seq

问题 So I've made some utility classes and implicit conversions for them. However, it works fine when converting from a Seq but not from a Set, although the code is the same, and those two traits seem rather similar at first sight. What could be the problem, and how would I fix it? import scala.collection.mutable.HashMap trait Indexed[T] { def index: T } class IndexMap[T, V <: Indexed[T]] extends HashMap[T, V] { override def put(key: T, value: V): Option[V] = { require(key == value.index) super

问题 I have a problem in generating proper time sequence using R: fivemin <- seq(as.POSIXct("2014/01/01 0:00:00"), as.POSIXct("2014/04/01 0:00:00"), by="5 mins",tz="EST") time <- data.frame(MasterTime=fivemin) Using above code, I can get a data frame with 25909 observations. However, under Eastern Standard Time (without daylight saving), the number of observations should be 25920. The difference is 1 hour from the transition of daylight saving time on 03/09/20014, because then the time would

问题 I know that this is a little bit tricky but i wonder why it doesn't work! module Main where sillyDebug :: Int -> Int -> Int sillyDebug x y = (print x) `seq` (x + y) main :: IO () main = do print (sillyDebug 1 2) while its ideal is the same as sillyDebug = (trace (show x) False) `seq` (x + y) Is it related to lazy evaluation or side effect in haskell? https://hackhands.com/lazy-evaluation-works-haskell/ 回答1: Merely evaluating some IO action doesn’t do anything at all. You can think of IO sort

问题 Stuck here, trying to convert a List of case class tuples to a tuple of sequences and multi-assign the result. val items = repo.foo.list // gives me a List[(A,B)] I can pull off multi-assignment like so: val(a,b) = (items.map(_._1).toSeq, items.map(_._2).toSeq) but it would be nicer to do in 1 step, along the lines of: val(a,b) = repo.foo.list.map{case(a,b) => (a,b)} 回答1: I am not sure if I understood the question correctly. Maybe unzip works for what you want? Here is a link with some

问题 I'm trying to create a list of words in Scala. I'm new to the language. I have read a lot of posts about how you can't edit immutable objects, but none that has managed to show me how to create the list I need in Scala. I am using var to initialise, but this isn't helping. var wordList = Seq.empty[String] for (x <- docSample.tokens) { wordList.++(x.word) } println(wordList.isEmpty) I would greatly appreciate some help with this. I understand that objects are immutable in Scala (although vars

问题 This question already has answers here : Numbering rows within groups in a data frame (6 answers) Closed 3 years ago . I have a df like ProjectID Dist 1 x 1 y 2 z 2 x 2 h 3 k .... .... I want to add a third column such that we have an incrementing counter for each ProjectID: ProjectID Dist counter 1 x 1 1 y 2 2 z 1 2 x 2 2 h 3 1 k 3 .... .... I've had a look at seq rank and a couple of other bits particularly looking to see if I could use ddply to help: df$counter <- ddply(df,.(projectID),

问题 I think I do not quite understand how F# infers types in sequence expressions and why types are not correctly recognized even if I specify the type of the elements directly from "seq". In the following F# code we have a base class A and two derived classes, B and C: type A(x) = member a.X = x type B(x) = inherit A(x) type C(x) = inherit A(x) If I try to "yield" their instances in a simple sequence expressions, I get two errors: // Doesn't work, but it makes sense. let testSeq = seq { yield A

问题 I have a data frame with 1666 rows. I would like to add a column with a repeating sequence of 1:5 to use with cut() to do cross validation. It would look like this: Y x1 x2 Id1 1 .15 3.6 1 0 1.1 2.2 2 0 .05 3.3 3 0 .45 2.8 4 1 .85 3.1 5 1 1.01 2.9 1 ... ... ... ... I've tried the following 2 ways but get an error message as it seems to only add numbers in increments of the full seq() argument: > tr2$Id1 <- rep(seq(1,5,1), (nrow(tr2)/5)) Error in `$<-.data.frame`(`*tmp*`, "Id", value = c(1, 2,

问题 We know that in F#, seq is lazy evaluated. My question is, if I have a seq with limited number of values, how to convert it into some data type that contains all its value evaluated? > seq { for i in 1 .. 10 do yield i * i };; val it : seq<int> = seq [1; 4; 9; 16; ...] Thanks a lot. 回答1: The answer from @Carsten is correct: you can use Seq.toArray or Seq.toList if you wish to convert lazily evaluated sequences to lists or arrays. Don't use these function to force evaluation, though. The most

问题 Given this code snippet: someFunction x = print x `seq` 1 main = do print (someFunction "test") why doesn't the print x print test when the code is executed? $./seq_test 1 If I replace it with error I can check that the left operand of seq is indeed evaluated. How could I achieve my expected output: test 1 modifying only someFunction ? 回答1: Evaluating an IO action does nothing whatsoever. That's right! If you like, values of IO type are merely "instruction lists". So all you do with that seq