Fast element-wise division of matrix, generated from vector with `Outer`, and another matrix












5












$begingroup$


m = {a, b, c};

n = {{e, r, t}, {y, u, i}, {g, h, j}};

k = Outer[Divide, m, m];

k/n


gives



{{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h), 

  1/j}}


I want to do this with very large matrices filled with numbers of arbitrary precision. Is there a faster way?



EDIT



The sizes I am looking at for my practical applications start at 20000 and 20000^2 for the vector and matrix, respectively (of course the examples don't have to be with that many).



I am also interested in any method that might parallelise well.






list-manipulation matrix performance-tuning array







share|improve this question











$endgroup$












  • $begingroup$
    What is the length of m in practical use?
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Dec 13 '18 at 3:03










  • $begingroup$
    You can try m/(n ConstantArray[m, Length[m]]) and see how fast it is.
    $endgroup$
    – C. E.
    Dec 13 '18 at 5:22










  • $begingroup$
    @ΑλέξανδροςΖεγγ I editted my question to include some information on that.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 6:46
















5












$begingroup$


m = {a, b, c};

n = {{e, r, t}, {y, u, i}, {g, h, j}};

k = Outer[Divide, m, m];

k/n


gives



{{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h), 

  1/j}}


I want to do this with very large matrices filled with numbers of arbitrary precision. Is there a faster way?



EDIT



The sizes I am looking at for my practical applications start at 20000 and 20000^2 for the vector and matrix, respectively (of course the examples don't have to be with that many).



I am also interested in any method that might parallelise well.






list-manipulation matrix performance-tuning array







share|improve this question











$endgroup$












  • $begingroup$
    What is the length of m in practical use?
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Dec 13 '18 at 3:03










  • $begingroup$
    You can try m/(n ConstantArray[m, Length[m]]) and see how fast it is.
    $endgroup$
    – C. E.
    Dec 13 '18 at 5:22










  • $begingroup$
    @ΑλέξανδροςΖεγγ I editted my question to include some information on that.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 6:46














5












5








5


1



$begingroup$


m = {a, b, c};

n = {{e, r, t}, {y, u, i}, {g, h, j}};

k = Outer[Divide, m, m];

k/n


gives



{{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h), 

  1/j}}


I want to do this with very large matrices filled with numbers of arbitrary precision. Is there a faster way?



EDIT



The sizes I am looking at for my practical applications start at 20000 and 20000^2 for the vector and matrix, respectively (of course the examples don't have to be with that many).



I am also interested in any method that might parallelise well.






list-manipulation matrix performance-tuning array







share|improve this question











$endgroup$




m = {a, b, c};

n = {{e, r, t}, {y, u, i}, {g, h, j}};

k = Outer[Divide, m, m];

k/n


gives



{{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h), 

  1/j}}


I want to do this with very large matrices filled with numbers of arbitrary precision. Is there a faster way?



EDIT



The sizes I am looking at for my practical applications start at 20000 and 20000^2 for the vector and matrix, respectively (of course the examples don't have to be with that many).



I am also interested in any method that might parallelise well.






list-manipulation matrix performance-tuning array




list-manipulation matrix performance-tuning array






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 13 '18 at 6:45







ThunderBiggi

















asked Dec 12 '18 at 23:45









ThunderBiggiThunderBiggi

388112




388112












  • $begingroup$
    What is the length of m in practical use?
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Dec 13 '18 at 3:03










  • $begingroup$
    You can try m/(n ConstantArray[m, Length[m]]) and see how fast it is.
    $endgroup$
    – C. E.
    Dec 13 '18 at 5:22










  • $begingroup$
    @ΑλέξανδροςΖεγγ I editted my question to include some information on that.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 6:46


















  • $begingroup$
    What is the length of m in practical use?
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Dec 13 '18 at 3:03










  • $begingroup$
    You can try m/(n ConstantArray[m, Length[m]]) and see how fast it is.
    $endgroup$
    – C. E.
    Dec 13 '18 at 5:22










  • $begingroup$
    @ΑλέξανδροςΖεγγ I editted my question to include some information on that.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 6:46
















$begingroup$
What is the length of m in practical use?
$endgroup$
– Αλέξανδρος Ζεγγ
Dec 13 '18 at 3:03




$begingroup$
What is the length of m in practical use?
$endgroup$
– Αλέξανδρος Ζεγγ
Dec 13 '18 at 3:03












$begingroup$
You can try m/(n ConstantArray[m, Length[m]]) and see how fast it is.
$endgroup$
– C. E.
Dec 13 '18 at 5:22




$begingroup$
You can try m/(n ConstantArray[m, Length[m]]) and see how fast it is.
$endgroup$
– C. E.
Dec 13 '18 at 5:22












$begingroup$
@ΑλέξανδροςΖεγγ I editted my question to include some information on that.
$endgroup$
– ThunderBiggi
Dec 13 '18 at 6:46




$begingroup$
@ΑλέξανδροςΖεγγ I editted my question to include some information on that.
$endgroup$
– ThunderBiggi
Dec 13 '18 at 6:46










2 Answers
2






active

oldest

votes


















6












$begingroup$

m = RandomReal[{-1, 1}, {2000}];

n = RandomReal[{-1, 1}, {2000, 2000}];

a = Outer[Divide, m, m]/n; // RepeatedTiming // First

b = Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]]; // 

  RepeatedTiming // First

c = m /(ConstantArray[m, Length[m]] n); // RepeatedTiming // First

d = KroneckerProduct[m, 1./m]/n; // RepeatedTiming // First

a == b == c == d



0.958



0.128



0.0281



0.0236



True




Edit



A parallelized version 



cf = Compile[{{x, _Real}, {y, _Real, 1}, {z, _Real, 1}},

   x/(y z),

   CompilationTarget -> "C",

   RuntimeAttributes -> {Listable},

   Parallelization -> True,

   RuntimeOptions -> "Speed"

   ];

e = cf[m, n, m]; // RepeatedTiming // First

a == e



0.0096



True




Timing has been measured on a Quad Core CPU which shows that this does not scale too well. Btw., the timing with CompilationTarget -> "C" is only 4% slower, so there is always no point to compile it into a library.






share|improve this answer











$endgroup$













  • $begingroup$
    I was just typing a comparison of the answers so far, but you were first. I wouldn't've expected that KroneckerProduct would be that quick. Any ideas on any way that might parallelise well? I will edit my question to include that as well.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 6:43










  • $begingroup$
    See my edit for a parallelized version.
    $endgroup$
    – Henrik Schumacher
    Dec 13 '18 at 7:03










  • $begingroup$
    I am using arbitrary precision numbers, so I guess 'Compile' is not really an option.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 7:38










  • $begingroup$
    Also, interestingly, but on my machine, with Mathematica 11.3, a is faster than b though still slower than the other two.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 7:42












  • $begingroup$
    Yeah, I was also surprised that a was so slow on my machine. I don't know what to think about it...
    $endgroup$
    – Henrik Schumacher
    Dec 13 '18 at 7:51



















4












$begingroup$

Pretty printing your result gives...



{{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h),1/j}}//MatrixForm


$left(
begin{array}{ccc}
frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
end{array}
right)$



Try this, it avoids constructing the huge Outer[ ] matrix



Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]] // MatrixForm


$left(
begin{array}{ccc}
frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
end{array}
right)$






share|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    m = RandomReal[{-1, 1}, {2000}];
    
n = RandomReal[{-1, 1}, {2000, 2000}];
    
a = Outer[Divide, m, m]/n; // RepeatedTiming // First
    
b = Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]]; // 
    
  RepeatedTiming // First
    
c = m /(ConstantArray[m, Length[m]] n); // RepeatedTiming // First
    
d = KroneckerProduct[m, 1./m]/n; // RepeatedTiming // First
    
a == b == c == d



    0.958



    0.128



    0.0281



    0.0236



    True




    Edit



    A parallelized version 



    cf = Compile[{{x, _Real}, {y, _Real, 1}, {z, _Real, 1}},
    
   x/(y z),
    
   CompilationTarget -> "C",
    
   RuntimeAttributes -> {Listable},
    
   Parallelization -> True,
    
   RuntimeOptions -> "Speed"
    
   ];
    
e = cf[m, n, m]; // RepeatedTiming // First
    
a == e



    0.0096



    True




    Timing has been measured on a Quad Core CPU which shows that this does not scale too well. Btw., the timing with CompilationTarget -> "C" is only 4% slower, so there is always no point to compile it into a library.






    share|improve this answer











    $endgroup$













    • $begingroup$
      I was just typing a comparison of the answers so far, but you were first. I wouldn't've expected that KroneckerProduct would be that quick. Any ideas on any way that might parallelise well? I will edit my question to include that as well.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 6:43










    • $begingroup$
      See my edit for a parallelized version.
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:03










    • $begingroup$
      I am using arbitrary precision numbers, so I guess 'Compile' is not really an option.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:38










    • $begingroup$
      Also, interestingly, but on my machine, with Mathematica 11.3, a is faster than b though still slower than the other two.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:42












    • $begingroup$
      Yeah, I was also surprised that a was so slow on my machine. I don't know what to think about it...
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:51
















    6












    $begingroup$

    m = RandomReal[{-1, 1}, {2000}];
    
n = RandomReal[{-1, 1}, {2000, 2000}];
    
a = Outer[Divide, m, m]/n; // RepeatedTiming // First
    
b = Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]]; // 
    
  RepeatedTiming // First
    
c = m /(ConstantArray[m, Length[m]] n); // RepeatedTiming // First
    
d = KroneckerProduct[m, 1./m]/n; // RepeatedTiming // First
    
a == b == c == d



    0.958



    0.128



    0.0281



    0.0236



    True




    Edit



    A parallelized version 



    cf = Compile[{{x, _Real}, {y, _Real, 1}, {z, _Real, 1}},
    
   x/(y z),
    
   CompilationTarget -> "C",
    
   RuntimeAttributes -> {Listable},
    
   Parallelization -> True,
    
   RuntimeOptions -> "Speed"
    
   ];
    
e = cf[m, n, m]; // RepeatedTiming // First
    
a == e



    0.0096



    True




    Timing has been measured on a Quad Core CPU which shows that this does not scale too well. Btw., the timing with CompilationTarget -> "C" is only 4% slower, so there is always no point to compile it into a library.






    share|improve this answer











    $endgroup$













    • $begingroup$
      I was just typing a comparison of the answers so far, but you were first. I wouldn't've expected that KroneckerProduct would be that quick. Any ideas on any way that might parallelise well? I will edit my question to include that as well.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 6:43










    • $begingroup$
      See my edit for a parallelized version.
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:03










    • $begingroup$
      I am using arbitrary precision numbers, so I guess 'Compile' is not really an option.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:38










    • $begingroup$
      Also, interestingly, but on my machine, with Mathematica 11.3, a is faster than b though still slower than the other two.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:42












    • $begingroup$
      Yeah, I was also surprised that a was so slow on my machine. I don't know what to think about it...
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:51














    6












    6








    6





    $begingroup$

    m = RandomReal[{-1, 1}, {2000}];
    
n = RandomReal[{-1, 1}, {2000, 2000}];
    
a = Outer[Divide, m, m]/n; // RepeatedTiming // First
    
b = Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]]; // 
    
  RepeatedTiming // First
    
c = m /(ConstantArray[m, Length[m]] n); // RepeatedTiming // First
    
d = KroneckerProduct[m, 1./m]/n; // RepeatedTiming // First
    
a == b == c == d



    0.958



    0.128



    0.0281



    0.0236



    True




    Edit



    A parallelized version 



    cf = Compile[{{x, _Real}, {y, _Real, 1}, {z, _Real, 1}},
    
   x/(y z),
    
   CompilationTarget -> "C",
    
   RuntimeAttributes -> {Listable},
    
   Parallelization -> True,
    
   RuntimeOptions -> "Speed"
    
   ];
    
e = cf[m, n, m]; // RepeatedTiming // First
    
a == e



    0.0096



    True




    Timing has been measured on a Quad Core CPU which shows that this does not scale too well. Btw., the timing with CompilationTarget -> "C" is only 4% slower, so there is always no point to compile it into a library.






    share|improve this answer











    $endgroup$



    m = RandomReal[{-1, 1}, {2000}];
    
n = RandomReal[{-1, 1}, {2000, 2000}];
    
a = Outer[Divide, m, m]/n; // RepeatedTiming // First
    
b = Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]]; // 
    
  RepeatedTiming // First
    
c = m /(ConstantArray[m, Length[m]] n); // RepeatedTiming // First
    
d = KroneckerProduct[m, 1./m]/n; // RepeatedTiming // First
    
a == b == c == d



    0.958



    0.128



    0.0281



    0.0236



    True




    Edit



    A parallelized version 



    cf = Compile[{{x, _Real}, {y, _Real, 1}, {z, _Real, 1}},
    
   x/(y z),
    
   CompilationTarget -> "C",
    
   RuntimeAttributes -> {Listable},
    
   Parallelization -> True,
    
   RuntimeOptions -> "Speed"
    
   ];
    
e = cf[m, n, m]; // RepeatedTiming // First
    
a == e



    0.0096



    True




    Timing has been measured on a Quad Core CPU which shows that this does not scale too well. Btw., the timing with CompilationTarget -> "C" is only 4% slower, so there is always no point to compile it into a library.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 13 '18 at 7:03

























    answered Dec 13 '18 at 6:32









    Henrik SchumacherHenrik Schumacher

    55.1k475154




    55.1k475154












    • $begingroup$
      I was just typing a comparison of the answers so far, but you were first. I wouldn't've expected that KroneckerProduct would be that quick. Any ideas on any way that might parallelise well? I will edit my question to include that as well.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 6:43










    • $begingroup$
      See my edit for a parallelized version.
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:03










    • $begingroup$
      I am using arbitrary precision numbers, so I guess 'Compile' is not really an option.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:38










    • $begingroup$
      Also, interestingly, but on my machine, with Mathematica 11.3, a is faster than b though still slower than the other two.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:42












    • $begingroup$
      Yeah, I was also surprised that a was so slow on my machine. I don't know what to think about it...
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:51


















    • $begingroup$
      I was just typing a comparison of the answers so far, but you were first. I wouldn't've expected that KroneckerProduct would be that quick. Any ideas on any way that might parallelise well? I will edit my question to include that as well.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 6:43










    • $begingroup$
      See my edit for a parallelized version.
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:03










    • $begingroup$
      I am using arbitrary precision numbers, so I guess 'Compile' is not really an option.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:38










    • $begingroup$
      Also, interestingly, but on my machine, with Mathematica 11.3, a is faster than b though still slower than the other two.
      $endgroup$
      – ThunderBiggi
      Dec 13 '18 at 7:42












    • $begingroup$
      Yeah, I was also surprised that a was so slow on my machine. I don't know what to think about it...
      $endgroup$
      – Henrik Schumacher
      Dec 13 '18 at 7:51
















    $begingroup$
    I was just typing a comparison of the answers so far, but you were first. I wouldn't've expected that KroneckerProduct would be that quick. Any ideas on any way that might parallelise well? I will edit my question to include that as well.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 6:43




    $begingroup$
    I was just typing a comparison of the answers so far, but you were first. I wouldn't've expected that KroneckerProduct would be that quick. Any ideas on any way that might parallelise well? I will edit my question to include that as well.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 6:43












    $begingroup$
    See my edit for a parallelized version.
    $endgroup$
    – Henrik Schumacher
    Dec 13 '18 at 7:03




    $begingroup$
    See my edit for a parallelized version.
    $endgroup$
    – Henrik Schumacher
    Dec 13 '18 at 7:03












    $begingroup$
    I am using arbitrary precision numbers, so I guess 'Compile' is not really an option.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 7:38




    $begingroup$
    I am using arbitrary precision numbers, so I guess 'Compile' is not really an option.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 7:38












    $begingroup$
    Also, interestingly, but on my machine, with Mathematica 11.3, a is faster than b though still slower than the other two.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 7:42






    $begingroup$
    Also, interestingly, but on my machine, with Mathematica 11.3, a is faster than b though still slower than the other two.
    $endgroup$
    – ThunderBiggi
    Dec 13 '18 at 7:42














    $begingroup$
    Yeah, I was also surprised that a was so slow on my machine. I don't know what to think about it...
    $endgroup$
    – Henrik Schumacher
    Dec 13 '18 at 7:51




    $begingroup$
    Yeah, I was also surprised that a was so slow on my machine. I don't know what to think about it...
    $endgroup$
    – Henrik Schumacher
    Dec 13 '18 at 7:51











    4












    $begingroup$

    Pretty printing your result gives...



    {{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h),1/j}}//MatrixForm


    $left(
    begin{array}{ccc}
    frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
    frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
    frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
    end{array}
    right)$



    Try this, it avoids constructing the huge Outer[ ] matrix



    Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]] // MatrixForm


    $left(
    begin{array}{ccc}
    frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
    frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
    frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
    end{array}
    right)$






    share|improve this answer









    $endgroup$


















      4












      $begingroup$

      Pretty printing your result gives...



      {{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h),1/j}}//MatrixForm


      $left(
      begin{array}{ccc}
      frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
      frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
      frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
      end{array}
      right)$



      Try this, it avoids constructing the huge Outer[ ] matrix



      Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]] // MatrixForm


      $left(
      begin{array}{ccc}
      frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
      frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
      frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
      end{array}
      right)$






      share|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Pretty printing your result gives...



        {{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h),1/j}}//MatrixForm


        $left(
        begin{array}{ccc}
        frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
        frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
        frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
        end{array}
        right)$



        Try this, it avoids constructing the huge Outer[ ] matrix



        Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]] // MatrixForm


        $left(
        begin{array}{ccc}
        frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
        frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
        frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
        end{array}
        right)$






        share|improve this answer









        $endgroup$



        Pretty printing your result gives...



        {{1/e, a/(b r), a/(c t)}, {b/(a y), 1/u, b/(c i)}, {c/(a g), c/(b h),1/j}}//MatrixForm


        $left(
        begin{array}{ccc}
        frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
        frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
        frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
        end{array}
        right)$



        Try this, it avoids constructing the huge Outer[ ] matrix



        Map[#/m &, MapThread[#1 #2 &, {m, 1/n}]] // MatrixForm


        $left(
        begin{array}{ccc}
        frac{1}{e} & frac{a}{b r} & frac{a}{c t} \
        frac{b}{a y} & frac{1}{u} & frac{b}{c i} \
        frac{c}{a g} & frac{c}{b h} & frac{1}{j} \
        end{array}
        right)$







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Dec 13 '18 at 0:43









        MikeYMikeY

        3,078413




        3,078413






























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