I have following expression
$$-3 b_{\sigma _d}-3 b_{\sigma _v}-b-b_2+2 j-2 j_2-2 j_3+2 j_6+12$$
Mathematica input
12-b+2 j-Subscript[b, 2]-3 Subscript[b, Subscript[\[Sigma], d]]-3 Subscript[b, Subscript[\[Sigma], v]]-2 Subscript[j, 2]-2 Subscript[j, 3]+2 Subscript[j, 6]
I want to reorder this expression by partitioning the occuring symbols. Additionally, common factors of the partitions should be factored out:
Partition (given by user): $$\{\{j,j_2,j_3,j_6\},\{b,b_2,b_{\sigma_d},b_{\sigma _v}\}\}$$
The order should also be respected. The end result would look like $$2(j-j_2-j_3+j_6)-(b+b_2+3 b_{\sigma _d}+3 b_{\sigma _v})+12$$
I have considered FactorTerms[poly,{x_1,x_2,...]
and Collect[expr,{x_1,x_2,...}]
but was not successful to achieve this.
Answer
expr = 12 - b + 2 j - Subscript[b, 2] -
3 Subscript[b, Subscript[σ, d]] -
3 Subscript[b, Subscript[σ, v]] - 2 Subscript[j, 2] -
2 Subscript[j, 3] + 2 Subscript[j, 6]
Inactive[Plus] @@
(Total /@ Join @@
GatherBy[List @@ expr, MatchQ[_ (# | Subscript[#, _])] & /@ {j, b}])
TeXForm @ %
$\Large 12+\left(-3 b_{\sigma _d}-3 b_{\sigma _v}-b-b_2\right)+\left(2 j-2 j_2-2 j_3+2 j_6\right)$
Update:
ClearAll[f]
f[e_] := Row @ Flatten @ Append[Reverse @ Values @
GroupBy[Transpose[{Coefficient[e, #], #}& @ Variables[e]] /.
{a_, b_Symbol} :> {a, Subscript[b, 0]}, #[[2,1]]&,
Row[{# /. { 1 -> " + ", -1 -> " - "}, "(", HoldForm @ #2, ")"}]& @@
FactorList[ Dot @@ Transpose[#]][[All, 1]]&],
If[# < 0, {" - ", -#}, {" + ", #}]&[e /.
(Alternatives@@Variables[e] -> 0)] /. {_, 0} -> Nothing] /.
Subscript[a_, 0] -> a
Examples:
f @ expr
System`Convert`CommonDump`templateBoxToDisplay = BoxForm`TemplateBoxToDisplayBoxes;
TeXForm @ f @ expr
$\Large 2(j-j_2-j_3+j_6)\text{ - }(b+b_2+3 b_{\sigma _d}+3 b_{\sigma _v})\text{ + }12$
f[- expr - 20] // TeXForm
$\Large -2(j-j_2-j_3+j_6)\text{ + }(b+b_2+3 b_{\sigma _d}+3 b_{\sigma _v})\text{ - }32$
Note: I used Carl's answer from this q/a to make TeXForm process Row
s properly.
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