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Replacing loop with functional style



Replacing loop with functional style


Functional style using lazy lists?Can this code be written in a more functional styleProgramming a numerical method in the functional styleGoing full functional (Haskell style)Replace For-loop with functional codeReplacing a for-loop with something functionalElegant functional equivalent to a nested loop?Delimited Continuations: Easy? or Fundamentally Difficult?RealDigits and functional styleReplacing procedural code with functional code






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








3












$begingroup$


I keep hearing that one should avoid loops in Mathematica and try to write everything in functional style. I've seen few examples of how this done on here, but I could't apply to my problem successfully. It would be great to pick up new ideas of doing it.



Some data for consistency:



X = 1, 2, 3, 4;
History = 1;
SomeVals = ;
SomeFunction[x_, y_] := x, x + y;


I'm trying to rewrite the following loop into functional style:



For[i = 1, i <= Length@X, i++,
val, History = SomeFunction[X[[i]]^2, History];
AppendTo[SomeVals, val];
]









share|improve this question









$endgroup$











  • $begingroup$
    I'm confused about what you're trying to do. Particularly, I don't understand what you intended History to be used for. The output of SomeVals appears to be 1,4,9,16, which is simply the square of the elements in X. That's trivial to implement (X^2), but I don't think that's what you intended.
    $endgroup$
    – Cassini
    9 hours ago

















3












$begingroup$


I keep hearing that one should avoid loops in Mathematica and try to write everything in functional style. I've seen few examples of how this done on here, but I could't apply to my problem successfully. It would be great to pick up new ideas of doing it.



Some data for consistency:



X = 1, 2, 3, 4;
History = 1;
SomeVals = ;
SomeFunction[x_, y_] := x, x + y;


I'm trying to rewrite the following loop into functional style:



For[i = 1, i <= Length@X, i++,
val, History = SomeFunction[X[[i]]^2, History];
AppendTo[SomeVals, val];
]









share|improve this question









$endgroup$











  • $begingroup$
    I'm confused about what you're trying to do. Particularly, I don't understand what you intended History to be used for. The output of SomeVals appears to be 1,4,9,16, which is simply the square of the elements in X. That's trivial to implement (X^2), but I don't think that's what you intended.
    $endgroup$
    – Cassini
    9 hours ago













3












3








3





$begingroup$


I keep hearing that one should avoid loops in Mathematica and try to write everything in functional style. I've seen few examples of how this done on here, but I could't apply to my problem successfully. It would be great to pick up new ideas of doing it.



Some data for consistency:



X = 1, 2, 3, 4;
History = 1;
SomeVals = ;
SomeFunction[x_, y_] := x, x + y;


I'm trying to rewrite the following loop into functional style:



For[i = 1, i <= Length@X, i++,
val, History = SomeFunction[X[[i]]^2, History];
AppendTo[SomeVals, val];
]









share|improve this question









$endgroup$




I keep hearing that one should avoid loops in Mathematica and try to write everything in functional style. I've seen few examples of how this done on here, but I could't apply to my problem successfully. It would be great to pick up new ideas of doing it.



Some data for consistency:



X = 1, 2, 3, 4;
History = 1;
SomeVals = ;
SomeFunction[x_, y_] := x, x + y;


I'm trying to rewrite the following loop into functional style:



For[i = 1, i <= Length@X, i++,
val, History = SomeFunction[X[[i]]^2, History];
AppendTo[SomeVals, val];
]






functions functional-style






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 9 hours ago









maxgo2maxgo2

583 bronze badges




583 bronze badges











  • $begingroup$
    I'm confused about what you're trying to do. Particularly, I don't understand what you intended History to be used for. The output of SomeVals appears to be 1,4,9,16, which is simply the square of the elements in X. That's trivial to implement (X^2), but I don't think that's what you intended.
    $endgroup$
    – Cassini
    9 hours ago
















  • $begingroup$
    I'm confused about what you're trying to do. Particularly, I don't understand what you intended History to be used for. The output of SomeVals appears to be 1,4,9,16, which is simply the square of the elements in X. That's trivial to implement (X^2), but I don't think that's what you intended.
    $endgroup$
    – Cassini
    9 hours ago















$begingroup$
I'm confused about what you're trying to do. Particularly, I don't understand what you intended History to be used for. The output of SomeVals appears to be 1,4,9,16, which is simply the square of the elements in X. That's trivial to implement (X^2), but I don't think that's what you intended.
$endgroup$
– Cassini
9 hours ago




$begingroup$
I'm confused about what you're trying to do. Particularly, I don't understand what you intended History to be used for. The output of SomeVals appears to be 1,4,9,16, which is simply the square of the elements in X. That's trivial to implement (X^2), but I don't think that's what you intended.
$endgroup$
– Cassini
9 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

This is precisely the function of FoldPairList:



SomeVals = FoldPairList[SomeFunction[#2^2, #1] &, 1, X]


Let's do this with a more complex example to show that it works in general:



X = Array[x, 10];
History = 1;
SomeVals = ;
SomeFunction[x_, y_] := a[x, y], b[x, y]

For[i = 1, i <= Length@X,
i++, val, History = SomeFunction[X[[i]]^2, History];
AppendTo[SomeVals, val];]

SomeVals == FoldPairList[SomeFunction[#2^2, #1] &, 1, X]
(* True *)





share|improve this answer











$endgroup$















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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    This is precisely the function of FoldPairList:



    SomeVals = FoldPairList[SomeFunction[#2^2, #1] &, 1, X]


    Let's do this with a more complex example to show that it works in general:



    X = Array[x, 10];
    History = 1;
    SomeVals = ;
    SomeFunction[x_, y_] := a[x, y], b[x, y]

    For[i = 1, i <= Length@X,
    i++, val, History = SomeFunction[X[[i]]^2, History];
    AppendTo[SomeVals, val];]

    SomeVals == FoldPairList[SomeFunction[#2^2, #1] &, 1, X]
    (* True *)





    share|improve this answer











    $endgroup$

















      4












      $begingroup$

      This is precisely the function of FoldPairList:



      SomeVals = FoldPairList[SomeFunction[#2^2, #1] &, 1, X]


      Let's do this with a more complex example to show that it works in general:



      X = Array[x, 10];
      History = 1;
      SomeVals = ;
      SomeFunction[x_, y_] := a[x, y], b[x, y]

      For[i = 1, i <= Length@X,
      i++, val, History = SomeFunction[X[[i]]^2, History];
      AppendTo[SomeVals, val];]

      SomeVals == FoldPairList[SomeFunction[#2^2, #1] &, 1, X]
      (* True *)





      share|improve this answer











      $endgroup$















        4












        4








        4





        $begingroup$

        This is precisely the function of FoldPairList:



        SomeVals = FoldPairList[SomeFunction[#2^2, #1] &, 1, X]


        Let's do this with a more complex example to show that it works in general:



        X = Array[x, 10];
        History = 1;
        SomeVals = ;
        SomeFunction[x_, y_] := a[x, y], b[x, y]

        For[i = 1, i <= Length@X,
        i++, val, History = SomeFunction[X[[i]]^2, History];
        AppendTo[SomeVals, val];]

        SomeVals == FoldPairList[SomeFunction[#2^2, #1] &, 1, X]
        (* True *)





        share|improve this answer











        $endgroup$



        This is precisely the function of FoldPairList:



        SomeVals = FoldPairList[SomeFunction[#2^2, #1] &, 1, X]


        Let's do this with a more complex example to show that it works in general:



        X = Array[x, 10];
        History = 1;
        SomeVals = ;
        SomeFunction[x_, y_] := a[x, y], b[x, y]

        For[i = 1, i <= Length@X,
        i++, val, History = SomeFunction[X[[i]]^2, History];
        AppendTo[SomeVals, val];]

        SomeVals == FoldPairList[SomeFunction[#2^2, #1] &, 1, X]
        (* True *)






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 9 hours ago

























        answered 9 hours ago









        RomanRoman

        12.7k1 gold badge19 silver badges50 bronze badges




        12.7k1 gold badge19 silver badges50 bronze badges



























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