I have 2 lists:
val list_1 = List((1, 11), (2, 12), (3, 13), (4, 14))
val list_2 = List((1, 111), (2, 122), (3, 133), (4, 144), (1, 123), (2, 234))
I want to replace key in the second list as value of first list, resulting in a new list that looks like:
List ((11, 111), (12, 122), (13, 133), (14, 144), (11, 123), (12, 234))
This is my attempt:
object UniqueTest {
def main(args: Array[String]){
val l_1 = List((1, 11), (2, 12), (3, 13), (4, 14))
val l_2 = List((1, 111), (2,122), (3, 133), (4, 144), (1, 123), (2, 234))
val l_3 = l_2.map(x => (f(x._1, l_1), x._2))
print(l_3)
}
def f(i: Int, list: List[(Int, Int)]): Int = {
for(pair <- list){
if(i == pair._1){
return pair._2
}
}
return 0
}
}
This results in:
((11, 111), (12, 122), (13, 133), (14, 144), (11, 123), (12, 234))
Is the program above a good way to do this? Are there built-in functions in Scala to handle this need, or another way to do this manipulation?
The only real over-complication you make is this line:
val l_3 = l_2.map(x => (f(x._1, l_1), x._2))
Your f function uses an imperative style to loop over a list to find a key. Any time you find yourself doing this, it's a good indication what you want is a map. By doing the for loop each time you're exploding the computational complexity: a map will allow you to fetch the corresponding value for a given key in O(1). With a map you first convert your list, which is a key-value pair, to a datastructure explicit about supporting the key-value pair relationship.
Thus, the first thing you should do is build your map. Scala provides a really easy way to do this with toMap:
val map_1 = list_1.toMap
Then it is just a matter of 'mapping':
val result = list_2.map { case (key, value) => map_1.getOrElse(key, 0), value) }
This takes each case in your list_2, matches the first value (key) to a key in your map_1, retrieves that value (or the default 0) and puts as the first value in a key-value tuple.
You can do:
val map = l_1.toMap // transform l_1 to a Map[Int, Int]
// for each (a, b) in l_2, retrieve the new value v of a and return (v, b)
val res = l_2.map { case (a, b) => (map.getOrElse(a, 0), b) }
The most idiomatic way is zipping them together and then transforming according to your needs:
(list_1 zip list_2) map { case ((k1, v1), (k2, v2)) => (v1, v2) }
Related
I am a newbie to Scala.
I have a Tuple[Int, String]
((1, "alpha"), (2, "beta"), (3, "gamma"), (4, "zeta"), (5, "omega"))
For the above list, I want to print all strings where the corresponding length is 4.
printing length of string of Tuples in Scala
val tuples = List((1, "alpha"), (2, "beta"), (3, "gamma"), (4, "zeta"), (5, "omega"))
println(tuples.map(x => (x._2, x._2.length)))
//List((alpha,5), (beta,4), (gamma,5), (zeta,4), (omega,5))
I want to print all strings where the corresponding length is 4
You can filter first and then print as
val tuples = List((1, "alpha"), (2, "beta"), (3, "gamma"), (4, "zeta"), (5, "omega"))
tuples.filter(_._2.length == 4).foreach(x => println(x._2))
it should print
beta
zeta
You can convert your Tuple to List and then map and filter as you need:
tuple.productIterator.toList
.map{case (a,b) => b.toString}
.filter(_.length==4)
Example:
For the given input:
val tuple = ((1, "alpha"), (2, "beta"), (3, "gamma"), (4, "zeta"), (5, "omega"))
tuple: ((Int, String), (Int, String), (Int, String), (Int, String), (Int, String)) = ((1,alpha),(2,beta),(3,gamma),(4,zeta),(5,omega))
Output:
List[String] = List(beta, zeta)
Let's suppose you have a list of Tuple, and you need all the values with string length equal to 4.
You can do a filter on the list:
val filteredList = list.filter(_._2.length == 4)
And then iterate over each element to print them:
filteredList.foreach(tuple => println(tuple._2))
Here is way to achieve this
scala> val x = ((1, "alpha"), (2, "beta"), (3, "gamma"), (4, "zeta"), (5, "omega"))
x: ((Int, String), (Int, String), (Int, String), (Int, String), (Int, String)) = ((1,alpha),(2,beta),(3,gamma),(4,zeta),(5,omega))
scala> val y = x.productIterator.toList.collect{
case ele : (Int, String) if ele._2.length == 4 => ele._2
}
y: List[String] = List(beta, zeta)
I have two RDDs.
rdd1 = (string, integer)
(a, 10)
(b, 15)
(c, 20)
(d, 20)
(e, 13)
rdd2 = (string, string)
(1, a)
(2, b)
(3, e)
When some of rdd2's values are the same with rdd1's keys, how can I merge them like below? Is it possible?
(1, 10)
(2, 15)
(3, 13)
Try:
rdd1.join(rdd2.map(_.swap)).values
I have an Array[(List(String)), Array[(Int, Int)]] like this
((123, 456, 789), (1, 24))
((89, 284), (2, 6))
((125, 173, 88, 222), (3, 4))
I would like to distribute each element of the first list to the second list, like this
(123, (1, 24))
(456, (1, 24))
(789, (1, 24))
(89, (2, 6))
(284, (2, 6))
(125, (3, 4))
(173, (3, 4))
(88, (3, 4))
(22, (3, 4))
Can anyone help me with this? Thank you very much.
For input data defined as follows:
val data = Array((List("123", "456", "789"), (1, 24)), (List("89", "284"), (2, 6)), (List("125", "173", "88", "222"), (3, 4)))
you can use:
data.flatMap { case (l, ii) => l.map((_, ii)) }
which yields:
Array[(String, (Int, Int))] = Array(("123", (1, 24)), ("456", (1, 24)), ("789", (1, 24)), ("89", (2, 6)), ("284", (2, 6)), ("125", (3, 4)), ("173", (3, 4)), ("88", (3, 4)), ("222", (3, 4)))
which I believe matches what you are looking for.
Based on your example, it seemed to me that you were using a single type.
scala> val xs: List[(List[Int], (Int, Int))] =
| List( ( List(123, 456, 789), (1, 24) ),
| ( List(89, 284), (2,6)),
| ( List(125, 173, 88, 222), (3, 4)) )
xs: List[(List[Int], (Int, Int))] = List((List(123, 456, 789), (1,24)),
(List(89, 284),(2,6)),
(List(125, 173, 88, 222),(3,4)))
Then I wrote this function:
scala> def f[A](xs: List[(List[A], (A, A))]): List[(A, (A, A))] =
| for {
| x <- xs
| head <- x._1
| } yield (head, x._2)
f: [A](xs: List[(List[A], (A, A))])List[(A, (A, A))]
Apply f to xs.
scala> f(xs)
res9: List[(Int, (Int, Int))] = List((123,(1,24)), (456,(1,24)),
(789,(1,24)), (89,(2,6)), (284,(2,6)), (125,(3,4)),
(173,(3,4)), (88,(3,4)), (222,(3,4)))
I have 2 types of array like this:
array one,
Array(productId, categoryId)
(2, 423)
(6, 859)
(3, 423)
(5, 859)
and another array Array((productId1, productId2), count)
((2, 6), 1), ((2, 3), 1), ((6, 5), 1), ((6, 3), 1)
I would like to filter the second array based on the first array,
firstly I want to check array 2 to see if productId1 and productId2 having the same category, if yes, will keep, otherwise will filter out this element.
So the list above will be filtered to remain:
( ((2, 3), 1), ((6, 5), 1) )
Can anybody help me with this? Thank you very much.
If you don't mind working with the first array as a map, ie:
scala> val categ_info = cats = Array((2, 423), (6, 859), (3, 423), (5, 859)).toMap
categ_info: Map[Int, Int] = Map(2 -> 423, 6 -> 859, 3 -> 423, 5 -> 859)
then we have (setting up example data as simple Ints for convenience):
val data = Array(((2, 6), 1), ((2, 3), 1), ((6, 5), 1), ((6, 3), 1))
data.filter { case ((prod1_id, prod2_id), _) =>
categ_info(prod1_id) == categ_info(prod2_id)
}
producing:
res2: Array[((Int, Int), Int)] = Array(((2, 3), 1), ((6, 5), 1))
as requested.
I have 2 arrays Array[(Int, Int)], and Array[(Int, List[String])],
for examples:
(1, 2) and (1, (123, 456, 789))
(2, 8) and (2, (678, 1000))
(3, 4) and (3, (587, 923, 168, 392))
I would like to merge these two arrays into one Array [(Int, List[String], Int)] like this:
(1, (123, 456, 789), 2)
(2, (678, 1000), 8)
(3, (587, 923, 168, 392), 4)
and would like scala still realize the second element is a List[String],
I tried many ways they can combine these 2 maps or arrays, but cannot realize the second element is a List[String], after merging, it treated the second element as Any or Some and cannot traverse it.
I found the solution:
array1.zip(array2).map {
case ((p1,count), (p2,categoryList)) => (p1,categoryList,count)
}