How get the permutation of two lists but the elements of each list remain in the same order?












3












$begingroup$


So if I have lst1 = [a, b] and lst2 = [x, y] the result would be:



[x, y, a, b]

[x, a, y, b]

[x, a, b, y]

[a, x, y, b]

[a, x, b, y]

[a, b, x, y]


I'm thinking about doing something recursive where I take the first element of a list, place it at the start, then recursively take the next element of that list and shift it through each position (and on each shift go through all remaining elements of that list).



But I'm wondering if there may be a nicer way to do this?



edit: Some more elaborate info here






algorithms







share|cite|improve this question









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Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    3












    $begingroup$


    So if I have lst1 = [a, b] and lst2 = [x, y] the result would be:



    [x, y, a, b]
    
[x, a, y, b]
    
[x, a, b, y]
    
[a, x, y, b]
    
[a, x, b, y]
    
[a, b, x, y]


    I'm thinking about doing something recursive where I take the first element of a list, place it at the start, then recursively take the next element of that list and shift it through each position (and on each shift go through all remaining elements of that list).



    But I'm wondering if there may be a nicer way to do this?



    edit: Some more elaborate info here






    algorithms







    share|cite|improve this question









    New contributor




    Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3


      1



      $begingroup$


      So if I have lst1 = [a, b] and lst2 = [x, y] the result would be:



      [x, y, a, b]
      
[x, a, y, b]
      
[x, a, b, y]
      
[a, x, y, b]
      
[a, x, b, y]
      
[a, b, x, y]


      I'm thinking about doing something recursive where I take the first element of a list, place it at the start, then recursively take the next element of that list and shift it through each position (and on each shift go through all remaining elements of that list).



      But I'm wondering if there may be a nicer way to do this?



      edit: Some more elaborate info here






      algorithms







      share|cite|improve this question









      New contributor




      Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      So if I have lst1 = [a, b] and lst2 = [x, y] the result would be:



      [x, y, a, b]
      
[x, a, y, b]
      
[x, a, b, y]
      
[a, x, y, b]
      
[a, x, b, y]
      
[a, b, x, y]


      I'm thinking about doing something recursive where I take the first element of a list, place it at the start, then recursively take the next element of that list and shift it through each position (and on each shift go through all remaining elements of that list).



      But I'm wondering if there may be a nicer way to do this?



      edit: Some more elaborate info here






      algorithms




      algorithms






      share|cite|improve this question









      New contributor




      Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 10 hours ago







      Nimitz14













      New contributor




      Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 11 hours ago









      Nimitz14Nimitz14

      1184




      1184




      New contributor




      Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Nimitz14 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Generate a string chi with len(lst1) 0s and len(lst2) 1s, e.g. for lst1 = [x, y] and lst2 = [a, b, c], you generate chi = [0, 0, 1, 1, 1]. Then you shuffle chi to obtain your "characteristic vector". This new list will dictate the order in which to output elements from lst1 and lst2.



          I hope this Python program speaks for itself:



          from random import shuffle
          

          
def permutation(lst1, lst2, chi):
          
    idx1 = 0
          
    idx2 = 0
          
    for i in range(len(lst1) + len(lst2)):
          
        if chi[i] == '0':
          
            yield lst1[idx1]
          
            idx1 += 1
          
        else:
          
            yield lst2[idx2]
          
            idx2 += 1
          

          
lst1 = 'xy'
          
lst2 = 'abc'
          

          
chi = list('0' * len(lst1) + '1' * len(lst2))
          
shuffle(chi)
          
print(''.join(chi))
          
print(''.join(list(permutation(lst1, lst2, chi))))


          Outputs:



          11001
          
abxyc
          

          
10011
          
axybc
          

          
00111
          
xyabc





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I like it. You view it as picking from two lists and have a binary vector which decides from which you pick. Obvious in hindsight!
            $endgroup$
            – Nimitz14
            10 hours ago












          • $begingroup$
            Doing a full shuffle requires distinct_permutations from more_itertools, just incase anyone else wonders how to do that
            $endgroup$
            – Nimitz14
            10 hours ago











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Generate a string chi with len(lst1) 0s and len(lst2) 1s, e.g. for lst1 = [x, y] and lst2 = [a, b, c], you generate chi = [0, 0, 1, 1, 1]. Then you shuffle chi to obtain your "characteristic vector". This new list will dictate the order in which to output elements from lst1 and lst2.



          I hope this Python program speaks for itself:



          from random import shuffle
          

          
def permutation(lst1, lst2, chi):
          
    idx1 = 0
          
    idx2 = 0
          
    for i in range(len(lst1) + len(lst2)):
          
        if chi[i] == '0':
          
            yield lst1[idx1]
          
            idx1 += 1
          
        else:
          
            yield lst2[idx2]
          
            idx2 += 1
          

          
lst1 = 'xy'
          
lst2 = 'abc'
          

          
chi = list('0' * len(lst1) + '1' * len(lst2))
          
shuffle(chi)
          
print(''.join(chi))
          
print(''.join(list(permutation(lst1, lst2, chi))))


          Outputs:



          11001
          
abxyc
          

          
10011
          
axybc
          

          
00111
          
xyabc





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I like it. You view it as picking from two lists and have a binary vector which decides from which you pick. Obvious in hindsight!
            $endgroup$
            – Nimitz14
            10 hours ago












          • $begingroup$
            Doing a full shuffle requires distinct_permutations from more_itertools, just incase anyone else wonders how to do that
            $endgroup$
            – Nimitz14
            10 hours ago
















          3












          $begingroup$

          Generate a string chi with len(lst1) 0s and len(lst2) 1s, e.g. for lst1 = [x, y] and lst2 = [a, b, c], you generate chi = [0, 0, 1, 1, 1]. Then you shuffle chi to obtain your "characteristic vector". This new list will dictate the order in which to output elements from lst1 and lst2.



          I hope this Python program speaks for itself:



          from random import shuffle
          

          
def permutation(lst1, lst2, chi):
          
    idx1 = 0
          
    idx2 = 0
          
    for i in range(len(lst1) + len(lst2)):
          
        if chi[i] == '0':
          
            yield lst1[idx1]
          
            idx1 += 1
          
        else:
          
            yield lst2[idx2]
          
            idx2 += 1
          

          
lst1 = 'xy'
          
lst2 = 'abc'
          

          
chi = list('0' * len(lst1) + '1' * len(lst2))
          
shuffle(chi)
          
print(''.join(chi))
          
print(''.join(list(permutation(lst1, lst2, chi))))


          Outputs:



          11001
          
abxyc
          

          
10011
          
axybc
          

          
00111
          
xyabc





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I like it. You view it as picking from two lists and have a binary vector which decides from which you pick. Obvious in hindsight!
            $endgroup$
            – Nimitz14
            10 hours ago












          • $begingroup$
            Doing a full shuffle requires distinct_permutations from more_itertools, just incase anyone else wonders how to do that
            $endgroup$
            – Nimitz14
            10 hours ago














          3












          3








          3





          $begingroup$

          Generate a string chi with len(lst1) 0s and len(lst2) 1s, e.g. for lst1 = [x, y] and lst2 = [a, b, c], you generate chi = [0, 0, 1, 1, 1]. Then you shuffle chi to obtain your "characteristic vector". This new list will dictate the order in which to output elements from lst1 and lst2.



          I hope this Python program speaks for itself:



          from random import shuffle
          

          
def permutation(lst1, lst2, chi):
          
    idx1 = 0
          
    idx2 = 0
          
    for i in range(len(lst1) + len(lst2)):
          
        if chi[i] == '0':
          
            yield lst1[idx1]
          
            idx1 += 1
          
        else:
          
            yield lst2[idx2]
          
            idx2 += 1
          

          
lst1 = 'xy'
          
lst2 = 'abc'
          

          
chi = list('0' * len(lst1) + '1' * len(lst2))
          
shuffle(chi)
          
print(''.join(chi))
          
print(''.join(list(permutation(lst1, lst2, chi))))


          Outputs:



          11001
          
abxyc
          

          
10011
          
axybc
          

          
00111
          
xyabc





          share|cite|improve this answer











          $endgroup$



          Generate a string chi with len(lst1) 0s and len(lst2) 1s, e.g. for lst1 = [x, y] and lst2 = [a, b, c], you generate chi = [0, 0, 1, 1, 1]. Then you shuffle chi to obtain your "characteristic vector". This new list will dictate the order in which to output elements from lst1 and lst2.



          I hope this Python program speaks for itself:



          from random import shuffle
          

          
def permutation(lst1, lst2, chi):
          
    idx1 = 0
          
    idx2 = 0
          
    for i in range(len(lst1) + len(lst2)):
          
        if chi[i] == '0':
          
            yield lst1[idx1]
          
            idx1 += 1
          
        else:
          
            yield lst2[idx2]
          
            idx2 += 1
          

          
lst1 = 'xy'
          
lst2 = 'abc'
          

          
chi = list('0' * len(lst1) + '1' * len(lst2))
          
shuffle(chi)
          
print(''.join(chi))
          
print(''.join(list(permutation(lst1, lst2, chi))))


          Outputs:



          11001
          
abxyc
          

          
10011
          
axybc
          

          
00111
          
xyabc






          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 11 hours ago

























          answered 11 hours ago









          Pål GDPål GD

          6,9452342




          6,9452342












          • $begingroup$
            I like it. You view it as picking from two lists and have a binary vector which decides from which you pick. Obvious in hindsight!
            $endgroup$
            – Nimitz14
            10 hours ago












          • $begingroup$
            Doing a full shuffle requires distinct_permutations from more_itertools, just incase anyone else wonders how to do that
            $endgroup$
            – Nimitz14
            10 hours ago


















          • $begingroup$
            I like it. You view it as picking from two lists and have a binary vector which decides from which you pick. Obvious in hindsight!
            $endgroup$
            – Nimitz14
            10 hours ago












          • $begingroup$
            Doing a full shuffle requires distinct_permutations from more_itertools, just incase anyone else wonders how to do that
            $endgroup$
            – Nimitz14
            10 hours ago
















          $begingroup$
          I like it. You view it as picking from two lists and have a binary vector which decides from which you pick. Obvious in hindsight!
          $endgroup$
          – Nimitz14
          10 hours ago






          $begingroup$
          I like it. You view it as picking from two lists and have a binary vector which decides from which you pick. Obvious in hindsight!
          $endgroup$
          – Nimitz14
          10 hours ago














          $begingroup$
          Doing a full shuffle requires distinct_permutations from more_itertools, just incase anyone else wonders how to do that
          $endgroup$
          – Nimitz14
          10 hours ago




          $begingroup$
          Doing a full shuffle requires distinct_permutations from more_itertools, just incase anyone else wonders how to do that
          $endgroup$
          – Nimitz14
          10 hours ago










          Nimitz14 is a new contributor. Be nice, and check out our Code of Conduct.










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