Dot products and For-loops Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm...

Why is my conclusion inconsistent with the van't Hoff equation?

ListPlot join points by nearest neighbor rather than order

How do pianists reach extremely loud dynamics?

Bete Noir -- no dairy

How come Sam didn't become Lord of Horn Hill?

How to react to hostile behavior from a senior developer?

Identifying polygons that intersect with another layer using QGIS?

What LEGO pieces have "real-world" functionality?

Why is "Consequences inflicted." not a sentence?

Why are there no cargo aircraft with "flying wing" design?

What does "fit" mean in this sentence?

English words in a non-english sci-fi novel

Can a USB port passively 'listen only'?

Fundamental Solution of the Pell Equation

Output the ŋarâþ crîþ alphabet song without using (m)any letters

Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?

Storing hydrofluoric acid before the invention of plastics

2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?

51k Euros annually for a family of 4 in Berlin: Is it enough?

Single word antonym of "flightless"

Is the Standard Deduction better than Itemized when both are the same amount?

How to run gsettings for another user Ubuntu 18.04.2 LTS

What's the purpose of writing one's academic biography in the third person?

Seeking colloquialism for “just because”



Dot products and For-loops



Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Why should I avoid the For loop in Mathematica?Maximizing functions with Which statementsNested For Loops that result in no outputHow do I write nested for-loops?How would I write iterative code to try multiple coefficients for an equation?Plotting Invariant Manifolds of the Henon MapVariable Increments in a TableWhat is p[a] forFormatting a list of values and dependent formatting of corresponding errors listSymmetry of a Dot with functions with argumentsFor Loop and Sum












4












$begingroup$


I'm taking a class where we're working with matrix applications inside of Mathematica. In this case, we're analyzing senatorial votes that have been numericized so that a value of 1 corresponds to "aye," -1 corresponds to "nay," and 0 corresponds to no vote/abstain.



Taking the dot products of these values, I'm trying to find, for each senator, the senator(s) with the most and least similar voting patterns (largest and smallest dot products) and plot these in a chart.



This may not be the most efficient code to accomplish this, but given that it's for a class, I am supposed to modify what's below:



closestvoter[n_] := 
Module[{max = 0, k = 0, close = 0},
For[i = 1, i < 100, i++,
k = senatorvote[n] . senatorvote[i];
If[k >= max && i != n, close = i; max = k];
];
close
]


The way the code works right now, it will only give me a single value of i when sometimes there is more than one senator that produces the same dot product. How can I get the loop to check and print all values of i that satisfy this?










share|improve this question











$endgroup$












  • $begingroup$
    Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
    $endgroup$
    – Roman
    19 hours ago
















4












$begingroup$


I'm taking a class where we're working with matrix applications inside of Mathematica. In this case, we're analyzing senatorial votes that have been numericized so that a value of 1 corresponds to "aye," -1 corresponds to "nay," and 0 corresponds to no vote/abstain.



Taking the dot products of these values, I'm trying to find, for each senator, the senator(s) with the most and least similar voting patterns (largest and smallest dot products) and plot these in a chart.



This may not be the most efficient code to accomplish this, but given that it's for a class, I am supposed to modify what's below:



closestvoter[n_] := 
Module[{max = 0, k = 0, close = 0},
For[i = 1, i < 100, i++,
k = senatorvote[n] . senatorvote[i];
If[k >= max && i != n, close = i; max = k];
];
close
]


The way the code works right now, it will only give me a single value of i when sometimes there is more than one senator that produces the same dot product. How can I get the loop to check and print all values of i that satisfy this?










share|improve this question











$endgroup$












  • $begingroup$
    Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
    $endgroup$
    – Roman
    19 hours ago














4












4








4





$begingroup$


I'm taking a class where we're working with matrix applications inside of Mathematica. In this case, we're analyzing senatorial votes that have been numericized so that a value of 1 corresponds to "aye," -1 corresponds to "nay," and 0 corresponds to no vote/abstain.



Taking the dot products of these values, I'm trying to find, for each senator, the senator(s) with the most and least similar voting patterns (largest and smallest dot products) and plot these in a chart.



This may not be the most efficient code to accomplish this, but given that it's for a class, I am supposed to modify what's below:



closestvoter[n_] := 
Module[{max = 0, k = 0, close = 0},
For[i = 1, i < 100, i++,
k = senatorvote[n] . senatorvote[i];
If[k >= max && i != n, close = i; max = k];
];
close
]


The way the code works right now, it will only give me a single value of i when sometimes there is more than one senator that produces the same dot product. How can I get the loop to check and print all values of i that satisfy this?










share|improve this question











$endgroup$




I'm taking a class where we're working with matrix applications inside of Mathematica. In this case, we're analyzing senatorial votes that have been numericized so that a value of 1 corresponds to "aye," -1 corresponds to "nay," and 0 corresponds to no vote/abstain.



Taking the dot products of these values, I'm trying to find, for each senator, the senator(s) with the most and least similar voting patterns (largest and smallest dot products) and plot these in a chart.



This may not be the most efficient code to accomplish this, but given that it's for a class, I am supposed to modify what's below:



closestvoter[n_] := 
Module[{max = 0, k = 0, close = 0},
For[i = 1, i < 100, i++,
k = senatorvote[n] . senatorvote[i];
If[k >= max && i != n, close = i; max = k];
];
close
]


The way the code works right now, it will only give me a single value of i when sometimes there is more than one senator that produces the same dot product. How can I get the loop to check and print all values of i that satisfy this?







conditional homework iterators






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 21 hours ago









m_goldberg

88.8k873200




88.8k873200










asked 22 hours ago









H. S.H. S.

261




261












  • $begingroup$
    Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
    $endgroup$
    – Roman
    19 hours ago


















  • $begingroup$
    Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
    $endgroup$
    – Roman
    19 hours ago
















$begingroup$
Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
$endgroup$
– Roman
19 hours ago




$begingroup$
Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
$endgroup$
– Roman
19 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

I recommend you make close a list in which you can accumulate values of i that give the same k. The test for updating close and max will have to more elaborate.



Clear[closestvoter]
closestvoter[n_, maxn_] :=
Module[{max = 0, k = 0, close = {}},
For[i = 1, i <= maxn, i++,
If[i != n,
k = senatorvote[n].senatorvote[i];
Which[
k == max, AppendTo[close, i],
k > max, close = {i}; max = k]]];
{max, close}]]


Note this version of closestvoter reports max as well as the all the indexes for which k == max.



Contrived test data.



SeedRandom[42]; Clear[senatorvote]; 
Do[senatorvote[i] = RandomInteger[{-1, 1}, 30], {i, 25}];


Here is the results for the contrived data.



Column @ Table[Join[{i}, closestvoter[i, 25]], {i, 25}]


results



We see that only the first senator has the same maximum score with more than one other senator.






share|improve this answer











$endgroup$





















    3












    $begingroup$

    Define the votes as lists instead of a function:



    SeedRandom[1234];
    senatorvote = RandomInteger[{-1, 1}, {25, 30}];

    closestvoters[n_] := MaximalBy[Delete[Range[Length[senatorvote]], n],
    senatorvote[[n]].senatorvote[[#]] &]

    farthestvoters[n_] := MinimalBy[Range[Length[senatorvote]],
    senatorvote[[n]].senatorvote[[#]] &]


    test:



    closestvoters[1]



    {13, 20}




    farthestvoters[1]



    {5}







    share|improve this answer









    $endgroup$





















      2












      $begingroup$

      In addition to Roman's post, here is a method to compute all closest and farthest voters to all senators at once as follows; this should be about to 2 orders of magnitude faster than Roman's code.



      SeedRandom[1234];
      senatorvote = RandomInteger[{-1, 1}, {2500, 3000}];

      A = senatorvote.senatorvote[Transpose];

      closestvoters = Flatten[Position[#, Max[#]]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
      farthestvoters = Flatten[Position[#, Min[#]]] & /@ A;


      By computing A first, you can take greatest advantage of the linear algebra capabilities of your hardware. Afterwards, you have to find only the minimum and maximum positions per row. So the closest voters to senator n will be written into closestvoters[[n]].



      Since Position is somewhat slow in finding positions of integers in a vector (notice that Position can do a lot more than that which comes at a cost), here a faster implementation utilizing an undocumented function:



      closestvoters = Random`Private`PositionsOf[#, Max[#]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
      farthestvoters = Random`Private`PositionsOf[#, Min[#]] & /@ A;





      share|improve this answer











      $endgroup$














        Your Answer








        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%2f195264%2fdot-products-and-for-loops%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        I recommend you make close a list in which you can accumulate values of i that give the same k. The test for updating close and max will have to more elaborate.



        Clear[closestvoter]
        closestvoter[n_, maxn_] :=
        Module[{max = 0, k = 0, close = {}},
        For[i = 1, i <= maxn, i++,
        If[i != n,
        k = senatorvote[n].senatorvote[i];
        Which[
        k == max, AppendTo[close, i],
        k > max, close = {i}; max = k]]];
        {max, close}]]


        Note this version of closestvoter reports max as well as the all the indexes for which k == max.



        Contrived test data.



        SeedRandom[42]; Clear[senatorvote]; 
        Do[senatorvote[i] = RandomInteger[{-1, 1}, 30], {i, 25}];


        Here is the results for the contrived data.



        Column @ Table[Join[{i}, closestvoter[i, 25]], {i, 25}]


        results



        We see that only the first senator has the same maximum score with more than one other senator.






        share|improve this answer











        $endgroup$


















          3












          $begingroup$

          I recommend you make close a list in which you can accumulate values of i that give the same k. The test for updating close and max will have to more elaborate.



          Clear[closestvoter]
          closestvoter[n_, maxn_] :=
          Module[{max = 0, k = 0, close = {}},
          For[i = 1, i <= maxn, i++,
          If[i != n,
          k = senatorvote[n].senatorvote[i];
          Which[
          k == max, AppendTo[close, i],
          k > max, close = {i}; max = k]]];
          {max, close}]]


          Note this version of closestvoter reports max as well as the all the indexes for which k == max.



          Contrived test data.



          SeedRandom[42]; Clear[senatorvote]; 
          Do[senatorvote[i] = RandomInteger[{-1, 1}, 30], {i, 25}];


          Here is the results for the contrived data.



          Column @ Table[Join[{i}, closestvoter[i, 25]], {i, 25}]


          results



          We see that only the first senator has the same maximum score with more than one other senator.






          share|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            I recommend you make close a list in which you can accumulate values of i that give the same k. The test for updating close and max will have to more elaborate.



            Clear[closestvoter]
            closestvoter[n_, maxn_] :=
            Module[{max = 0, k = 0, close = {}},
            For[i = 1, i <= maxn, i++,
            If[i != n,
            k = senatorvote[n].senatorvote[i];
            Which[
            k == max, AppendTo[close, i],
            k > max, close = {i}; max = k]]];
            {max, close}]]


            Note this version of closestvoter reports max as well as the all the indexes for which k == max.



            Contrived test data.



            SeedRandom[42]; Clear[senatorvote]; 
            Do[senatorvote[i] = RandomInteger[{-1, 1}, 30], {i, 25}];


            Here is the results for the contrived data.



            Column @ Table[Join[{i}, closestvoter[i, 25]], {i, 25}]


            results



            We see that only the first senator has the same maximum score with more than one other senator.






            share|improve this answer











            $endgroup$



            I recommend you make close a list in which you can accumulate values of i that give the same k. The test for updating close and max will have to more elaborate.



            Clear[closestvoter]
            closestvoter[n_, maxn_] :=
            Module[{max = 0, k = 0, close = {}},
            For[i = 1, i <= maxn, i++,
            If[i != n,
            k = senatorvote[n].senatorvote[i];
            Which[
            k == max, AppendTo[close, i],
            k > max, close = {i}; max = k]]];
            {max, close}]]


            Note this version of closestvoter reports max as well as the all the indexes for which k == max.



            Contrived test data.



            SeedRandom[42]; Clear[senatorvote]; 
            Do[senatorvote[i] = RandomInteger[{-1, 1}, 30], {i, 25}];


            Here is the results for the contrived data.



            Column @ Table[Join[{i}, closestvoter[i, 25]], {i, 25}]


            results



            We see that only the first senator has the same maximum score with more than one other senator.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 21 hours ago

























            answered 21 hours ago









            m_goldbergm_goldberg

            88.8k873200




            88.8k873200























                3












                $begingroup$

                Define the votes as lists instead of a function:



                SeedRandom[1234];
                senatorvote = RandomInteger[{-1, 1}, {25, 30}];

                closestvoters[n_] := MaximalBy[Delete[Range[Length[senatorvote]], n],
                senatorvote[[n]].senatorvote[[#]] &]

                farthestvoters[n_] := MinimalBy[Range[Length[senatorvote]],
                senatorvote[[n]].senatorvote[[#]] &]


                test:



                closestvoters[1]



                {13, 20}




                farthestvoters[1]



                {5}







                share|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Define the votes as lists instead of a function:



                  SeedRandom[1234];
                  senatorvote = RandomInteger[{-1, 1}, {25, 30}];

                  closestvoters[n_] := MaximalBy[Delete[Range[Length[senatorvote]], n],
                  senatorvote[[n]].senatorvote[[#]] &]

                  farthestvoters[n_] := MinimalBy[Range[Length[senatorvote]],
                  senatorvote[[n]].senatorvote[[#]] &]


                  test:



                  closestvoters[1]



                  {13, 20}




                  farthestvoters[1]



                  {5}







                  share|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Define the votes as lists instead of a function:



                    SeedRandom[1234];
                    senatorvote = RandomInteger[{-1, 1}, {25, 30}];

                    closestvoters[n_] := MaximalBy[Delete[Range[Length[senatorvote]], n],
                    senatorvote[[n]].senatorvote[[#]] &]

                    farthestvoters[n_] := MinimalBy[Range[Length[senatorvote]],
                    senatorvote[[n]].senatorvote[[#]] &]


                    test:



                    closestvoters[1]



                    {13, 20}




                    farthestvoters[1]



                    {5}







                    share|improve this answer









                    $endgroup$



                    Define the votes as lists instead of a function:



                    SeedRandom[1234];
                    senatorvote = RandomInteger[{-1, 1}, {25, 30}];

                    closestvoters[n_] := MaximalBy[Delete[Range[Length[senatorvote]], n],
                    senatorvote[[n]].senatorvote[[#]] &]

                    farthestvoters[n_] := MinimalBy[Range[Length[senatorvote]],
                    senatorvote[[n]].senatorvote[[#]] &]


                    test:



                    closestvoters[1]



                    {13, 20}




                    farthestvoters[1]



                    {5}








                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 19 hours ago









                    RomanRoman

                    5,49111131




                    5,49111131























                        2












                        $begingroup$

                        In addition to Roman's post, here is a method to compute all closest and farthest voters to all senators at once as follows; this should be about to 2 orders of magnitude faster than Roman's code.



                        SeedRandom[1234];
                        senatorvote = RandomInteger[{-1, 1}, {2500, 3000}];

                        A = senatorvote.senatorvote[Transpose];

                        closestvoters = Flatten[Position[#, Max[#]]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                        farthestvoters = Flatten[Position[#, Min[#]]] & /@ A;


                        By computing A first, you can take greatest advantage of the linear algebra capabilities of your hardware. Afterwards, you have to find only the minimum and maximum positions per row. So the closest voters to senator n will be written into closestvoters[[n]].



                        Since Position is somewhat slow in finding positions of integers in a vector (notice that Position can do a lot more than that which comes at a cost), here a faster implementation utilizing an undocumented function:



                        closestvoters = Random`Private`PositionsOf[#, Max[#]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                        farthestvoters = Random`Private`PositionsOf[#, Min[#]] & /@ A;





                        share|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          In addition to Roman's post, here is a method to compute all closest and farthest voters to all senators at once as follows; this should be about to 2 orders of magnitude faster than Roman's code.



                          SeedRandom[1234];
                          senatorvote = RandomInteger[{-1, 1}, {2500, 3000}];

                          A = senatorvote.senatorvote[Transpose];

                          closestvoters = Flatten[Position[#, Max[#]]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                          farthestvoters = Flatten[Position[#, Min[#]]] & /@ A;


                          By computing A first, you can take greatest advantage of the linear algebra capabilities of your hardware. Afterwards, you have to find only the minimum and maximum positions per row. So the closest voters to senator n will be written into closestvoters[[n]].



                          Since Position is somewhat slow in finding positions of integers in a vector (notice that Position can do a lot more than that which comes at a cost), here a faster implementation utilizing an undocumented function:



                          closestvoters = Random`Private`PositionsOf[#, Max[#]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                          farthestvoters = Random`Private`PositionsOf[#, Min[#]] & /@ A;





                          share|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            In addition to Roman's post, here is a method to compute all closest and farthest voters to all senators at once as follows; this should be about to 2 orders of magnitude faster than Roman's code.



                            SeedRandom[1234];
                            senatorvote = RandomInteger[{-1, 1}, {2500, 3000}];

                            A = senatorvote.senatorvote[Transpose];

                            closestvoters = Flatten[Position[#, Max[#]]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                            farthestvoters = Flatten[Position[#, Min[#]]] & /@ A;


                            By computing A first, you can take greatest advantage of the linear algebra capabilities of your hardware. Afterwards, you have to find only the minimum and maximum positions per row. So the closest voters to senator n will be written into closestvoters[[n]].



                            Since Position is somewhat slow in finding positions of integers in a vector (notice that Position can do a lot more than that which comes at a cost), here a faster implementation utilizing an undocumented function:



                            closestvoters = Random`Private`PositionsOf[#, Max[#]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                            farthestvoters = Random`Private`PositionsOf[#, Min[#]] & /@ A;





                            share|improve this answer











                            $endgroup$



                            In addition to Roman's post, here is a method to compute all closest and farthest voters to all senators at once as follows; this should be about to 2 orders of magnitude faster than Roman's code.



                            SeedRandom[1234];
                            senatorvote = RandomInteger[{-1, 1}, {2500, 3000}];

                            A = senatorvote.senatorvote[Transpose];

                            closestvoters = Flatten[Position[#, Max[#]]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                            farthestvoters = Flatten[Position[#, Min[#]]] & /@ A;


                            By computing A first, you can take greatest advantage of the linear algebra capabilities of your hardware. Afterwards, you have to find only the minimum and maximum positions per row. So the closest voters to senator n will be written into closestvoters[[n]].



                            Since Position is somewhat slow in finding positions of integers in a vector (notice that Position can do a lot more than that which comes at a cost), here a faster implementation utilizing an undocumented function:



                            closestvoters = Random`Private`PositionsOf[#, Max[#]] & /@ (A - 2 DiagonalMatrix[Diagonal[A]]);
                            farthestvoters = Random`Private`PositionsOf[#, Min[#]] & /@ A;






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 18 hours ago

























                            answered 18 hours ago









                            Henrik SchumacherHenrik Schumacher

                            60.3k583169




                            60.3k583169






























                                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.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195264%2fdot-products-and-for-loops%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

                                Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

                                VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...

                                Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...