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












5














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
























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










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










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
















5














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
























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










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










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














5












5








5


1





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















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 at 6:45

























asked Dec 12 at 23:45









ThunderBiggi

383112




383112












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










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










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


















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










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










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
















What is the length of m in practical use?
– Αλέξανδρος Ζεγγ
Dec 13 at 3:03




What is the length of m in practical use?
– Αλέξανδρος Ζεγγ
Dec 13 at 3:03












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




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












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




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










2 Answers
2






active

oldest

votes


















6














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























  • 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.
    – ThunderBiggi
    Dec 13 at 6:43










  • See my edit for a parallelized version.
    – Henrik Schumacher
    Dec 13 at 7:03










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










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












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



















4














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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "387"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f187804%2ffast-element-wise-division-of-matrix-generated-from-vector-with-outer-and-an%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    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























    • 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.
      – ThunderBiggi
      Dec 13 at 6:43










    • See my edit for a parallelized version.
      – Henrik Schumacher
      Dec 13 at 7:03










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










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












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
















    6














    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























    • 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.
      – ThunderBiggi
      Dec 13 at 6:43










    • See my edit for a parallelized version.
      – Henrik Schumacher
      Dec 13 at 7:03










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










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












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














    6












    6








    6






    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














    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 at 7:03

























    answered Dec 13 at 6:32









    Henrik Schumacher

    48.1k467136




    48.1k467136












    • 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.
      – ThunderBiggi
      Dec 13 at 6:43










    • See my edit for a parallelized version.
      – Henrik Schumacher
      Dec 13 at 7:03










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










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












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


















    • 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.
      – ThunderBiggi
      Dec 13 at 6:43










    • See my edit for a parallelized version.
      – Henrik Schumacher
      Dec 13 at 7:03










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










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












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
















    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.
    – ThunderBiggi
    Dec 13 at 6:43




    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.
    – ThunderBiggi
    Dec 13 at 6:43












    See my edit for a parallelized version.
    – Henrik Schumacher
    Dec 13 at 7:03




    See my edit for a parallelized version.
    – Henrik Schumacher
    Dec 13 at 7:03












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




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












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






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














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




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











    4














    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


























      4














      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
























        4












        4








        4






        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












        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 at 0:43









        MikeY

        1,952410




        1,952410






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematica Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f187804%2ffast-element-wise-division-of-matrix-generated-from-vector-with-outer-and-an%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Morgemoulin

            Image
            Morgemoulin — miejscowość i gmina — Państwo   Francja Region Grand Est Departament Moza Okręg Verdun Kod INSEE 55357 Powierzchnia 6,82 km² Populacja  (1990) • liczba ludności 75 • gęstość 11 os./km² Kod pocztowy 55400 Położenie na mapie Francji Morgemoulin 49°14′N   5°35′E / 49,233333   5,583333 Portal Francja Morgemoulin – miejscowość i gmina we Francji, w regionie Grand Est, w departamencie Moza. Według danych na rok 1990 gminę zamieszkiwało 75 osób, a gęstość zaludnienia wynosiła 11 osób/km² (wśród 2335 gmin Lotaryngii Morgemoulin plasuje się na 970. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 848.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Verdun Abaucourt-Hautecourt • Aincreville • Ambly-sur-Meuse • Amel-sur-l'Étang • Ancemont • Arrancy-sur-Crusne • Aubréville • Autré...
            Read more

            Scott Moir

            Image
            Scott Moir Tessa Virtue i Scott Moir w 2016 r. Reprezentacja   Kanada Data i miejsce urodzenia 2 września 1987 London Wzrost 180 cm Konkurencja Pary taneczne Partner sportowy Tessa Virtue Trener Marie-France Dubreuil, Patrice Lauzon, Romain Haguenauer Poprzednio: Marina Zujewa, Oleg Epstein, Igor Szpilband, Johnny Johns, Carol Moir, Paul MacIntosh, Suzanne Killing Klub Montreal International School of Skating Poprzednio: Arctic Edge FSC Ilderton Skating Club Lokalizacja treningowa Montreal Poprzednio: Canton Kitchener Rekordy życiowe ISU przed sezonem 2018/2019 Nota łączna 206,07 Igrzyska Olimpijskie 2018 Taniec krótki 83,67 Igrzyska Olimpijskie 2018 Taniec dowolny 122,40 Igrzyska Olimpijskie 2018 Dorobek medalowy Reprezentacja   Kanada Igrzyska olimpijskie Złoto Vancouver 2010 pary taneczne Złoto Pjongczang 2018 pary taneczne Złoto Pjongczang 2018 drużynowo Srebro S...
            Read more

            Souastre

            Image
            Souastre — miejscowość i gmina — Herb Państwo   Francja Region Hauts-de-France Departament Pas-de-Calais Okręg Arras Kod INSEE 62800 Powierzchnia 7,18 km² Populacja  (1990) • liczba ludności 352 • gęstość 49 os./km² Kod pocztowy 62111 Położenie na mapie Francji Souastre 50°09′N   2°34′E / 50,150000   2,566667 Portal Francja Souastre – miejscowość i gmina we Francji, w regionie Hauts-de-France, w departamencie Pas-de-Calais. Według danych na rok 1990 gminę zamieszkiwały 352 osoby, a gęstość zaludnienia wynosiła 49 osób/km² (wśród 1549 gmin regionu Nord-Pas-de-Calais Souastre plasuje się na 887. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 507.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Arras Ablain-Saint-Nazaire Ablainzevelle Acheville Achicourt Achiet-le-Grand Achiet-le-Pe...
            Read more