I'm trying to find the intersection of two strings in order with Scala. I'm pretty new to Scala, but I feel like this should be a one-liner. I've tried using both map and foldLeft, and have yet to attain the correct answer.
Given two strings, return a list of characters that are the same in order. For instance, "abcd", "acc" should return "a", and "abcd", "abc" should return "abc".
Here are the two functions I've tried so far
(str1 zip str2).map{ case(a, b) => if (a == b) a else ""}
and
(str1 zip str2).foldLeft(""){case(acc,n) => if (n._1 == n._2) acc+n._1.toString else ""}
What I want to do is something like this
(str1 zip str2).map{ case(a, b) => if (a == b) a else break}
but that doesn't work.
I know that I can do this with multiple lines and a for loop, but this feels like a one liner. Can anyone help?
Thanks
(str1 zip str2).takeWhile( pair => pair._1 == pair._2).map( _._1).mkString
Testing it out in the scala REPL:
scala> val str1 = "abcd"
str1: String = abcd
scala> val str2 = "abc"
str2: String = abc
scala> (str1 zip str2).takeWhile( pair => pair._1 == pair._2).map( _._1).mkString
res26: String = abc
Edited to pass both test cases
scala> (str1 zip "acc").takeWhile( pair => pair._1 == pair._2).map( _._1).mkString
res27: String = a
This is not at all efficient, but it is obvious:
def lcp(str1:String, str2:String) =
(str1.inits.toSet intersect str2.inits.toSet).maxBy(_.length)
lcp("abce", "abcd") //> res0: String = abc
lcp("abcd", "bcd") //> res1: String = ""
(take the longest of the intersection of all of the prefixes of string 1 with all of the prefixes of string 2)
Alternatively, to avoid zipping the entire strings:
(s1, s2).zipped.takeWhile(Function.tupled(_ == _)).unzip._1.mkString
Here it is:
scala> val (s1, s2) = ("abcd", "bcd")
s1: String = abcd
s2: String = bcd
scala> Iterator.iterate(s1)(_.init).find(s2.startsWith).get
res1: String = ""
scala> val (s1, s2) = ("abcd", "abc")
s1: String = abcd
s2: String = abc
scala> Iterator.iterate(s1)(_.init).find(s2.startsWith).get
res2: String = abc
Related
I have some strings that have a specific duplicate word between them. I neet to join all these strings keeping the first "duplicated" word and remove the others.
Ex:
val p1 = "Hello John Doe"
val p2 = "Hello Jane Doe"
val p3 = "Hello Mary"
For the output, I'd pass the string to be removed, in this case "Hello" and would like to get a string (or a seq that I can later transform with mkString(" ")) with the contents Hello John Doe Jane Doe Mary.
The tricky part is that .distinct can't be used because it would remove the second "Doe" and it's not desired.
Just scan the words, counting occurrences. Boring.
#tailrec
def keepN(words: List[String], toDedup: String, toKeep: Int = 1, acc: List[String] = Nil) = words match {
case Nil => acc.reverse
case `toDedup` :: tail if toKeep > 0 => keepN(tail, toDedup, toKeep - 1, toDedup::acc)
case `toDedup` :: tail => keepN(tail, toDedup, 0, acc)
case head :: tail => keepN(tail, toDedup, toKeep, head :: acc)
}
You mention a "duplicate word BETWEEN them" which implies that the duplicate is always the first word of each String. So why not just dropping the first few characters?
val pList = p1 +: List(p2, p3).map(_.drop(5).trim)
Or if you want to refer a specific word...
val myWord = "Hello"
val pList = p1 +: List(p2, p3).map(_.drop(myWord.length).trim)
Let's say I have this string:
val mainString: String = "AAA.BBB"
And now I define two children strings:
val firstString: String = ""
val secondString: String = ""
What code should I write to make firstString equals to "AAA", and secondString equals to "BBB"?
The below code works for any amount of strings separated by delimiters
val texto = "111.222.333"
val vet = texto.split(".")
for (st in vet) println(st)
It prints
111
222
333
I am writing a program in which I need to filter a string. So I have a map of characters, and I want the string to filter out all characters that are not in the map. Is there a way for me to do this?
Let's say we have the string and map:
str = "ABCDABCDABCDABCDABCD"
Map('A' -> "A", 'D' -> "D")
Then I want the string to be filtered down to:
str = "BCBCBCBCBC"
Also, if I find a given substring in the string, is there a way I can replace that with a different substring?
So for example, if we have the string:
"The number ten is even"
Could we replace that with:
"The number 10 is even"
To filter the String with the map is just a filter command:
val str = "ABCDABCDABCDABCDABCD"
val m = Map('A' -> "A", 'D' -> "D")
str.filterNot(elem => m.contains(elem))
A more functional alternative as recommended in comments
str.filterNot(m.contains)
Output
scala> str.filterNot(elem => m.contains(elem))
res3: String = BCBCBCBCBC
To replace elements in the String:
string.replace("ten", "10")
Output
scala> val s = "The number ten is even"
s: String = The number ten is even
scala> s.replace("ten", "10")
res4: String = The number 10 is even
I am writing a small program in Spark using Scala, and came across a problem. I have a List/RDD of single word strings and a List/RDD of sentences which might or might not contain words from the list of single words. i.e.
val singles = Array("this", "is")
val sentence = Array("this Date", "is there something", "where are something", "this is a string")
and I want to select the sentences that contains one or more of the words from singles such that the result should be something like:
output[(this, Array(this Date, this is a String)),(is, Array(is there something, this is a string))]
I thought about two approaches, one by splitting the sentence and filtering using .contains. The other is to split and format sentence into a RDD and use the .join for RDD intersection. I am looking at around 50 single words and 5 million sentences, which method would be faster? Are there any other solutions? Could you also help me with the coding, I seem to get no results with my code (although it compiles and run without error)
You can create a set of required keys, look up the keys in sentences and group by keys.
val singles = Array("this", "is")
val sentences = Array("this Date",
"is there something",
"where are something",
"this is a string")
val rdd = sc.parallelize(sentences) // create RDD
val keys = singles.toSet // words required as keys.
val result = rdd.flatMap{ sen =>
val words = sen.split(" ").toSet;
val common = keys & words; // intersect
common.map(x => (x, sen)) // map as key -> sen
}
.groupByKey.mapValues(_.toArray) // group values for a key
.collect // get rdd contents as array
// result:
// Array((this, Array(this Date, this is a string)),
// (is, Array(is there something, this is a string)))
I've just tried to solve your problem and I've ended up with this code:
def check(s:String, l: Array[String]): Boolean = {
var temp:Int = 0
for (element <- l) {
if (element.equals(s)) {temp = temp +1}
}
var result = false
if (temp > 0) {result = true}
result
}
val singles = sc.parallelize(Array("this", "is"))
val sentence = sc.parallelize(Array("this Date", "is there something", "where are something", "this is a string"))
val result = singles.cartesian(sentence)
.filter(x => check(x._1,x._2.split(" ")) == true )
.groupByKey()
.map(x => (x._1,x._2.mkString(", ") )) // pay attention here(*)
result.foreach(println)
The last map line (*) is there just beacause without it I get something with CompactBuffer, like this:
(is,CompactBuffer(is there something, this is a string))
(this,CompactBuffer(this Date, this is a string))
With that map line (with a mkString command) I get a more readable output like this:
(is,is there something, this is a string)
(this,this Date, this is a string)
Hope it could help in some way.
FF
I want to split a String into alternating words. There will always be an even number.
e.g.
val text = "this here is a test sentence"
should transform to some ordered collection type containing
"this", "is", "test"
and
"here", "a", "sentence"
I've come up with
val (l1, l2) = text.split(" ").zipWithIndex.partition(_._2 % 2 == 0) match {
case (a,b) => (a.map(_._1), b.map(_._1))}
which gives me the right results as two Arrays.
Can this be done more elegantly?
scala> val s = "this here is a test sentence"
s: java.lang.String = this here is a test sentence
scala> val List(l1, l2) = s.split(" ").grouped(2).toList.transpose
l1: List[java.lang.String] = List(this, is, test)
l2: List[java.lang.String] = List(here, a, sentence)
So, how about this:
scala> val text = "this here is a test sentence"
text: java.lang.String = this here is a test sentence
scala> val Reg = """\s*(\w+)\s*(\w+)""".r
Reg: scala.util.matching.Regex = \s*(\w+)\s*(\w+)
scala> (for(Reg(x,y) <- Reg.findAllIn(text)) yield(x,y)).toList.unzip
res8: (List[String], List[String]) = (List(this, is, test),List(here, a, sentence))
scala>