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Dot products and For-loops


Why should I avoid the For loop in Mathematica?Maximizing functions with Which statementsHow do I produce iterative equations and bifurcation diagrams?How 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 MapTrouble defining function for homeworkSymmetry of a Dot with functions with argumentsFor Loop and Sum






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

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








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
    Apr 16 at 6:49

















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
    Apr 16 at 6:49













4












4








4


1



$begingroup$


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 Apr 16 at 4:43









m_goldberg

91.8k8 gold badges75 silver badges210 bronze badges




91.8k8 gold badges75 silver badges210 bronze badges










asked Apr 16 at 3:25









H. S.H. S.

261 bronze badge




261 bronze badge














  • $begingroup$
    Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
    $endgroup$
    – Roman
    Apr 16 at 6:49
















  • $begingroup$
    Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
    $endgroup$
    – Roman
    Apr 16 at 6:49















$begingroup$
Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
$endgroup$
– Roman
Apr 16 at 6:49




$begingroup$
Why should I avoid the For loop in Mathematica? mathematica.stackexchange.com/questions/134609/…
$endgroup$
– Roman
Apr 16 at 6:49










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











      $endgroup$
















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        3 Answers
        3






        active

        oldest

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        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 Apr 16 at 4:45

























            answered Apr 16 at 4:40









            m_goldbergm_goldberg

            91.8k8 gold badges75 silver badges210 bronze badges




            91.8k8 gold badges75 silver badges210 bronze badges


























                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 Apr 16 at 6:46









                    RomanRoman

                    15.9k1 gold badge22 silver badges55 bronze badges




                    15.9k1 gold badge22 silver badges55 bronze badges
























                        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 Apr 16 at 7:37

























                            answered Apr 16 at 7:27









                            Henrik SchumacherHenrik Schumacher

                            68.7k5 gold badges98 silver badges191 bronze badges




                            68.7k5 gold badges98 silver badges191 bronze badges































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                                Training a classifier when some of the features are unknownWhy does Gradient Boosting regression predict negative values when there are no negative y-values in my training set?How to improve an existing (trained) classifier?What is effect when I set up some self defined predisctor variables?Why Matlab neural network classification returns decimal values on prediction dataset?Fitting and transforming text data in training, testing, and validation setsHow to quantify the performance of the classifier (multi-class SVM) using the test data?How do I control for some patients providing multiple samples in my training data?Training and Test setTraining a convolutional neural network for image denoising in MatlabShouldn't an autoencoder with #(neurons in hidden layer) = #(neurons in input layer) be “perfect”?