Sorting a list according to some pre-specified rules

Question

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. Here I will try to clarify the ordering I want to achieve:the integers to the right of 3n are smaller than 3n and those to the left of 3n are larger than 3n. Do not change the ordering of integers that are not multiple of 3. Thank you again. I am reading all your answers and testing them so I understand how they work.

Answer 1

Answer

Depending what you mean one of these has your answer.

This matches your example, requiring any of the two numbers to be Divisible by 3 in order to sort them.

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

This matches better your original text description, requires both numbers to be Divisible to sort them. Otherwise they remain in place.

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. Number that are not sorted remain in their place.

Answer 2

Update: A more convenient function that takes a criterion and a sort-order as arguments:

ClearAll[f]
f[criterion_, sortorder_: Greater] := sortorder[##] Boole[AnyTrue[{##}, criterion]] &

Examples:

lst = {12, 9, 3, 42, 30, 10, 11, 1, 2};
sorted = {42, 30, 12, 10, 11, 9, 3, 1, 2}; (* desired output from OP *)   
Sort[lst, f[Divisible[#, 3] &]] == sorted

True

Sort in increasing order with the same criterion:

Sort[lst, f[Divisible[#, 3]&, Less]]

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

Use Divisible[#, 2]& as the criterion to specify the sublist that should be kept in its original order (those elements of lst that does not satisfy the criterion):

Sort[lst, f[Divisible[#, 2]&]]

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

Original answer:

lst = {12, 9, 3, 42, 30, 10, 11, 1, 2};
sorted = {42, 30, 12, 10, 11, 9, 3, 1, 2}; (* desired output from OP *)

ClearAll[f1]
f1[x_] := Boole[Or @@ Divisible[{##}, x]]Greater[##] &;

Sort[lst, f1 @ 3] == sorted

True

Sort[lst, f1 @ 2]

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

Answer 3

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}