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Changing iteration variable in Do loop


Changing a Part of a variable within ParallelDoIterating FindRoot to solve a differential equationBeginners problem, Do Loop, Eigenfunction iterationExecute a notebook with a changing variablesuccessive iterationHow do I use the values generated in a for-loop in my first iteration to my next iteration?While loop with changing variable , NDSolve and an IntegralHow do I repeat the number of times a nested for loop does an iteration?Evaluating number of iteration with a certain map with WhileThe variable in the While loop does not change






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








2












$begingroup$


I want to generate random positive hermitian matrices, I start with such a code



rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 -
rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0,
Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1,
4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]],
1 - rho11R[[i]], i, 1, 4]


This does not work, as the matrices I get are in general not positive.
The problem is my If function, which does not seem to change the iteration variable i to i-1, when the condition for positivity is not fulfilled (so that I can get new random variable for rho10R[[I]] and again check the condition for positivity). How can I reach this?










share|improve this question









$endgroup$









  • 1




    $begingroup$
    Have you tried a While[] loop instead?
    $endgroup$
    – Somos
    Jul 19 at 15:53










  • $begingroup$
    I assume that by positive you mean positive semi-definite, right?
    $endgroup$
    – Roman
    Jul 19 at 17:04

















2












$begingroup$


I want to generate random positive hermitian matrices, I start with such a code



rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 -
rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0,
Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1,
4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]],
1 - rho11R[[i]], i, 1, 4]


This does not work, as the matrices I get are in general not positive.
The problem is my If function, which does not seem to change the iteration variable i to i-1, when the condition for positivity is not fulfilled (so that I can get new random variable for rho10R[[I]] and again check the condition for positivity). How can I reach this?










share|improve this question









$endgroup$









  • 1




    $begingroup$
    Have you tried a While[] loop instead?
    $endgroup$
    – Somos
    Jul 19 at 15:53










  • $begingroup$
    I assume that by positive you mean positive semi-definite, right?
    $endgroup$
    – Roman
    Jul 19 at 17:04













2












2








2





$begingroup$


I want to generate random positive hermitian matrices, I start with such a code



rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 -
rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0,
Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1,
4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]],
1 - rho11R[[i]], i, 1, 4]


This does not work, as the matrices I get are in general not positive.
The problem is my If function, which does not seem to change the iteration variable i to i-1, when the condition for positivity is not fulfilled (so that I can get new random variable for rho10R[[I]] and again check the condition for positivity). How can I reach this?










share|improve this question









$endgroup$




I want to generate random positive hermitian matrices, I start with such a code



rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 -
rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0,
Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1,
4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]],
1 - rho11R[[i]], i, 1, 4]


This does not work, as the matrices I get are in general not positive.
The problem is my If function, which does not seem to change the iteration variable i to i-1, when the condition for positivity is not fulfilled (so that I can get new random variable for rho10R[[I]] and again check the condition for positivity). How can I reach this?







procedural-programming






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jul 19 at 15:25









AgnieszkaAgnieszka

10610 bronze badges




10610 bronze badges










  • 1




    $begingroup$
    Have you tried a While[] loop instead?
    $endgroup$
    – Somos
    Jul 19 at 15:53










  • $begingroup$
    I assume that by positive you mean positive semi-definite, right?
    $endgroup$
    – Roman
    Jul 19 at 17:04












  • 1




    $begingroup$
    Have you tried a While[] loop instead?
    $endgroup$
    – Somos
    Jul 19 at 15:53










  • $begingroup$
    I assume that by positive you mean positive semi-definite, right?
    $endgroup$
    – Roman
    Jul 19 at 17:04







1




1




$begingroup$
Have you tried a While[] loop instead?
$endgroup$
– Somos
Jul 19 at 15:53




$begingroup$
Have you tried a While[] loop instead?
$endgroup$
– Somos
Jul 19 at 15:53












$begingroup$
I assume that by positive you mean positive semi-definite, right?
$endgroup$
– Roman
Jul 19 at 17:04




$begingroup$
I assume that by positive you mean positive semi-definite, right?
$endgroup$
– Roman
Jul 19 at 17:04










3 Answers
3






active

oldest

votes


















4












$begingroup$

This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.



Example:



For[
i = 1,
i <= 4,
i++,
If[
i == 2,
Print[i++],
Print[i]
]
]


This prints 1, 2, 4.



Another option is to use While, as Somos wrote in a comment.






share|improve this answer









$endgroup$






















    4












    $begingroup$

    A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:



    randommatrix[n_Integer?Positive] := 
    RandomVariate[NormalDistribution[], n, n, 2].1, I

    randomHermitian[n_Integer?Positive] :=
    (# + ConjugateTranspose[#])/2 &[ConjugateTranspose[#].# &[randommatrix[n]]]


    In this way you don't need to reject anything.



    Test:



    randomHermitian[10] // Eigenvalues
    (* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
    12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)


    There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.



    Thanks to @mikado for pointing out that Hermitian symmetrization is needed to avoid generating matrices that are almost-but-not-quite Hermitian because of numerical precision limits.






    share|improve this answer











    $endgroup$










    • 1




      $begingroup$
      While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
      $endgroup$
      – mikado
      Jul 19 at 22:14


















    0












    $begingroup$

    The following is a more "functional" way of doing acceptance/rejection



    mtx := RandomComplex[1+I, 2, 2]

    pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]];
    If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]


    You can then generate a list of positive definite matrices of any length e.g.



    Table[pmtx, 20] 


    (Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)






    share|improve this answer









    $endgroup$

















      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.



      Example:



      For[
      i = 1,
      i <= 4,
      i++,
      If[
      i == 2,
      Print[i++],
      Print[i]
      ]
      ]


      This prints 1, 2, 4.



      Another option is to use While, as Somos wrote in a comment.






      share|improve this answer









      $endgroup$



















        4












        $begingroup$

        This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.



        Example:



        For[
        i = 1,
        i <= 4,
        i++,
        If[
        i == 2,
        Print[i++],
        Print[i]
        ]
        ]


        This prints 1, 2, 4.



        Another option is to use While, as Somos wrote in a comment.






        share|improve this answer









        $endgroup$

















          4












          4








          4





          $begingroup$

          This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.



          Example:



          For[
          i = 1,
          i <= 4,
          i++,
          If[
          i == 2,
          Print[i++],
          Print[i]
          ]
          ]


          This prints 1, 2, 4.



          Another option is to use While, as Somos wrote in a comment.






          share|improve this answer









          $endgroup$



          This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.



          Example:



          For[
          i = 1,
          i <= 4,
          i++,
          If[
          i == 2,
          Print[i++],
          Print[i]
          ]
          ]


          This prints 1, 2, 4.



          Another option is to use While, as Somos wrote in a comment.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jul 19 at 16:17









          C. E.C. E.

          53.4k3 gold badges102 silver badges210 bronze badges




          53.4k3 gold badges102 silver badges210 bronze badges


























              4












              $begingroup$

              A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:



              randommatrix[n_Integer?Positive] := 
              RandomVariate[NormalDistribution[], n, n, 2].1, I

              randomHermitian[n_Integer?Positive] :=
              (# + ConjugateTranspose[#])/2 &[ConjugateTranspose[#].# &[randommatrix[n]]]


              In this way you don't need to reject anything.



              Test:



              randomHermitian[10] // Eigenvalues
              (* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
              12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)


              There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.



              Thanks to @mikado for pointing out that Hermitian symmetrization is needed to avoid generating matrices that are almost-but-not-quite Hermitian because of numerical precision limits.






              share|improve this answer











              $endgroup$










              • 1




                $begingroup$
                While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
                $endgroup$
                – mikado
                Jul 19 at 22:14















              4












              $begingroup$

              A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:



              randommatrix[n_Integer?Positive] := 
              RandomVariate[NormalDistribution[], n, n, 2].1, I

              randomHermitian[n_Integer?Positive] :=
              (# + ConjugateTranspose[#])/2 &[ConjugateTranspose[#].# &[randommatrix[n]]]


              In this way you don't need to reject anything.



              Test:



              randomHermitian[10] // Eigenvalues
              (* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
              12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)


              There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.



              Thanks to @mikado for pointing out that Hermitian symmetrization is needed to avoid generating matrices that are almost-but-not-quite Hermitian because of numerical precision limits.






              share|improve this answer











              $endgroup$










              • 1




                $begingroup$
                While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
                $endgroup$
                – mikado
                Jul 19 at 22:14













              4












              4








              4





              $begingroup$

              A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:



              randommatrix[n_Integer?Positive] := 
              RandomVariate[NormalDistribution[], n, n, 2].1, I

              randomHermitian[n_Integer?Positive] :=
              (# + ConjugateTranspose[#])/2 &[ConjugateTranspose[#].# &[randommatrix[n]]]


              In this way you don't need to reject anything.



              Test:



              randomHermitian[10] // Eigenvalues
              (* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
              12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)


              There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.



              Thanks to @mikado for pointing out that Hermitian symmetrization is needed to avoid generating matrices that are almost-but-not-quite Hermitian because of numerical precision limits.






              share|improve this answer











              $endgroup$



              A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:



              randommatrix[n_Integer?Positive] := 
              RandomVariate[NormalDistribution[], n, n, 2].1, I

              randomHermitian[n_Integer?Positive] :=
              (# + ConjugateTranspose[#])/2 &[ConjugateTranspose[#].# &[randommatrix[n]]]


              In this way you don't need to reject anything.



              Test:



              randomHermitian[10] // Eigenvalues
              (* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
              12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)


              There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.



              Thanks to @mikado for pointing out that Hermitian symmetrization is needed to avoid generating matrices that are almost-but-not-quite Hermitian because of numerical precision limits.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Jul 20 at 18:18

























              answered Jul 19 at 17:02









              RomanRoman

              14.9k1 gold badge20 silver badges51 bronze badges




              14.9k1 gold badge20 silver badges51 bronze badges










              • 1




                $begingroup$
                While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
                $endgroup$
                – mikado
                Jul 19 at 22:14












              • 1




                $begingroup$
                While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
                $endgroup$
                – mikado
                Jul 19 at 22:14







              1




              1




              $begingroup$
              While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
              $endgroup$
              – mikado
              Jul 19 at 22:14




              $begingroup$
              While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
              $endgroup$
              – mikado
              Jul 19 at 22:14











              0












              $begingroup$

              The following is a more "functional" way of doing acceptance/rejection



              mtx := RandomComplex[1+I, 2, 2]

              pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]];
              If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]


              You can then generate a list of positive definite matrices of any length e.g.



              Table[pmtx, 20] 


              (Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)






              share|improve this answer









              $endgroup$



















                0












                $begingroup$

                The following is a more "functional" way of doing acceptance/rejection



                mtx := RandomComplex[1+I, 2, 2]

                pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]];
                If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]


                You can then generate a list of positive definite matrices of any length e.g.



                Table[pmtx, 20] 


                (Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)






                share|improve this answer









                $endgroup$

















                  0












                  0








                  0





                  $begingroup$

                  The following is a more "functional" way of doing acceptance/rejection



                  mtx := RandomComplex[1+I, 2, 2]

                  pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]];
                  If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]


                  You can then generate a list of positive definite matrices of any length e.g.



                  Table[pmtx, 20] 


                  (Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)






                  share|improve this answer









                  $endgroup$



                  The following is a more "functional" way of doing acceptance/rejection



                  mtx := RandomComplex[1+I, 2, 2]

                  pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]];
                  If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]


                  You can then generate a list of positive definite matrices of any length e.g.



                  Table[pmtx, 20] 


                  (Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Jul 19 at 22:03









                  mikadomikado

                  7,3301 gold badge9 silver badges29 bronze badges




                  7,3301 gold badge9 silver badges29 bronze badges






























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