Count how many times numbers from a list occur in a list of tupled intervals in Scala

Say I have a list of tuples:

 val ranges= List((1,4), (5,8), (9,10))

and a list of numbers

val nums = List(2,2,3,7,8,9)

I want to make a map from tuple in ranges to how many times a given number from nums fall into the interval of that tuple.

Output:

Map ((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

What is the best way to go about it in Scala

I have been trying to use for loops and keeping a counter but am falling short. Any help would be very much appreciated.

Best Regards

asked Jan 18 at 14:19

Arun Shyam

2251417

add a comment |

Say I have a list of tuples:

 val ranges= List((1,4), (5,8), (9,10))

and a list of numbers

val nums = List(2,2,3,7,8,9)

I want to make a map from tuple in ranges to how many times a given number from nums fall into the interval of that tuple.

Output:

Map ((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

What is the best way to go about it in Scala

I have been trying to use for loops and keeping a counter but am falling short. Any help would be very much appreciated.

Best Regards

asked Jan 18 at 14:19

Arun Shyam

2251417

add a comment |

Say I have a list of tuples:

 val ranges= List((1,4), (5,8), (9,10))

and a list of numbers

val nums = List(2,2,3,7,8,9)

I want to make a map from tuple in ranges to how many times a given number from nums fall into the interval of that tuple.

Output:

Map ((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

What is the best way to go about it in Scala

I have been trying to use for loops and keeping a counter but am falling short. Any help would be very much appreciated.

Best Regards

asked Jan 18 at 14:19

Arun Shyam

2251417

Say I have a list of tuples:

 val ranges= List((1,4), (5,8), (9,10))

and a list of numbers

val nums = List(2,2,3,7,8,9)

I want to make a map from tuple in ranges to how many times a given number from nums fall into the interval of that tuple.

Output:

Map ((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

What is the best way to go about it in Scala

I have been trying to use for loops and keeping a counter but am falling short. Any help would be very much appreciated.

Best Regards

scala tuples accumulator

asked Jan 18 at 14:19

Arun Shyam

2251417

asked Jan 18 at 14:19

Arun Shyam

2251417

asked Jan 18 at 14:19

Arun Shyam

2251417

asked Jan 18 at 14:19

Arun Shyam

2251417

asked Jan 18 at 14:19

Arun Shyam

2251417

add a comment |

2 Answers
2

active

oldest

votes

Something like this:

val ranges = List((1, 4), (5, 8), (9, 10))

val nums = List(2, 2, 3, 7, 8, 9)



val occurences = ranges.map { case (l, r) => nums.count((l to r) contains _) }

val map = (ranges zip occurences).toMap



println(map) // Map((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

Basically it first calculates the number of occurrences, [3, 2, 1]. From there it's easy to construct a map. And the way it calculates the occurrences is:

go through the list of ranges

transform each range into number of occurrences for that range, which is done like this :
- how many numbers from the list nums are contained in that range?

EDIT: Not sure why this answer is getting downvoted, perhaps I misunderstood the question somehow.

edited Jan 18 at 15:01

answered Jan 18 at 14:43

slouc

5,6062729

3

I don't know who downvoted you, it's the good answer and very scala-ish.

– Etienne Herlaut
Jan 18 at 14:58

No you didn’t misunderstand it, I believe on a large (l, r) range, contains can be a bottleneck. But I corrected for that myself 👍🏽. This was definitely right

– Arun Shyam
Jan 18 at 20:19

@Arun Shyam Yes this is inefficient, but I wanted to point you in the direction of thinking about the solution in a different way. BTW if you get a question like this on something like HackerRank, it might even be outside of time constraints and you would in that case be forced to an iterative approach with loops and counters. But whenever I can, I prefer FP solutions.

– slouc
Jan 19 at 14:51

add a comment |

Here is an efficient single-pass solution:

ranges

  .map(r => r -> nums.count(n => n >= r._1 && n <= r._2))

  .toMap

This avoids the overhead of creating a list of numbers and then zipping them with the ranges in a separate step.

This is a version that uses more Scala features but is a bit too fancy:

(for {

  r <- ranges

  range = r._1 to r._2

} yield r -> nums.count(range.contains)

).toMap

This is also less efficient because contains has to allow for ranges with a step value and is therefore more complicated.

And here is an even more efficient version that avoids any temporary data structures:

val result: Map[(Int, Int), Int] =

  ranges.map(r => r -> nums.count(n => n >= r._1 && n <= r._2))(collection.breakOut)

See this explanation of breakOut if you are not familiar with it. Using breakOut means that the map call will build the Map directly rather than creating a List that has to be converted to a Map using toMap.

edited Jan 19 at 12:21

answered Jan 18 at 15:09

Tim

5,0791317

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Something like this:

val ranges = List((1, 4), (5, 8), (9, 10))

val nums = List(2, 2, 3, 7, 8, 9)



val occurences = ranges.map { case (l, r) => nums.count((l to r) contains _) }

val map = (ranges zip occurences).toMap



println(map) // Map((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

Basically it first calculates the number of occurrences, [3, 2, 1]. From there it's easy to construct a map. And the way it calculates the occurrences is:

go through the list of ranges

transform each range into number of occurrences for that range, which is done like this :
- how many numbers from the list nums are contained in that range?

EDIT: Not sure why this answer is getting downvoted, perhaps I misunderstood the question somehow.

edited Jan 18 at 15:01

answered Jan 18 at 14:43

slouc

5,6062729

3

I don't know who downvoted you, it's the good answer and very scala-ish.

– Etienne Herlaut
Jan 18 at 14:58

No you didn’t misunderstand it, I believe on a large (l, r) range, contains can be a bottleneck. But I corrected for that myself 👍🏽. This was definitely right

– Arun Shyam
Jan 18 at 20:19

@Arun Shyam Yes this is inefficient, but I wanted to point you in the direction of thinking about the solution in a different way. BTW if you get a question like this on something like HackerRank, it might even be outside of time constraints and you would in that case be forced to an iterative approach with loops and counters. But whenever I can, I prefer FP solutions.

– slouc
Jan 19 at 14:51

add a comment |

Something like this:

val ranges = List((1, 4), (5, 8), (9, 10))

val nums = List(2, 2, 3, 7, 8, 9)



val occurences = ranges.map { case (l, r) => nums.count((l to r) contains _) }

val map = (ranges zip occurences).toMap



println(map) // Map((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

Basically it first calculates the number of occurrences, [3, 2, 1]. From there it's easy to construct a map. And the way it calculates the occurrences is:

go through the list of ranges

transform each range into number of occurrences for that range, which is done like this :
- how many numbers from the list nums are contained in that range?

EDIT: Not sure why this answer is getting downvoted, perhaps I misunderstood the question somehow.

edited Jan 18 at 15:01

answered Jan 18 at 14:43

slouc

5,6062729

3

I don't know who downvoted you, it's the good answer and very scala-ish.

– Etienne Herlaut
Jan 18 at 14:58

No you didn’t misunderstand it, I believe on a large (l, r) range, contains can be a bottleneck. But I corrected for that myself 👍🏽. This was definitely right

– Arun Shyam
Jan 18 at 20:19

@Arun Shyam Yes this is inefficient, but I wanted to point you in the direction of thinking about the solution in a different way. BTW if you get a question like this on something like HackerRank, it might even be outside of time constraints and you would in that case be forced to an iterative approach with loops and counters. But whenever I can, I prefer FP solutions.

– slouc
Jan 19 at 14:51

add a comment |

Something like this:

val ranges = List((1, 4), (5, 8), (9, 10))

val nums = List(2, 2, 3, 7, 8, 9)



val occurences = ranges.map { case (l, r) => nums.count((l to r) contains _) }

val map = (ranges zip occurences).toMap



println(map) // Map((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

Basically it first calculates the number of occurrences, [3, 2, 1]. From there it's easy to construct a map. And the way it calculates the occurrences is:

go through the list of ranges

transform each range into number of occurrences for that range, which is done like this :
- how many numbers from the list nums are contained in that range?

EDIT: Not sure why this answer is getting downvoted, perhaps I misunderstood the question somehow.

edited Jan 18 at 15:01

answered Jan 18 at 14:43

slouc

5,6062729

Something like this:

val ranges = List((1, 4), (5, 8), (9, 10))

val nums = List(2, 2, 3, 7, 8, 9)



val occurences = ranges.map { case (l, r) => nums.count((l to r) contains _) }

val map = (ranges zip occurences).toMap



println(map) // Map((1,4) -> 3, (5,8) -> 2, (9,10) -> 1)

Basically it first calculates the number of occurrences, [3, 2, 1]. From there it's easy to construct a map. And the way it calculates the occurrences is:

go through the list of ranges

transform each range into number of occurrences for that range, which is done like this :
- how many numbers from the list nums are contained in that range?

EDIT: Not sure why this answer is getting downvoted, perhaps I misunderstood the question somehow.

edited Jan 18 at 15:01

answered Jan 18 at 14:43

slouc

5,6062729

edited Jan 18 at 15:01

answered Jan 18 at 14:43

slouc

5,6062729

answered Jan 18 at 14:43

slouc

5,6062729

answered Jan 18 at 14:43

slouc

5,6062729

3

I don't know who downvoted you, it's the good answer and very scala-ish.

– Etienne Herlaut
Jan 18 at 14:58

No you didn’t misunderstand it, I believe on a large (l, r) range, contains can be a bottleneck. But I corrected for that myself 👍🏽. This was definitely right

– Arun Shyam
Jan 18 at 20:19

@Arun Shyam Yes this is inefficient, but I wanted to point you in the direction of thinking about the solution in a different way. BTW if you get a question like this on something like HackerRank, it might even be outside of time constraints and you would in that case be forced to an iterative approach with loops and counters. But whenever I can, I prefer FP solutions.

– slouc
Jan 19 at 14:51

add a comment |

3

I don't know who downvoted you, it's the good answer and very scala-ish.

– Etienne Herlaut
Jan 18 at 14:58

No you didn’t misunderstand it, I believe on a large (l, r) range, contains can be a bottleneck. But I corrected for that myself 👍🏽. This was definitely right

– Arun Shyam
Jan 18 at 20:19

@Arun Shyam Yes this is inefficient, but I wanted to point you in the direction of thinking about the solution in a different way. BTW if you get a question like this on something like HackerRank, it might even be outside of time constraints and you would in that case be forced to an iterative approach with loops and counters. But whenever I can, I prefer FP solutions.

– slouc
Jan 19 at 14:51

I don't know who downvoted you, it's the good answer and very scala-ish.

– Etienne Herlaut
Jan 18 at 14:58

No you didn’t misunderstand it, I believe on a large (l, r) range, contains can be a bottleneck. But I corrected for that myself 👍🏽. This was definitely right

– Arun Shyam
Jan 18 at 20:19

@Arun Shyam Yes this is inefficient, but I wanted to point you in the direction of thinking about the solution in a different way. BTW if you get a question like this on something like HackerRank, it might even be outside of time constraints and you would in that case be forced to an iterative approach with loops and counters. But whenever I can, I prefer FP solutions.

– slouc
Jan 19 at 14:51

add a comment |

Here is an efficient single-pass solution:

ranges

  .map(r => r -> nums.count(n => n >= r._1 && n <= r._2))

  .toMap

This avoids the overhead of creating a list of numbers and then zipping them with the ranges in a separate step.

This is a version that uses more Scala features but is a bit too fancy:

(for {

  r <- ranges

  range = r._1 to r._2

} yield r -> nums.count(range.contains)

).toMap

This is also less efficient because contains has to allow for ranges with a step value and is therefore more complicated.

And here is an even more efficient version that avoids any temporary data structures:

val result: Map[(Int, Int), Int] =

  ranges.map(r => r -> nums.count(n => n >= r._1 && n <= r._2))(collection.breakOut)

edited Jan 19 at 12:21

answered Jan 18 at 15:09

Tim

5,0791317

add a comment |

Here is an efficient single-pass solution:

ranges

  .map(r => r -> nums.count(n => n >= r._1 && n <= r._2))

  .toMap

This avoids the overhead of creating a list of numbers and then zipping them with the ranges in a separate step.

This is a version that uses more Scala features but is a bit too fancy:

(for {

  r <- ranges

  range = r._1 to r._2

} yield r -> nums.count(range.contains)

).toMap

This is also less efficient because contains has to allow for ranges with a step value and is therefore more complicated.

And here is an even more efficient version that avoids any temporary data structures:

val result: Map[(Int, Int), Int] =

  ranges.map(r => r -> nums.count(n => n >= r._1 && n <= r._2))(collection.breakOut)

edited Jan 19 at 12:21

answered Jan 18 at 15:09

Tim

5,0791317

add a comment |

Here is an efficient single-pass solution:

ranges

  .map(r => r -> nums.count(n => n >= r._1 && n <= r._2))

  .toMap

This avoids the overhead of creating a list of numbers and then zipping them with the ranges in a separate step.

This is a version that uses more Scala features but is a bit too fancy:

(for {

  r <- ranges

  range = r._1 to r._2

} yield r -> nums.count(range.contains)

).toMap

This is also less efficient because contains has to allow for ranges with a step value and is therefore more complicated.

And here is an even more efficient version that avoids any temporary data structures:

val result: Map[(Int, Int), Int] =

  ranges.map(r => r -> nums.count(n => n >= r._1 && n <= r._2))(collection.breakOut)

edited Jan 19 at 12:21

answered Jan 18 at 15:09

Tim

5,0791317

Here is an efficient single-pass solution:

ranges

  .map(r => r -> nums.count(n => n >= r._1 && n <= r._2))

  .toMap

This avoids the overhead of creating a list of numbers and then zipping them with the ranges in a separate step.

This is a version that uses more Scala features but is a bit too fancy:

(for {

  r <- ranges

  range = r._1 to r._2

} yield r -> nums.count(range.contains)

).toMap

This is also less efficient because contains has to allow for ranges with a step value and is therefore more complicated.

And here is an even more efficient version that avoids any temporary data structures:

val result: Map[(Int, Int), Int] =

  ranges.map(r => r -> nums.count(n => n >= r._1 && n <= r._2))(collection.breakOut)

edited Jan 19 at 12:21

answered Jan 18 at 15:09

Tim

5,0791317

edited Jan 19 at 12:21

answered Jan 18 at 15:09

Tim

5,0791317

answered Jan 18 at 15:09

Tim

5,0791317

answered Jan 18 at 15:09

Tim

5,0791317

add a comment |

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