How to remake function that group list to simple recursion?

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I write function that group list elements by index, with odd index in first list, even in second. But I don not know how to make it with simple recursion and don not get type mismatch.

Here is the code:

// Simple recursion
def group1(list: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case head :: tail => // how can I make this case?


group1(List(2, 6, 7, 9, 0, 4, 1))

// Tail recursion
def group2(list: List[Int]): (List[Int], List[Int]) = 
 def group2Helper(list: List[Int], listA: List[Int], listB: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (listA.reverse, listB.reverse)
 case head :: Nil => ((head :: listA).reverse, listB.reverse)
 case head :: headNext :: tail => group2Helper(tail, head :: listA, headNext :: listB)
 
 group2Helper(list, Nil, Nil)


group2(List(2, 6, 7, 9, 0, 4, 1))

asked Nov 12 at 7:31

Ilya

335

What did you try that caused a type mismatch?
– Bergi
Nov 12 at 7:38

Don't forget to add headNext (as in head :: next :: tail) to the simple recursion as well
– Bergi
Nov 12 at 7:38

@Bergi if I trying to do this function like another with simple recursion like this: case head :: headnext :: tail => (head :: group1(List(tail.head)), headnext :: group1(tail.tail)) It gives me mismatch because I can not add head element to (_, _), sorry I don not know how it calls on English. This function is not correct, here it is going about idea
– Ilya
Nov 12 at 7:48

2

You don't want to group1 the .head and .tail of tail - you just want to group1(tail). And then group1(tail) will return a tuple of two lists. You will want to prepend head to the first and headNext to the second one. Use temporary variables so that you need to call group1(tail) only once and so that you can break up the result into the two lists.
– Bergi
Nov 12 at 8:02

add a comment |

up vote
0
down vote

favorite

I write function that group list elements by index, with odd index in first list, even in second. But I don not know how to make it with simple recursion and don not get type mismatch.

Here is the code:

// Simple recursion
def group1(list: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case head :: tail => // how can I make this case?


group1(List(2, 6, 7, 9, 0, 4, 1))

// Tail recursion
def group2(list: List[Int]): (List[Int], List[Int]) = 
 def group2Helper(list: List[Int], listA: List[Int], listB: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (listA.reverse, listB.reverse)
 case head :: Nil => ((head :: listA).reverse, listB.reverse)
 case head :: headNext :: tail => group2Helper(tail, head :: listA, headNext :: listB)
 
 group2Helper(list, Nil, Nil)


group2(List(2, 6, 7, 9, 0, 4, 1))

asked Nov 12 at 7:31

Ilya

335

What did you try that caused a type mismatch?
– Bergi
Nov 12 at 7:38

Don't forget to add headNext (as in head :: next :: tail) to the simple recursion as well
– Bergi
Nov 12 at 7:38

@Bergi if I trying to do this function like another with simple recursion like this: case head :: headnext :: tail => (head :: group1(List(tail.head)), headnext :: group1(tail.tail)) It gives me mismatch because I can not add head element to (_, _), sorry I don not know how it calls on English. This function is not correct, here it is going about idea
– Ilya
Nov 12 at 7:48

2

You don't want to group1 the .head and .tail of tail - you just want to group1(tail). And then group1(tail) will return a tuple of two lists. You will want to prepend head to the first and headNext to the second one. Use temporary variables so that you need to call group1(tail) only once and so that you can break up the result into the two lists.
– Bergi
Nov 12 at 8:02

add a comment |

up vote
0
down vote

favorite

I write function that group list elements by index, with odd index in first list, even in second. But I don not know how to make it with simple recursion and don not get type mismatch.

Here is the code:

// Simple recursion
def group1(list: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case head :: tail => // how can I make this case?


group1(List(2, 6, 7, 9, 0, 4, 1))

// Tail recursion
def group2(list: List[Int]): (List[Int], List[Int]) = 
 def group2Helper(list: List[Int], listA: List[Int], listB: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (listA.reverse, listB.reverse)
 case head :: Nil => ((head :: listA).reverse, listB.reverse)
 case head :: headNext :: tail => group2Helper(tail, head :: listA, headNext :: listB)
 
 group2Helper(list, Nil, Nil)


group2(List(2, 6, 7, 9, 0, 4, 1))

asked Nov 12 at 7:31

Ilya

335

I write function that group list elements by index, with odd index in first list, even in second. But I don not know how to make it with simple recursion and don not get type mismatch.

Here is the code:

// Simple recursion
def group1(list: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case head :: tail => // how can I make this case?


group1(List(2, 6, 7, 9, 0, 4, 1))

// Tail recursion
def group2(list: List[Int]): (List[Int], List[Int]) = 
 def group2Helper(list: List[Int], listA: List[Int], listB: List[Int]): (List[Int], List[Int]) = list match 
 case Nil => (listA.reverse, listB.reverse)
 case head :: Nil => ((head :: listA).reverse, listB.reverse)
 case head :: headNext :: tail => group2Helper(tail, head :: listA, headNext :: listB)
 
 group2Helper(list, Nil, Nil)


group2(List(2, 6, 7, 9, 0, 4, 1))

scala function functional-programming

asked Nov 12 at 7:31

Ilya

335

asked Nov 12 at 7:31

Ilya

335

asked Nov 12 at 7:31

Ilya

335

asked Nov 12 at 7:31

Ilya

335

asked Nov 12 at 7:31

Ilya

335

What did you try that caused a type mismatch?
– Bergi
Nov 12 at 7:38

Don't forget to add headNext (as in head :: next :: tail) to the simple recursion as well
– Bergi
Nov 12 at 7:38

@Bergi if I trying to do this function like another with simple recursion like this: case head :: headnext :: tail => (head :: group1(List(tail.head)), headnext :: group1(tail.tail)) It gives me mismatch because I can not add head element to (_, _), sorry I don not know how it calls on English. This function is not correct, here it is going about idea
– Ilya
Nov 12 at 7:48

2

You don't want to group1 the .head and .tail of tail - you just want to group1(tail). And then group1(tail) will return a tuple of two lists. You will want to prepend head to the first and headNext to the second one. Use temporary variables so that you need to call group1(tail) only once and so that you can break up the result into the two lists.
– Bergi
Nov 12 at 8:02

add a comment |

What did you try that caused a type mismatch?
– Bergi
Nov 12 at 7:38

Don't forget to add headNext (as in head :: next :: tail) to the simple recursion as well
– Bergi
Nov 12 at 7:38

@Bergi if I trying to do this function like another with simple recursion like this: case head :: headnext :: tail => (head :: group1(List(tail.head)), headnext :: group1(tail.tail)) It gives me mismatch because I can not add head element to (_, _), sorry I don not know how it calls on English. This function is not correct, here it is going about idea
– Ilya
Nov 12 at 7:48

2

You don't want to group1 the .head and .tail of tail - you just want to group1(tail). And then group1(tail) will return a tuple of two lists. You will want to prepend head to the first and headNext to the second one. Use temporary variables so that you need to call group1(tail) only once and so that you can break up the result into the two lists.
– Bergi
Nov 12 at 8:02

What did you try that caused a type mismatch?
– Bergi
Nov 12 at 7:38

Don't forget to add headNext (as in head :: next :: tail) to the simple recursion as well
– Bergi
Nov 12 at 7:38

@Bergi if I trying to do this function like another with simple recursion like this: case head :: headnext :: tail => (head :: group1(List(tail.head)), headnext :: group1(tail.tail)) It gives me mismatch because I can not add head element to (_, _), sorry I don not know how it calls on English. This function is not correct, here it is going about idea
– Ilya
Nov 12 at 7:48

You don't want to group1 the .head and .tail of tail - you just want to group1(tail). And then group1(tail) will return a tuple of two lists. You will want to prepend head to the first and headNext to the second one. Use temporary variables so that you need to call group1(tail) only once and so that you can break up the result into the two lists.
– Bergi
Nov 12 at 8:02

add a comment |

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

You have to invoke the next recursion, unpack the result tuple, pre-pend each head element to the proper List, and repackage the new result tuple.

def group1(list: List[Int]) :(List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case hdA :: hdB :: tail => val (lstA, lstB) = group1(tail)
 (hdA :: lstA, hdB :: lstB)

answered Nov 12 at 8:16

jwvh

25.3k52038

Thanks, now I understand how to use variable in case. But I don not know is it still functional way programming? Or this variable does not like a statement or imperative style?
– Ilya
Nov 12 at 8:43

1

This group1() method creates 7 variables. They are (in order) list, head, hdA, hdB, tail, lstA, and lstB. Every one of them is a val (i.e. immutable) so it doesn't conflict with the tenets of Functional Programming.
– jwvh
Nov 12 at 9:46

Thanks a lot for explaining
– Ilya
Nov 12 at 11:10

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

You have to invoke the next recursion, unpack the result tuple, pre-pend each head element to the proper List, and repackage the new result tuple.

def group1(list: List[Int]) :(List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case hdA :: hdB :: tail => val (lstA, lstB) = group1(tail)
 (hdA :: lstA, hdB :: lstB)

answered Nov 12 at 8:16

jwvh

25.3k52038

Thanks, now I understand how to use variable in case. But I don not know is it still functional way programming? Or this variable does not like a statement or imperative style?
– Ilya
Nov 12 at 8:43

1

This group1() method creates 7 variables. They are (in order) list, head, hdA, hdB, tail, lstA, and lstB. Every one of them is a val (i.e. immutable) so it doesn't conflict with the tenets of Functional Programming.
– jwvh
Nov 12 at 9:46

Thanks a lot for explaining
– Ilya
Nov 12 at 11:10

add a comment |

up vote
2
down vote

accepted

You have to invoke the next recursion, unpack the result tuple, pre-pend each head element to the proper List, and repackage the new result tuple.

def group1(list: List[Int]) :(List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case hdA :: hdB :: tail => val (lstA, lstB) = group1(tail)
 (hdA :: lstA, hdB :: lstB)

answered Nov 12 at 8:16

jwvh

25.3k52038

Thanks, now I understand how to use variable in case. But I don not know is it still functional way programming? Or this variable does not like a statement or imperative style?
– Ilya
Nov 12 at 8:43

1

This group1() method creates 7 variables. They are (in order) list, head, hdA, hdB, tail, lstA, and lstB. Every one of them is a val (i.e. immutable) so it doesn't conflict with the tenets of Functional Programming.
– jwvh
Nov 12 at 9:46

Thanks a lot for explaining
– Ilya
Nov 12 at 11:10

add a comment |

up vote
2
down vote

accepted

You have to invoke the next recursion, unpack the result tuple, pre-pend each head element to the proper List, and repackage the new result tuple.

def group1(list: List[Int]) :(List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case hdA :: hdB :: tail => val (lstA, lstB) = group1(tail)
 (hdA :: lstA, hdB :: lstB)

answered Nov 12 at 8:16

jwvh

25.3k52038

You have to invoke the next recursion, unpack the result tuple, pre-pend each head element to the proper List, and repackage the new result tuple.

def group1(list: List[Int]) :(List[Int], List[Int]) = list match 
 case Nil => (Nil, Nil)
 case head :: Nil => (List(head), Nil)
 case hdA :: hdB :: tail => val (lstA, lstB) = group1(tail)
 (hdA :: lstA, hdB :: lstB)

answered Nov 12 at 8:16

jwvh

25.3k52038

answered Nov 12 at 8:16

jwvh

25.3k52038

answered Nov 12 at 8:16

jwvh

25.3k52038

answered Nov 12 at 8:16

jwvh

25.3k52038

Thanks, now I understand how to use variable in case. But I don not know is it still functional way programming? Or this variable does not like a statement or imperative style?
– Ilya
Nov 12 at 8:43

1

This group1() method creates 7 variables. They are (in order) list, head, hdA, hdB, tail, lstA, and lstB. Every one of them is a val (i.e. immutable) so it doesn't conflict with the tenets of Functional Programming.
– jwvh
Nov 12 at 9:46

Thanks a lot for explaining
– Ilya
Nov 12 at 11:10

add a comment |

Thanks, now I understand how to use variable in case. But I don not know is it still functional way programming? Or this variable does not like a statement or imperative style?
– Ilya
Nov 12 at 8:43

1

This group1() method creates 7 variables. They are (in order) list, head, hdA, hdB, tail, lstA, and lstB. Every one of them is a val (i.e. immutable) so it doesn't conflict with the tenets of Functional Programming.
– jwvh
Nov 12 at 9:46

Thanks a lot for explaining
– Ilya
Nov 12 at 11:10

Thanks, now I understand how to use variable in case. But I don not know is it still functional way programming? Or this variable does not like a statement or imperative style?
– Ilya
Nov 12 at 8:43

This group1() method creates 7 variables. They are (in order) list, head, hdA, hdB, tail, lstA, and lstB. Every one of them is a val (i.e. immutable) so it doesn't conflict with the tenets of Functional Programming.
– jwvh
Nov 12 at 9:46

Thanks a lot for explaining
– Ilya
Nov 12 at 11:10

add a comment |

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