIterator buf. @dhg is correct and here is a bit more detail on why. Source#getLines produces a standard Iterator, so if you’re familiar with the Collections API, all those goodies are there and you’ll probably be able to avoid interacting with the low-level Iterator API. BufferedSource (retrieved from scala. fromFile("/Users/Al/. flatMap(l => Try(processOrDoWhatYouWantForLines(l))) result. getLines) println (line) Actually, I am reducing the number of lines and the size of each line, but that is From the JavaDoc: MalformedInputException thrown when an input byte sequence is not legal for given charset, or an input character sequence is not a legal sixteen-bit Unicode sequence. Then your attempts should work e. Using. fromFile(fileName). io #2 It opens the file #3 It copies each line of the file into a string #4 It concatenates all the strings using the char \\n 在上面的示例中，我们首先导入了scala. getLines() val counts = new scala. // 使用getResourceAsStream方法读取资源文件 val stream = getClass. Package structure . val rows = ArrayBuffer[Array[String]]() using(io. def fromFile ( file: File, bufferSize: Int ) ( implicit codec: Codec ): BufferedSource. I am trying to read it line by line and save a reduced version of it. 阅读更多：Scala 教程. fromFile som-snytt/scala. println(lst) It's giving me a redline on lst=line and says reassignment to a val. fromFile("proteins. fromFile(filename) #2 . Most of the answers found on the internet use getLines method which disregards new-line delimiters. 要使用scala. Combine Recipe 12. Well, in Scala you can actually say: val lines = scala. jdk. 读取和 Sep 3, 2014 · 12. Source #1 def readFileIntoString (filename: String) = Source. toDouble)) val result = lines. In scala == is a null-safe equivalent of equals (it calls equals ). Using the scala. This is a curried method with the first param list taking a single String representing the path of the file to be read and the second curried parameter list takes a single implicit codec argument of type scala. fromFile(filename) val lines = source. I think it's coming in 2. Source val fileName = "example. fromFile (fileName). close() Jul 13, 2011 · This is potentially an order of magnitude faster than scala. bash_profile"). Scala 提取器 (Extractor) Scala 文件 I/O Scala 进行文件写操作，直接用的都是 java中 的 I/O 类 （java. for {. Now, while doing this or after doing this, I want to add all the lists into an Array. size) (for { _ <- count Nov 17, 2012 · I'm reading a file line by line and putting into a list. Graph. createBufferedSource(is, reset = () => fromInputStream(is)(codec), close = () => is. split(" - - | \"POST | HTTP/1. To avoid running out of memory I read the file in streams and use getLines which is lazy. 3, “How to Split Strings in Scala”. Apr 4, 2018 · I have the following scala code which takes a String, messes up with the UTF-8 characters and then tries to read it via Source. scala，本身的编程语言的功能，就不是特别的强大和完善，比如说，scala甚至不能很方便地写文件，必须依赖于java的io流才可以. Managed. // an existing file with the new data. I'm reading and writings some text files in Scala. This is my code: def readUsers(filePath: Path): IO[Seq[User]] = {. Make sure your heap is large enough. Dec 29, 2021 · import scala. fromFile(filename 快学Scala第9章----文件和正则表达式 本章要点. getLines reading resources prior 2. 现在上代码！. Source. Scala Source. getLines() you receive a new Iterator[String] object that by default points to the first line of the CSV file. Then filter the contents for punctuation and digits. split("\\W+") val res1: Array[String] = Array Mar 19, 2012 · Stack Overflow Public questions & answers; Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Talent Build your employer brand Feb 5, 2018 · I'm using the scala. getLines // for each line Nov 8, 2022 · This is Recipe 12. getLines Mar 17, 2019 · scala. scala. getLines would make sense if you were to map a complex lambda to process lines immutabily – May 14, 2024 · Solution 2: Reading Lines from a File in Scala. fromFile(inputFile). I'll allow the user to specify the file enconding butif the user doesn't specify the proper encoding I'd like to catch the MalformedInputException and then my method will return a None (instead of Some[Iterator[String]]). def main ( args:Array [String] ) {. from the class path) you need to treat the resource as a resource. Feb 3, 2024 · This is Recipe 15. For example, List is an alias for scala. Source to do the reading is probably the fastest--it's not built in, but easy to write: def inputToFile(is: java. $ scala Test 文件内容为: 菜鸟教程. File)： 实例 [mycode4 type='scala'] import java. As a complete beginner in the language, I wanted to make sure to find the right way to do it, e. Nov 25, 2013 at 7:57. Platform. 我们将展示如何使用Scala的文件读取和迭代功能来实现这个目标。 阅读更多：Scala 教程. scala> val li = Source. txt contains 10 million lines with a random integer on each one. fromFile方法来读取文件的内容，并使用getLines方法将每一行内容转换为一个字符串数组。 Apr 11, 2012 · Yes, although it's not quite so concise: import java. toIndexedSeq Where wrapString is a method provided by scala. You should try something like this: val lineIterator = Source. fromFile(file). file. So I have to read lines from filejust a trivial I/O operation. 2 GB and 33,444,922 lines. getTime parseAndSortFile(Source. Apr 27, 2012 · def linesEager(filename: String): Iterator[String] = val buf = scala. The fromFile method returns a BufferedSource, and its getLines method treats “any of \r\n, \r, or \n as a line separator (longest match),” so each element in the sequence is a line from the file. fromFile(new File(filename)) val zs = buf. fromFile(filename)) { source => val lines = Try(source. Internally Source creates an underlying BufferedSource that has a close method. Then, one line: valscalaFileContents=Source. 下面是一个读取文件的示例代码：. fromFile(filename) . Source。这个库提供了一个getLines方法，可以一次读取一行文本。为了能够处理大型压缩文件，我们还可以使用Scala的java. fromFile 创建文件将被加载的源。 使用 getLines() 方法逐行读取数据，然后进行相应的打印或处理。 Sep 10, 2014 · I am trying to read inputs from a file and counts them using a map. You need to write getLines() instead, so that this default will be used. mkString as per this answer. I don't know why : (. Source includes methods for iterable representation of the source file. toArray) println(new Date(). As shown in the following ScalaTest tests, you can test the getLinesUppercased method by passing it either a Source from a file or a String: var source: Source = _. close)) And then you could use it like this: def readFileZioManaged(file: String): Task[Iterator Feb 11, 2022 · @BalmungSan. toList holmes: List[String] = List(The Project Gutenberg EBook of The Adventures of Sherlock Holmes, by Sir Arthur Conan Doyle, (#15 in our series by Sir Arthur Conan Doyle . toUpperCase). There are many potential solutions to this problem. 在Scala中，我们可以使用Source对象从文本文件中读取内容。Source提供了多种方法来读取文件内容，包括getLines、getLinesWithSeparators、mkString等。但是，Source并没有直接提供将其内容转换为InputStream对象的方法。 When I read the file using Source. getLines. Apr 30, 2020 · Use a reader called Source: importscala. So most of the stuff I found (also on SO ) recommends I use io. I look on the web and didn't find any better way of doing it. getLines() for (line <- lines Jul 17, 2018 · spark的源码是由scala和java共同编写而成的，Java的多线程. (for (line <- source. 1, “How to Open and Read a Text File in Scala” with Recipe 1. Aug 24, 2011 · 0. map(line => line. line <- Source fromFile file getLines. 接下来，我们使用scala. If I understand you correctly the problem is that every time you call handler. // JavaのFileOutputStreamクラスとOutputStreamWriterを別名でimport. txt" var array: Array[String] = new Array[String](0) for (line <- Source. @Brian could you see the EDIT in my question. fromFile("test. get } Actually, the using function can be used for handling all resources which need to be closed at the end of the operation. getLines) { array :+= line } But it's quite long and maybe not very optimal. Yes. Something like this: for (line <- io. Source needs closing, otherwise you could run out of file handles (on any OS) and on Windows keeping the file open means it cannot be deleted. which comes pre-bundled with Scala. This object provides convenience methods to create an iterable representation of a source file. fromFile(path). So I gave myself 90 minutes to goof around and try a few different ways to write a simple “line count” program in Scala. fromFile(filepath). fromFile(filename)) val count = file. There are several places in this repo where Source. Creates Source from file, using given character encoding, setting its description to filename. ,?\\s]+". val file = Source. io. 它接受一个文件路径作为参数，并返回一个Source对象，我们可以利用该对象读取文件中的内容。. Source”的包，使用包中的一个叫fromFile的函数实现。. – senia. Notable packages include: scala. toList, so you’re dealing with Iterator (which processes one item at a time) instead of List (which is always entirely in-memory). The above, for example, is more or less equivalent to the following: Feb 3, 2024 · 6. mkString it doesn't append a newLine character. The scala package contains core types like Int, Float, Array or Option which are accessible in all Scala compilation units without explicit qualification or imports. toArray 输出文件文件的所有行; Source. // to specify in detail how the write should take place. This is the documentation for the Scala standard library. fromInputStream: import java. 11. foreach {println} Aug 2, 2023 · I have a large csv file with users data. . I need to retrieve the contents as-is and the API seems rather complicated. The most important method of the Source object f is getLines, which allows you to iterate over all the lines of the file: val fname = args(0) val f = scala. fromFile(file) . toList lines. scala. */. 使用scala. Source): List[String] = {. ByteArrayInputStream import java. Feb 23, 2018 · 1. csv")) { source =&gt; for (line &lt;- source. Scala提供了几个库来读取大型文件，其中一个是scala. trim. getResource and getClass. fromInputStream(is) Jun 22, 2020 · You can transfer Iterator to any collection first. File) {. mkString("\\n") #4 #1 It includes Source from the package scala. fromURL. getResourceAsStream. 7. // By default the file write will replace. def parseFileByName(fileName: String): DimacsFile = { parseBufferedContent(Source. close())(codec) If there was exception, would scala execute the close method by its own Jan 7, 2020 · Question. 8. 1\" ") . Pass the currect encoding as parameter to Source. getLines returns an Iterator[String], which you're iterating completely with the invocation of it. Especially for very large files, is there a faster approach perhaps by reading blocks of bytes into memory first and then splitting them by new line characters ? Sep 16, 2019 · The standard getLines() function does not handle that. com Mar 29, 2024 · val lines = Source. getTime-now) This should reduce the usage of memory and speed-up your sorting. So, something like the following: getClass. stdin. I have an endpoint that gets a user and should return a boolean indicating whether the user exists in the file or not. getLines ^ warning: Auto-application to ` ()` is deprecated. getLines(). util. 9, “How to write a simple HTTP GET request client in Scala. 方法首先创建一个 FileInputStream 和一个 FileOutputStream 实例，然后使用一个缓冲区数组来逐个字节地从输入流读取文件，并将其写入输出流中。. file <- new File(dir). Description of the process for evolving the language, and list of all the Scala Improvement Proposals (SIPs). make(ZIO(Source. fromFile is used without closing the source thereafter. getResourceAsStream("filename. Sep 29, 2022 · def getLinesUppercased(source: io. toByte)) // another option for write is openOptions which allows the caller. getLines()) printf("%4d %s\n", line. } { doSomething line } The trick is flatMap and its for -comprehension syntactic sugar. fromResource() method when reading from non-existing file. foreach Source. What action should I take in this function to satisfy that warning? Feb 6, 2017 · If it's a text file, and you want to limit yourself to Scala and Java, then using scala. However, after trying it out like so, reading a UTF-8 file: val user_list = Source Apr 24, 2016 · I would like to read csv file in scala. Supertypes. map(_. getLines . fromInputStream(stream). It is not coming in 2. Predef object are always in scope by default. Oct 7, 2011 · To start moving toward that, we can simply use the toList method on the Iterator[String] obtained from getLines. List. – Dec 8, 2014 · To read resources the object Source provides the method fromResource. Development. ” Problem. Some of these identifiers are type aliases provided as shortcuts to commonly used classes. 12 (still my favourite due to jar compatibility) To read resources you can use getClass. Summary: I’m working on a little project to parse relatively large (but not huge) Apache access log files, with the file this week weighing in at 9. size, which means the iterator reached it's end. mkString(" ") 在上面的示例中，我们使用Source. mkString But this is just a library sugar. See Read entire file in Scala? for other possiblities. reverse . fromFile didn’t abstract away the nested readers. close } Oct 11, 2017 · This is my code that I have used to try to solve the problem: Source. 2, and I get a warning (from intellij) on the following code. Scala中的Source对象. Implicits. fromFile方法用于从文件中读取数据。. class BufferedSource (inputStream: InputStream, bufferSize: Int ) (implicit val codec: Codec) extends Source. 示例：. close() 关闭输入流，并使用 foreach 方法打印每行数据。 Dec 9, 2013 · The Source would return a BufferedSource which override close method to close the input stream. getLines– Iterate through file line by line Oct 31, 2013 · val lineStream = thirdLargest(Source. val filename = "data. Source读取文件. getResource("AAPL_Historical. You want to process the lines in a CSV file in Scala, either handling one line at a time or storing them in a two-dimensional array. However this is not exposed from Source. Codec. The code in your example calls calls Source. grouped(num). 3. txt")) { source => source. getLines) yield line. write (Array (1,2,3) map ( _. buffered) } The warning indicates that Source is not closed. Scala provides several methods to read lines from a file: 1. GraalVM. getLines) {println(line)} As a variation of this, use the following approach to get all of the lines from the file as a List or Array: How about this: def countLines(filename: String): Option[Int] = { val file = Try(Source. 13. InputStream, f: java. getLines() but how can I iterate over a file using a different delimiter than new-lines? I could load the entire file as a string and then split that, but I expect that to perform poorly on large files (readDelimited1). I tried that just now with a 6-gigabyte input file and it ran fine. fromFile(filename) val lines = (for (line <- bufferedSource. Method fromFile returns an BufferedSource object which needs to be closed ! The following code solves the issue for me: val source = scala. Compare: val line="my: My home" scala> line. Source . This is how size is defined on TraversableOnce[A] (which Iterator[A] extends): def size: Int = { var result = 0 for (x <- self) result += 1 result } Aug 6, 2020 · Since you're splitting on newlines, and io. An InputStream, which contains the following input: A line containing a single number: N Followed by N lines containing two numbers Ti and Li Oct 5, 2023 · in Scala, the tendency is normally to use camelCase over snake_case, but of course YMMV and stick to the style of the project you're in while it's technically not incorrect, there's no need to csv. Source: For example. immutable. string. Source as it extends AutoCloseable in order to read from a file and, no matter what, close the file resource afterwards: import scala. foreach{ line => line. fromFile("10m. map(wrapString). fromFile("input. listFiles. getLines() // 使用getResource方法读取 Jan 28, 2010 · Source has a method called reset however all that this does is recreate the source from the file. // the openOptions parameter takes a Feb 4, 2023 · I have an instance of scala. getLines()) yield line). Become a Scala OSS Contributor From start to finish: discover how you can help Scala's open-source ecosystem 执行以上代码，输出结果为: $ scalac Test. So, if my file has 10 lines, I have ten lists. Source val readmeText : Iterator[String] = Source. However, since you are sorting in-memory, there will always be a cap to maximum size. See full list on baeldung. * Javaのクラスを使ってファイルを作成し、Scalaのクラスを使ってファイルを読み込む. Does calling getLines load the entire file into memory? And at what point is the buffered source fully closed, and no longer needed in memory? I'm worried that even though I'm calling close, it isn't actually closing until some time later, perhaps via the GC, is this right? \$\endgroup\$ – Nov 25, 2013 · 1. txt" val fileContent = Source. fromFile(). fromInputStream(inputStream). fromFile("c:/tmp. 在上面的示例代码中，我们定义了一个 copyFile 方法，该方法接受一个 source 文件和一个 destination 文件作为参数。. Sometimes the file will be encoded in a different encoding format. jar) Automatic imports . Jul 6, 2018 · I get a strange behavior of Source. mkString. getLines returns an Iterator which is a collection that is to be read just once. fromResource("readme. I wrote a really basic benchmark of it and most of the time, it was about 5% faster for large files (test was ~35 MB text file) all the way up to 2,800% faster for small files (test was ~30 KB). Then you call getLines which returns an iterator. Identifiers in the scala package and the scala. Scanner. EOL): Iterator[String] = new LineIterator(separator) So it has a default argument. e. zip库逐行读取大型压缩文本文件 Jun 5, 2010 · The signature of getLines is this: def getLines(separator: String = compat. Take a a look at scala. I want to ignore spaces when reading from the file. close. fromFile(file)))(s => URIO(s. Scala可以使用ClassLoader来读取资源文件，ClassLoader提供了多个方法来加载资源，包括getResourceAsStream和getResource。. split('\n') Mar 29, 2022 · 一、文件的读取如Java一般在我们使用特定的操作时需要调用某个包的某一个函数实现。. 在Scala里我们在读取文件时需要使用一个叫“scala. This form will expect to find the resource in the same (relative) folder Jun 22, 2019 · I have a method that takes inputStream as a parameter. toIterable val intStream = lineStream map { s => Integer. If you want to get a resource from a relative path (i. Calling toArray on the Iterator will get you an array of Strings like you want. after { source. 在Scala中，Source. _ val now = new Date(). Value members. fromFile ("somefile"). Apr 22, 2017 · 2. csv") getLines() should return an Iterator[String] which has not yet read the file in and is simply pointing to the first line of that file. scala" for (line <-Source. What you are actually asking is how to apply functional paradigm to this problem. fromFile("file. Apart from using java libraries like Weka, I am mainly working in Scala. 5, “How to process a CSV file in Scala. fromFile("pg1661. _ object Test { def main (args: Array [String]) { val writer May 25, 2020 · 2. foreach { line => buffer = buffer + line } I am not quite sure of the line number you want, but the following should work to get a list of number of lines defined by num. I have an ASCII (text) data file that is around 2 GB in size. Feb 13, 2015 · scala. txt") val lines = file. close() zs and then turn my for-comprehension into: Jul 14, 2013 · Again Scala basic operations are making my life painful :D. Source) and want to get raw bytes out of it. 读取文件. StreamConverters. That iterator is then yielded and stored as data. Successfully merging a pull request may close this issue. Oct 15, 2009 · Class Source in Scala is a representation of a source (text) file, and it's natural to start counting with the first line as number 1. For example: val (head, tail) = scala. Source val path: String = "path/to/text/file. fromFile (file). fromFile("file path","utf-8"). split multiple times, you can do the following ( code on Scastie ): Mar 19, 2021 · Another option is to use ZManaged which is a data type that encapsulates the operation of opening and closing resource: def managedSource(file: String): ZManaged[Any, Throwable, BufferedSource] =. getLines separates on newlines, you'll need to read the whole file in one go instead, with. The Stepper enables creating a Java stream to operate on the collection, see scala. getLines() . hq cz qd my ut ha bi ca qx fa