Can you form a list with elements which are terms from an equation?

Question

Is there a command that takes an equation as an input and creates an output with a list of terms in the original equation? Something like:

Input: Foo[ (1+a) x+ (2.5-b) y^2+ 3 c z^{1.3}]

Output: { (1+a) x ,  (2.5-b) y^2 ,  3 c z^{1.3} }

Variables, e.g. Variables[x^3 + 6 x^2 y + 3 x y z + x z^2 + 1] yields {x, y, z} — Artes, May 16 '13 at 17:32
If you want monomials there is e.g. MonomialList[a x + b y^2 + 3 c z^{1.3}] yielding {{1. a x, 1. b y^2, 3. c z^(13/10)}} — Artes, May 16 '13 at 17:38
Hi guys, thank you very much for your help so far. Unfortunately I have had to update the question as I try your suggestions and see what additional flexibility I else I need. — DJBunk, May 16 '13 at 17:41
So you need List @@ ((1 + a) x + (2.5 - b) y^2 + 3 c z^1.3) returning {(1 + a) x, (2.5 - b) y^2, 3 c z^1.3} — Artes, May 16 '13 at 17:44
@Spawn1701D - I see what you are saying now. Apply[List,%] does the trick. If you want to write this as an answer I will accept it. Thanks for the help! — DJBunk, May 16 '13 at 17:45
Huh, I can't seem to see an == sign anywhere. Where's the "equation"? — J. M.'s missing motivation, May 17 '13 at 07:35

score 7 · Answer 1 · answered May 16 '13 at 17:43

7

What you are asking to do, it seems, is to replace the Plus Head, with the List Head. The Apply function, shorthanded as @@, will do what you want:

Input: expr = Foo[a + b + c];

Now we can get just the a+b+c with First:

Input: expr2 = First@expr;

Check out FullForm to get rid of shorthanded notation:

Input: FullForm[expr2]
Output: Plus[a,b,c]

And finally, we can turn Plus into List with Apply:

Input: List@@expr2
Output: {a,b,c}

All in one line:

Input: List@@First[Foo[a+b+c]]

answered May 16 '13 at 17:43

Ian Hincks

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I am fairly certain that Foo was given as the example of the function itself, meaning that Foo[a + b + c] should directly evaluate to {a, b, c}. Nevertheless +1. – Mr.Wizard Oct 08 '14 at 16:11

Aky · Answer 2 · 2013-05-17T07:08:39.403

5

Examining the structure of the expression with TreeForm

TreeForm@Foo[(1 + a) x + (2.5 - b) y^2 + 3 c z^{1.3}]

$TreeForm@Foo[(1 + a) x + (2.5 - b) y^2 + 3 c z^{1.3}]$

shows us another way:

Level[Foo[(1 + a) x + (2.5 - b) y^2 + 3 c z^{1.3}], {3}]

(* {(1 + a) x, (2.5 - b) y^2, 3 c z^1.3} *)

(Simply count the depth of the function arguments from the Head of the expression.)

edited May 17 '13 at 07:08

answered May 17 '13 at 07:01

Aky

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12
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Can you form a list with elements which are terms from an equation?

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