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3

$begingroup$

Given a list such as

12,9,3,42,30,10,11,1,2

I would like to place only the multiples of 3 in the natural order without changing the order of the other integers, so I would like to get as a result

42,30,12,10,11,9,3,1,2

What is a (simple) way to do this?

Thank you!
This is my first post and I tried to read related questions before posting.
I am just learning my way around in Mathematica. I am trying to write a little program.

list-manipulation

share|improve this question

edited 7 hours ago

rhermans

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi StefanoK, welcome to Mma.SE. Start by taking the tour now and learning about asking and what's on-topic. Always edit if improvable, show due diligence, give brief context, include minimal working example of code and data in formatted form. By doing all this you help us to help you and likely you will inspire great answers. The site depends on participation, as you receive give back: vote and answer questions, keep the site useful, be kind, correct mistakes and share what you have learned.
  $endgroup$
  – rhermans
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please notice that accepting is one of the things to do after your question is answered, but we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers.
  $endgroup$
  – rhermans
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Given a list such as

12,9,3,42,30,10,11,1,2

I would like to place only the multiples of 3 in the natural order without changing the order of the other integers, so I would like to get as a result

42,30,12,10,11,9,3,1,2

What is a (simple) way to do this?

Thank you!
This is my first post and I tried to read related questions before posting.
I am just learning my way around in Mathematica. I am trying to write a little program.

list-manipulation

share|improve this question

edited 7 hours ago

rhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

asked 8 hours ago

StefanoKStefanoK

1611 bronze badge

New contributor

StefanoK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi StefanoK, welcome to Mma.SE. Start by taking the tour now and learning about asking and what's on-topic. Always edit if improvable, show due diligence, give brief context, include minimal working example of code and data in formatted form. By doing all this you help us to help you and likely you will inspire great answers. The site depends on participation, as you receive give back: vote and answer questions, keep the site useful, be kind, correct mistakes and share what you have learned.
  $endgroup$
  – rhermans
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please notice that accepting is one of the things to do after your question is answered, but we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers.
  $endgroup$
  – rhermans
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Given a list such as

12,9,3,42,30,10,11,1,2

I would like to place only the multiples of 3 in the natural order without changing the order of the other integers, so I would like to get as a result

42,30,12,10,11,9,3,1,2

What is a (simple) way to do this?

Thank you!
This is my first post and I tried to read related questions before posting.
I am just learning my way around in Mathematica. I am trying to write a little program.

list-manipulation

share|improve this question

edited 7 hours ago

rhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

asked 8 hours ago

StefanoKStefanoK

1611 bronze badge

New contributor

StefanoK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

12,9,3,42,30,10,11,1,2

42,30,12,10,11,9,3,1,2

share|improve this question

edited 7 hours ago

rhermans

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share|improve this question

edited 7 hours ago

rhermans

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edited 7 hours ago

rhermans

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edited 7 hours ago

rhermans

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asked 8 hours ago

StefanoKStefanoK

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asked 8 hours ago

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2 Answers
2

active

oldest

votes

3

$begingroup$

Answer

Sort[
 12, 9, 3, 42, 30, 10, 11, 1, 2
 , If[
 Divisible[#1, 3] && Divisible[#2, 3]
 , #2 < #1
 , 0
 ] &
 ]
(* 42, 30, 12, 9, 3, 10, 11, 1, 2 *)

Explanation

Sort allows a second argument to define the ordering function.

The ordering function takes two arguments #1 and #2 to be compared and normally should return either True or False. Sort also accepts it to return 0 when two elements should be treated as identical.

So the strategy is to ask first If the numbers should be sorted using Divisible, and then return True or False if they should be sorted, or 0 otherwise.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

rhermansrhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

This would be how I would think of it

Extract multiples of 3

Sort into decreasing order

Extract non-multiples of 3

Join those two results together

and translate that into Mathematica

v=12,9,3,42,30,10,11,1,2;
mult3[x_]:=Mod[x,3]==0;
notmult3[x_]:=Mod[x,3]!=0;
Join[
 Sort[Select[v,mult3],Greater],
 Select[v,notmult3]
]

giving

42, 30, 12, 9, 3, 10, 11, 1, 2

Look up each of those functions in the help system and see how much you can make of this.

The usual Mathematica culture would turn all that into a single line and replace some of those names with abbreviations that are made up of a few punctuation characters. You can start learning that after you have figured out a little more about how to think in Mathematica.

There are always multiple ways of doing anything in Mathematica. Pick a few simple ways to do things that you can remember how to correctly use.

share|improve this answer

answered 7 hours ago

BillBill

6,58566 silver badges99 bronze badges

$endgroup$

add a comment | 

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3

$begingroup$

Answer

Sort[
 12, 9, 3, 42, 30, 10, 11, 1, 2
 , If[
 Divisible[#1, 3] && Divisible[#2, 3]
 , #2 < #1
 , 0
 ] &
 ]
(* 42, 30, 12, 9, 3, 10, 11, 1, 2 *)

Explanation

Sort allows a second argument to define the ordering function.

The ordering function takes two arguments #1 and #2 to be compared and normally should return either True or False. Sort also accepts it to return 0 when two elements should be treated as identical.

So the strategy is to ask first If the numbers should be sorted using Divisible, and then return True or False if they should be sorted, or 0 otherwise.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

rhermansrhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Answer

Sort[
 12, 9, 3, 42, 30, 10, 11, 1, 2
 , If[
 Divisible[#1, 3] && Divisible[#2, 3]
 , #2 < #1
 , 0
 ] &
 ]
(* 42, 30, 12, 9, 3, 10, 11, 1, 2 *)

Explanation

Sort allows a second argument to define the ordering function.

The ordering function takes two arguments #1 and #2 to be compared and normally should return either True or False. Sort also accepts it to return 0 when two elements should be treated as identical.

So the strategy is to ask first If the numbers should be sorted using Divisible, and then return True or False if they should be sorted, or 0 otherwise.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

rhermansrhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

$endgroup$

add a comment | 

$begingroup$

Answer

Sort[
 12, 9, 3, 42, 30, 10, 11, 1, 2
 , If[
 Divisible[#1, 3] && Divisible[#2, 3]
 , #2 < #1
 , 0
 ] &
 ]
(* 42, 30, 12, 9, 3, 10, 11, 1, 2 *)

Explanation

Sort allows a second argument to define the ordering function.

The ordering function takes two arguments #1 and #2 to be compared and normally should return either True or False. Sort also accepts it to return 0 when two elements should be treated as identical.

So the strategy is to ask first If the numbers should be sorted using Divisible, and then return True or False if they should be sorted, or 0 otherwise.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

rhermansrhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

$endgroup$

Sort[
 12, 9, 3, 42, 30, 10, 11, 1, 2
 , If[
 Divisible[#1, 3] && Divisible[#2, 3]
 , #2 < #1
 , 0
 ] &
 ]
(* 42, 30, 12, 9, 3, 10, 11, 1, 2 *)

share|improve this answer

edited 7 hours ago

answered 7 hours ago

rhermansrhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

answered 7 hours ago

rhermansrhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

answered 7 hours ago

rhermansrhermans

22.4k44 gold badges4141 silver badges107107 bronze badges

1

$begingroup$

This would be how I would think of it

Extract multiples of 3

Sort into decreasing order

Extract non-multiples of 3

Join those two results together

and translate that into Mathematica

v=12,9,3,42,30,10,11,1,2;
mult3[x_]:=Mod[x,3]==0;
notmult3[x_]:=Mod[x,3]!=0;
Join[
 Sort[Select[v,mult3],Greater],
 Select[v,notmult3]
]

giving

42, 30, 12, 9, 3, 10, 11, 1, 2

Look up each of those functions in the help system and see how much you can make of this.

The usual Mathematica culture would turn all that into a single line and replace some of those names with abbreviations that are made up of a few punctuation characters. You can start learning that after you have figured out a little more about how to think in Mathematica.

There are always multiple ways of doing anything in Mathematica. Pick a few simple ways to do things that you can remember how to correctly use.

share|improve this answer

answered 7 hours ago

BillBill

6,58566 silver badges99 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

This would be how I would think of it

Extract multiples of 3

Sort into decreasing order

Extract non-multiples of 3

Join those two results together

and translate that into Mathematica

v=12,9,3,42,30,10,11,1,2;
mult3[x_]:=Mod[x,3]==0;
notmult3[x_]:=Mod[x,3]!=0;
Join[
 Sort[Select[v,mult3],Greater],
 Select[v,notmult3]
]

giving

42, 30, 12, 9, 3, 10, 11, 1, 2

Look up each of those functions in the help system and see how much you can make of this.

The usual Mathematica culture would turn all that into a single line and replace some of those names with abbreviations that are made up of a few punctuation characters. You can start learning that after you have figured out a little more about how to think in Mathematica.

There are always multiple ways of doing anything in Mathematica. Pick a few simple ways to do things that you can remember how to correctly use.

share|improve this answer

answered 7 hours ago

BillBill

6,58566 silver badges99 bronze badges

$endgroup$

add a comment | 

$begingroup$

This would be how I would think of it

Extract multiples of 3

Sort into decreasing order

Extract non-multiples of 3

Join those two results together

and translate that into Mathematica

v=12,9,3,42,30,10,11,1,2;
mult3[x_]:=Mod[x,3]==0;
notmult3[x_]:=Mod[x,3]!=0;
Join[
 Sort[Select[v,mult3],Greater],
 Select[v,notmult3]
]

giving

42, 30, 12, 9, 3, 10, 11, 1, 2

Look up each of those functions in the help system and see how much you can make of this.

The usual Mathematica culture would turn all that into a single line and replace some of those names with abbreviations that are made up of a few punctuation characters. You can start learning that after you have figured out a little more about how to think in Mathematica.

There are always multiple ways of doing anything in Mathematica. Pick a few simple ways to do things that you can remember how to correctly use.

share|improve this answer

answered 7 hours ago

BillBill

6,58566 silver badges99 bronze badges

$endgroup$

v=12,9,3,42,30,10,11,1,2;
mult3[x_]:=Mod[x,3]==0;
notmult3[x_]:=Mod[x,3]!=0;
Join[
 Sort[Select[v,mult3],Greater],
 Select[v,notmult3]
]

42, 30, 12, 9, 3, 10, 11, 1, 2

share|improve this answer

answered 7 hours ago

BillBill

6,58566 silver badges99 bronze badges

answered 7 hours ago

BillBill

6,58566 silver badges99 bronze badges

answered 7 hours ago

BillBill

6,58566 silver badges99 bronze badges