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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;
$begingroup$
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
asked 9 hours ago
maxgo2maxgo2
583 bronze badges
$endgroup$
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
asked 9 hours ago
maxgo2maxgo2
583 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
asked 9 hours ago
maxgo2maxgo2
583 bronze badges
$endgroup$
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
functions functional-style
asked 9 hours ago
maxgo2maxgo2
583 bronze badges
asked 9 hours ago
maxgo2maxgo2
583 bronze badges
asked 9 hours ago
maxgo2maxgo2
583 bronze badges
asked 9 hours ago
maxgo2maxgo2
583 bronze badges
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 *)
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 bronze badges
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add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
oldest
votes
$begingroup$
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 *)
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 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 *)
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 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 *)
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 bronze badges
$endgroup$
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 *)
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 bronze badges
edited 9 hours ago
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 bronze badges
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 bronze badges
answered 9 hours ago
RomanRoman
12.7k1 gold badge19 silver badges50 bronze badges
12.7k1 gold badge19 silver badges50 bronze badges
add a comment |
add a comment |
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$begingroup$
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