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How to convert diagonal matrix to rectangular matrix


Is there a built in function to obtain the back diagonal of a matrix?How to select all elements above the main diagonal of matrix?Diagonal times dense matrix, high precisionHow to extract the list of all matrices from a Block Diagonal Matrix?Get rid of infinity in matrix elements (by separate definition of diagonal and off-diagonal elements)Summation over diagonal blocksHow to transform symmetric matrix to diagonal?Sum of elements in a diagonalReplace diagonal elements in sparse matrixConstructing a block diagonal matrix






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








5












$begingroup$


Suppose you have the following diagonal matrix:



DiagonalMatrix[Hold/@a, b, c]//ReleaseHold



a, 0, 0, b, c




How can the above matrix be converted to the following rectangular one?



a, 0, 0, 0, b, c









share|improve this question









New contributor



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






$endgroup$











  • $begingroup$
    Could you clarify what should happen for e.g. the matrix a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
    $endgroup$
    – C. E.
    5 mins ago


















5












$begingroup$


Suppose you have the following diagonal matrix:



DiagonalMatrix[Hold/@a, b, c]//ReleaseHold



a, 0, 0, b, c




How can the above matrix be converted to the following rectangular one?



a, 0, 0, 0, b, c









share|improve this question









New contributor



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






$endgroup$











  • $begingroup$
    Could you clarify what should happen for e.g. the matrix a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
    $endgroup$
    – C. E.
    5 mins ago














5












5








5





$begingroup$


Suppose you have the following diagonal matrix:



DiagonalMatrix[Hold/@a, b, c]//ReleaseHold



a, 0, 0, b, c




How can the above matrix be converted to the following rectangular one?



a, 0, 0, 0, b, c









share|improve this question









New contributor



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






$endgroup$




Suppose you have the following diagonal matrix:



DiagonalMatrix[Hold/@a, b, c]//ReleaseHold



a, 0, 0, b, c




How can the above matrix be converted to the following rectangular one?



a, 0, 0, 0, b, c






list-manipulation matrix






share|improve this question









New contributor



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










share|improve this question









New contributor



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








share|improve this question




share|improve this question








edited 8 hours ago







aleksander_si













New contributor



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








asked 8 hours ago









aleksander_sialeksander_si

284 bronze badges




284 bronze badges




New contributor



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




New contributor




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













  • $begingroup$
    Could you clarify what should happen for e.g. the matrix a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
    $endgroup$
    – C. E.
    5 mins ago

















  • $begingroup$
    Could you clarify what should happen for e.g. the matrix a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
    $endgroup$
    – C. E.
    5 mins ago
















$begingroup$
Could you clarify what should happen for e.g. the matrix a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
$endgroup$
– C. E.
5 mins ago





$begingroup$
Could you clarify what should happen for e.g. the matrix a, d, e, 0, 0, b, c as in my answer? Whether my answer is correct or not hinges on this, because it gives a different answer than kglr.
$endgroup$
– C. E.
5 mins ago











2 Answers
2






active

oldest

votes


















1












$begingroup$

kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.



For example:



PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm


Mathematica graphics



m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
PadRight[Flatten /@ m] // MatrixForm


Mathematica graphics



Here is a different solution:



diag = Flatten[#] & /@ Diagonal[m];
ncols = Length@Flatten[diag];
offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]

m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
matrix[diag, offsets, ncols] // MatrixForm


Mathematica graphics






share|improve this answer











$endgroup$




















    6












    $begingroup$

    PadRight[Flatten /@ a, 0, 0, b, c]



    a, 0, 0, 0, b, c







    share|improve this answer









    $endgroup$












    • $begingroup$
      Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
      $endgroup$
      – aleksander_si
      10 mins ago










    • $begingroup$
      Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
      $endgroup$
      – aleksander_si
      2 mins ago














    Your Answer








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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.



    For example:



    PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm


    Mathematica graphics



    m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
    PadRight[Flatten /@ m] // MatrixForm


    Mathematica graphics



    Here is a different solution:



    diag = Flatten[#] & /@ Diagonal[m];
    ncols = Length@Flatten[diag];
    offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
    row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
    matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]

    m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
    matrix[diag, offsets, ncols] // MatrixForm


    Mathematica graphics






    share|improve this answer











    $endgroup$

















      1












      $begingroup$

      kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.



      For example:



      PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm


      Mathematica graphics



      m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
      PadRight[Flatten /@ m] // MatrixForm


      Mathematica graphics



      Here is a different solution:



      diag = Flatten[#] & /@ Diagonal[m];
      ncols = Length@Flatten[diag];
      offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
      row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
      matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]

      m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
      matrix[diag, offsets, ncols] // MatrixForm


      Mathematica graphics






      share|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.



        For example:



        PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm


        Mathematica graphics



        m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
        PadRight[Flatten /@ m] // MatrixForm


        Mathematica graphics



        Here is a different solution:



        diag = Flatten[#] & /@ Diagonal[m];
        ncols = Length@Flatten[diag];
        offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
        row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
        matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]

        m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
        matrix[diag, offsets, ncols] // MatrixForm


        Mathematica graphics






        share|improve this answer











        $endgroup$



        kglr's solution fails if the number of elements on the diagonal in any of the preceding rows is larger than the number of elements on the diagonal of the last row.



        For example:



        PadRight[Flatten /@ a, d, e, 0, 0, b, c] // MatrixForm


        Mathematica graphics



        m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
        PadRight[Flatten /@ m] // MatrixForm


        Mathematica graphics



        Here is a different solution:



        diag = Flatten[#] & /@ Diagonal[m];
        ncols = Length@Flatten[diag];
        offsets = Most@Prepend[Accumulate[Length /@ diag], 0];
        row[values_, offset_, ncols_] := PadRight[ArrayPad[values, offset, 0], ncols]
        matrix[diag_, offsets_, ncols_] := MapThread[row[#, #2, ncols] &, diag, offsets]

        m = a, 0, 0, 0, b, c, d, e, f, 0, 0, 0, b, c
        matrix[diag, offsets, ncols] // MatrixForm


        Mathematica graphics







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 3 mins ago

























        answered 12 mins ago









        C. E.C. E.

        52.9k3 gold badges102 silver badges210 bronze badges




        52.9k3 gold badges102 silver badges210 bronze badges























            6












            $begingroup$

            PadRight[Flatten /@ a, 0, 0, b, c]



            a, 0, 0, 0, b, c







            share|improve this answer









            $endgroup$












            • $begingroup$
              Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
              $endgroup$
              – aleksander_si
              10 mins ago










            • $begingroup$
              Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
              $endgroup$
              – aleksander_si
              2 mins ago
















            6












            $begingroup$

            PadRight[Flatten /@ a, 0, 0, b, c]



            a, 0, 0, 0, b, c







            share|improve this answer









            $endgroup$












            • $begingroup$
              Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
              $endgroup$
              – aleksander_si
              10 mins ago










            • $begingroup$
              Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
              $endgroup$
              – aleksander_si
              2 mins ago














            6












            6








            6





            $begingroup$

            PadRight[Flatten /@ a, 0, 0, b, c]



            a, 0, 0, 0, b, c







            share|improve this answer









            $endgroup$



            PadRight[Flatten /@ a, 0, 0, b, c]



            a, 0, 0, 0, b, c








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 7 hours ago









            kglrkglr

            203k10 gold badges232 silver badges463 bronze badges




            203k10 gold badges232 silver badges463 bronze badges











            • $begingroup$
              Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
              $endgroup$
              – aleksander_si
              10 mins ago










            • $begingroup$
              Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
              $endgroup$
              – aleksander_si
              2 mins ago

















            • $begingroup$
              Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
              $endgroup$
              – aleksander_si
              10 mins ago










            • $begingroup$
              Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
              $endgroup$
              – aleksander_si
              2 mins ago
















            $begingroup$
            Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
            $endgroup$
            – aleksander_si
            10 mins ago




            $begingroup$
            Very neat solution!! I just hope I stated the question in a form which makes the problem searchable.
            $endgroup$
            – aleksander_si
            10 mins ago












            $begingroup$
            Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
            $endgroup$
            – aleksander_si
            2 mins ago





            $begingroup$
            Yes, in a situation where the diagonal matrix is as per the below example, then this approach should not be applied:
            $endgroup$
            – aleksander_si
            2 mins ago











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









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












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














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