Message Boards

WOLFRAM COMMUNITY

9330 Views

6 Replies

2 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Data Science Engineering Mathematics Mathematica Optimization Numerical Computation

Efficient method to cyclically swap items in a list

Paul Cleary

Posted 10 years ago

Hi Everyone Lets say I have a list a = {15, 5, 14, 1, 6, 10, 9, 2, 13, 7, 16, 17, 11, 20, 8, 19, 18, 12, 3, 4} and I wish to swap the numbers {1, 2, 3, 4} or more specifically rotate that sub list left or right within the larger list, of course the sub list could be any 4 numbers or more than 4. My method has been to find the position of the 4 numbers in the main list and then working with a temporary variable for the first value, do a replace procedure and then the last number takes the value of the temp variable. It works ok but seems clumsy and not very expandable in that it needs a rewrite if I alter the number of items to swap. can any one see a more efficient method? Thanking everyone in advance P.

POSTED BY: Paul Cleary

6 Replies

Sort By:

Wiel Aerts

Posted 10 years ago

You could also use Permutations. pos = Map[Position[a, x_ /; x == #] &, {1, 2, 3, 4}] // Flatten Then just permute your list, 4 times: Table[a = Permute[a, Cycles[{pos}]] , {i, 4}] {{15, 5, 14, 3, 6, 10, 9, 4, 13, 7, 16, 17, 11, 20, 8, 19, 18, 12, 1, 2}, {15, 5, 14, 2, 6, 10, 9, 3, 13, 7, 16, 17, 11, 20, 8, 19, 18, 12, 4, 1}, {15, 5, 14, 1, 6, 10, 9, 2, 13, 7, 16, 17, 11, 20, 8, 19, 18, 12, 3, 4}, {15, 5, 14, 4, 6, 10, 9, 1, 13, 7, 16, 17, 11, 20, 8, 19, 18, 12, 2, 3}}

You could also use Permutations.

pos = Map[Position[a, x_ /; x == #] &, {1, 2, 3, 4}] // Flatten

Then just permute your list, 4 times:

Table[a = Permute[a, Cycles[{pos}]] , {i, 4}]
{{15, 5, 14, 3, 6, 10, 9, 4, 13, 7, 16, 17, 11, 20, 8, 19, 18, 12, 1, 
  2}, {15, 5, 14, 2, 6, 10, 9, 3, 13, 7, 16, 17, 11, 20, 8, 19, 18, 
  12, 4, 1}, {15, 5, 14, 1, 6, 10, 9, 2, 13, 7, 16, 17, 11, 20, 8, 19,
   18, 12, 3, 4}, {15, 5, 14, 4, 6, 10, 9, 1, 13, 7, 16, 17, 11, 20, 
  8, 19, 18, 12, 2, 3}}

POSTED BY: Wiel Aerts

Udo Krause

Udo Krause, NOrganization

Posted 10 years ago

Another way Clear[clearySubRotate] clearySubRotate[l_List?VectorQ, n_Integer (* length sublist ), m_Integer ( left rotate, negative: right *)] := Flatten[RotateLeft[#, m] & /@ Partition[l, n], 1] /; n > 0 && Mod[Length[l], n] == 0 In[30]:= clearySubRotate[Join @@ (Array[#, 7] & /@ {a, b, c, d}), 7, 3] Out[30]= {a[4], a[5], a[6], a[7], a[1], a[2], a[3], b[4], b[5], b[6], b[7], b[1], b[2], b[3], c[4], c[5], c[6], c[7], c[1], c[2], c[3], d[4], d[5], d[6], d[7], d[1], d[2], d[3]}

Another way

Clear[clearySubRotate]
clearySubRotate[l_List?VectorQ, 
  n_Integer (* length sublist *),
  m_Integer (* left rotate, negative: right *)] :=  Flatten[RotateLeft[#, m] & /@ Partition[l, n], 1] /; n > 0 && 
                                                                    Mod[Length[l], n] == 0

In[30]:= clearySubRotate[Join @@ (Array[#, 7] & /@ {a, b, c, d}), 7, 3]
Out[30]= {a[4], a[5], a[6], a[7], a[1], a[2], a[3], 
          b[4], b[5], b[6], b[7], b[1], b[2], b[3], 
          c[4], c[5], c[6], c[7], c[1], c[2], c[3], 
          d[4], d[5], d[6], d[7], d[1], d[2], d[3]}

POSTED BY: Udo Krause

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 10 years ago

a/.{1->2, 2->3, 3->4, 4->1} rotates left: as the arrows indicate, it replaces 1 with 2, 2 with 3, 3 with 4 and 4 with 1. a/.{1 -> 4, 2 -> 1, 3 -> 2, 4 -> 3} rotates right: it replaces 1 with 4, 2 with 1, 3 with 2 and 4 with 3. a/.Thread[Range[4] -> RotateRight[Range[4]]] does the exact same thing, because Thread[Range[4] -> RotateRight[Range[4]]] evaluates to {1 -> 4, 2 -> 1, 3 -> 2, 4 -> 3}

POSTED BY: Gianluca Gorni

Paul Cleary

Posted 10 years ago

Gianluca, will definitely play with that method, even though it didn't see to behave as expected, However, Bill absolutely perfect, that does exactly what I need, thank you.