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Replacing loop with functional style

Replacing loop with functional style

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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];
 ]

functions functional-style

share|improve this question

asked 9 hours ago

maxgo2maxgo2

5833 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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];
 ]

functions functional-style

share|improve this question

asked 9 hours ago

maxgo2maxgo2

5833 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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];
 ]

functions functional-style

share|improve this question

asked 9 hours ago

maxgo2maxgo2

5833 bronze badges

$endgroup$

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

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

share|improve this question

asked 9 hours ago

maxgo2maxgo2

5833 bronze badges

share|improve this question

asked 9 hours ago

maxgo2maxgo2

5833 bronze badges

asked 9 hours ago

maxgo2maxgo2

5833 bronze badges

asked 9 hours ago

maxgo2maxgo2

5833 bronze badges

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

edited 9 hours ago

answered 9 hours ago

RomanRoman

12.7k11 gold badge1919 silver badges5050 bronze badges

$endgroup$

add a comment | 

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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

edited 9 hours ago

answered 9 hours ago

RomanRoman

12.7k11 gold badge1919 silver badges5050 bronze badges

$endgroup$

add a comment | 

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

edited 9 hours ago

answered 9 hours ago

RomanRoman

12.7k11 gold badge1919 silver badges5050 bronze badges

$endgroup$

add a comment | 

$begingroup$

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

edited 9 hours ago

answered 9 hours ago

RomanRoman

12.7k11 gold badge1919 silver badges5050 bronze badges

$endgroup$

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

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

edited 9 hours ago

answered 9 hours ago

RomanRoman

12.7k11 gold badge1919 silver badges5050 bronze badges

answered 9 hours ago

RomanRoman

12.7k11 gold badge1919 silver badges5050 bronze badges

answered 9 hours ago

RomanRoman

12.7k11 gold badge1919 silver badges5050 bronze badges