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Scala group by list of list and subtracts grouped values


Scala group by list of list and subtracts grouped values

By : leeflow
Date : November 14 2020, 04:51 PM
Hope that helps Below sample data , Consider
code :
combineList.groupBy(_._1).mapValues(xs => xs.map(v => X-v._2))
Map(E -> List(75, 77, 77, 84, 97), F -> List(86, 86, 86, 83, 84), A -> List(86, 76, 64, 87, 85), G -> List(31, 80), B -> List(87, 86, 64, 57), C -> List(86, 75, 77, 88), D -> List(84, 45, 75, 67, 89))


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How can I group the a list into tuple grouped items in Scala?

How can I group the a list into tuple grouped items in Scala?


By : Jonathan A.
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You may use grouped method for the List class: http://www.scala-lang.org/api/current/index.html#scala.collection.immutable.List
code :
list.grouped(2).toList.collect { case a :: b :: Nil => (a,b) }
res1: List[(Int, Int)] = List((1,2), (3,4), (5,6), (7,8), (9,10))
How to sum values and group them by a key value in Scala's List of Map?

How to sum values and group them by a key value in Scala's List of Map?


By : user3103323
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I have a List of Map: , Also using foldLeft:
code :
list.foldLeft(Map[String, Int]().withDefaultValue(0))((res, v) => {
  val key = v("id").toString
  res + (key -> (res(key) + v("value").asInstanceOf[Int]))
})
(Map[String, Any]().withDefaultValue(0) :: list).reduceLeft((res, v) => {
  val key = v("id").toString
  res + (key -> (res(key).asInstanceOf[Int] + v("value").asInstanceOf[Int]))
})
  def reduceLeft[B >: A](f: (B, A) => B): B =
    if (isEmpty) throw new UnsupportedOperationException("empty.reduceLeft")
    else tail.foldLeft[B](head)(f)
list.par.reduce((a, b) => {
  def toResultMap(m: Map[String, Any]) =
    if (m.contains("id"))
      Map(m("id").toString -> m("value")).withDefaultValue(0)
    else m
  val aM = toResultMap(a)
  val bM = toResultMap(b)
  aM.foldLeft(bM)((res, v) =>
    res + (v._1 -> (res(v._1).asInstanceOf[Int] + v._2.asInstanceOf[Int])))
})
Calculate average of list values grouped by second list

Calculate average of list values grouped by second list


By : Jose Castaneda
Date : March 29 2020, 07:55 AM
hope this fix your issue You can first perform a group by. We can do this for instance with a defaultdict:
code :
from collections import defaultdict

col = defaultdict(list)

for ai,bi in zip(a,b):
    col[ai].append(bi)
>>> col
defaultdict(<class 'list'>, {'C': [3, 2], 'B': [4], 'D': [5], 'A': [2, 8, 1]})
>>> {key:sum(vals)/len(vals) for key,vals in col.items()}
{'C': 2.5, 'B': 4.0, 'D': 5.0, 'A': 3.6666666666666665}
a,b = zip(*[(key,sum(vals)/len(vals)) for key,vals in col.items()])
>>> a,b = zip(*[(key,sum(vals)/len(vals)) for key,vals in col.items()])
>>> a
('C', 'B', 'D', 'A')
>>> b
(2.5, 4.0, 5.0, 3.6666666666666665)
a,b = map(list,zip(*[(key,sum(vals)/len(vals)) for key,vals in col.items()]))
>>> a,b = map(list,zip(*[(key,sum(vals)/len(vals)) for key,vals in col.items()]))
>>> a
['C', 'B', 'D', 'A']
>>> b
[2.5, 4.0, 5.0, 3.6666666666666665]
Subtracts values one list from another in javascript

Subtracts values one list from another in javascript


By : user7622480
Date : March 29 2020, 07:55 AM
this one helps. A simple solution is to use Array#filter, and check if the value exists using Array#indexOf:
code :
var a = [1, 2, 2, 2, 5];
var b = [2];

var result = a.filter(function(n) {
  return b.indexOf(n) === -1;
});

console.log(result);
var a = [1, 2, 2, 2, 5];
var b = [2];
var bDict = b.reduce(function(d, n) {
  d[n] = true;
  
  return d;
}, Object.create(null));

var result = a.filter(function(n) {
  return !bDict[n];
});

console.log(result);
How to convert a list to list of lists by group the elements when an element repeats in Scala idiomatic way

How to convert a list to list of lists by group the elements when an element repeats in Scala idiomatic way


By : clempat
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Using the solution provided for below question, I have come up with the below solution
https://stackoverflow.com/a/52976957/1696418
code :
mySeq.foldLeft(List.empty[List[Int]]) {
  case (acc, i) if acc.isEmpty => List(List(i))
  case (acc, i) if acc.last.contains(i) => acc :+ List(i)
  case (acc, i) => acc.init :+ (acc.last :+ i)
}
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