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18

One way would be to redirect all messages issued by ToExpression to a string-stream. Here is an example of that approach, with minimal error-checking: Needs["Developer`"] interpret[str_String] := Module[{s = StreamToString[], r, m} , Block[{$Messages = {s}}, r = ToExpression[str, InputForm, HoldComplete]] ; m = StringFromStream[s] ; Close[s] ; ...


13

Yes. It turns out that Normal accepts an undocumented second parameter which may be a Symbol or a list of Symbols and it will only affect those forms. Plain application converts all these forms (and more): abby = { SparseArray[{3 -> "a", 5 -> "b"}], <|1 -> "a", 2 -> "b", 3 -> "c"|>, Series[Exp[x], {x, 0, 5}], Quantity[1, ...


12

UPDATE: quite interesting parallel discussion and solutions (see Emerson Willard answer) can be found HERE. Maybe this is not exactly what you are looking for, but at least this gives you a very close guess and it is easy to figure out the rest. dis = ProbabilityDistribution[ 1/(2*E^((-m + Log[5])^2/8)*Sqrt[2*Pi]), {m, -Infinity, Infinity}]; PDF[dis, ...


10

As far as I know, there is no easy, general way to handle this kind of algebra with Sum expressions. What follows is an attempt to use replacement rules to handle a wider range of cases than chris's example. I don't consider it to be the canonical answer that is required, but perhaps someone might be able to use it as a starting point. I use Inactive on ...


9

res1 = data /. Plus -> List {a, b c, -d e f} res1[[2]] /. Times -> List {b, c} As a function: explodeOp[expr_, op_] := expr /. op -> List Then: res1 = explodeOp[data, Plus]; explodeOp[res1[[2]], Times]


9

f[m_] = 1/(2*E^((-m + Log[5])^2/8)*Sqrt[2*Pi]); Integrate[f[m], {m, -Infinity, Infinity}] 1 dist = ProbabilityDistribution[f[m], {m, -Infinity, Infinity}]; Since the integral of f[m] is unity, f[m] does not have to be scaled to be a distribution. A candidate distribution will probably have two parameters and must be defined on the interval ...


8

Yes we can ! MapAt[Integrate[#, {x, -Infinity, Infinity}] &, f[x], 1] // PowerExpand (* n *) tt = f[x]^2 /. Power[Sum[a__, b__], 2] :> sum[a (a /. i -> j) // Release, b, b /. {i -> j}] MapAt[Integrate[#, {x, -Infinity, Infinity}] &, tt, 1] /. sum -> Sum // PowerExpand


8

You can start by simply creating a function that tests, whether a string is a list of rules or not isTransformable[str_String] := SyntaxQ[str] && MatchQ[MakeExpression[str], HoldComplete[{_Rule ..}]]; isTransformable[___] := False; Note that this function does much more that search for a "->" inside a string. First, it tests, whether the ...


8

I think you've found a bug in pattern matcher. This problem can be reduced to matching sequence of length one with named BlankSequence patterns in Orderless functions, it stopped working in v10.1. In previous versions your replacement rule works (as noted by belisarius). Minimal example of this behavior is: ClearAll[f, a] SetAttributes[f, {Orderless}] ...


7

Looks like StringForm can achieve this: Cp = 1.5; deltastar = 0.123; Then: StringForm["The value for `1` is `2` and the value for `3` is `4`.", HoldForm @ Subscript[C, p], Cp, HoldForm @ Superscript[\[Delta], "*"], deltastar]


7

It sounds like you're merely looking for Row: Cp = 1.5; deltastar = 0.123; Row[{ "The value for ", HoldForm[Subscript[C, p]], " is ", Cp, " and the value for ", HoldForm[Superscript[\[Delta], "*"]], " is ", deltastar, "." }] If this does not work for you please clearly state how it fails so that those issues can be directly addressed.


7

Here is how I would do it: expr = 1/((x + a + b) (c + d)); Limit[ ϵ expr /. Thread[# -> ϵ #] &@Select[Variables[expr], # =!= x &], ϵ -> 0 ] (* ==> 1/(c x + d x) *) I replaced all variables except x by ϵ times themselves and took the limit of ϵ times the original expression as ϵ goes to zero. This will leave only terms linear in the ...


7

You mean like this? Times @@ Map[ Beta[First[#] + a, Last[#] + b] &, {{1, 1}, {12, 1}, {7, 9}, {10, 14}}]


7

A pattern matching method: fn[x_, {var__}] := List @@ Pick[x, x, Alternatives[var]^_.] fn[a x^2 + b x y + c y^2, {x, y}] {x^2, x y, y^2} But a better approach I believe is (hopefully now corrected at last): fn2[x_, var_] := Collect[List @@ Expand @ x, var, 1 &] fn2[a x^2 + b x y + c y^2, {x, y}] {x^2, x y, y^2} fn2[x (x^2 + y^2), {x, y}] ...


7

You can generate the monomials by using CoefficientRules, like this In[55]:= monomialList[poly_, vars_] := Times @@ (vars^#) & /@ CoefficientRules[poly, vars][[All, 1]] monomialList[a x^2 + b x y + c y^2, {x, y}] Out[56]= {x^2, x y, y^2}


6

As Mr.Wizard indicated, you can also reconstruct the plot using the data. Here is an example: restylePlot2[p_, op : OptionsPattern[ListLinePlot]] := ListLinePlot[Cases[Normal@p, Line[x__] :> x, ∞], op, Options[p]] then we can set the style as we do in plot. For example restylePlot2[myplot2, PlotStyle -> {{Green, Thick, Dashed}, ...


6

Use the Variables command: Variables[a x^2 + b x + c] (* {a,b,c,x} *) You can easily remove the x. Alternatively: CoefficientList[a x^2 + b x + c, x] (* {a, b, c} *)


6

I believe Block will work for you here: mem : averageFreeEnergyDensityCompiled["square", n_] := mem = Block[{Part}, Compile[{{j, _Real}, {d, _Real}, {a, _Real}, {h, _Real}, {dx, _Real}, {th, _Real, 2}, {ph, _Real, 2}}, Evaluate[(*apply boundary conditions*) s[i1_, i2_] := Which[i1 == n + 1 && i2 == n + 1, s[1, 1], i1 == ...


6

The following appears to be what you want. repl[expr_, max_] := With[{largest = Max[Cases[expr, C[n_] :> n, Infinity]]}, If[largest <= max, expr /. (R[C[largest], anything__] :> R[C[largest], anything].(II + R[C[largest + 1], C[largest]])) // repl, expr] ] repl[myExpr, 7] Alternative formulation if recursion doesn't float ...


6

Update In the interest of simplifying the code somewhat, I've modified one of the replacements. For instance, we can do expr2 = Thread[expr1, Plus] /. Plus -> Times or epxr2 = expr1 /. expT[Plus[a__]] :> Times @@ expT /@ a rather than expr2 = expr1 //. {expT[a_ + b_] :> expT[a] expT[b]} So: f[expr_] := Thread[expr /. Power[E, a_] :> ...


5

The existing answers work but I offer two improvements: terse code via pattern replacement making an actual formatting wrapper This replacement rule strips the heads of any expressions with only one argument: foo[1] /. _[x_] :> x foo[1, 2] /. _[x_] :> x 1 foo[1, 2] Format is used to describe the output format without losing the original ...


5

primeFactorForm[n_Integer] := Module[{fact = FactorInteger[n]}, If[Length[fact] == 1, Superscript @@ fact[[1]], CenterDot @@ Superscript @@@ fact]] primeFactorForm[9] primeFactorForm[72]


5

While I was working on alternative TeX export, I had similar requirement. I wanted to export annotated Mathematica code to TeX, with annotations reflecting FrontEnd's syntax highlighting. Since I couldn't find a way to use front end itself to do it, I decided to write my own package. My SyntaxAnnotations package is now available on GitHub. It works by ...


5

Given expression = (20*t^2*Erf[b + t])/(E^t^2*Sqrt[a + t^2]) can be achieved via replacement of the Head List @@ expression To find constant factors: GatherBy[List @@ expression, NumberQ] It is so because Times here is single Head for all first-level factors: TreeForm[expression] This also can prove to be useful: FactorList[expression] ...


5

f[a_, b_] := Fold[#1*Beta[#2[[1]] + a, #2[[2]] + b] &, 1, {{1, 1}, {12, 1}, {7, 9}, {10, 14}}] f[a, b] Beta[1 + a, 1 + b] Beta[7 + a, 9 + b] Beta[10 + a, 14 + b] Beta[12 + a, 1 + b]


5

res = Rest@Select[Subsets@pol, PolynomialMod[#, factor] == 0 &] So, out of the 8192 (== 2^13) possible "sub-polys", only these ones are divisible by factor: Grid[Join[{{"pol == Non-factored", "Plus (factor by ...)"}}, {pol - # factor // Expand, #} & /@ (PolynomialQuotient[#, factor, z] & /@ res)], Frame -> All] ...


4

Not that I'm aware of but perhaps this can work for you: x[a] Subscript[q, a] Superscript[m, a] d[a, b] /. y_ d[z_, p_] :> (y /. z -> p) d[z, p]


4

PaddedForm[Range[-2, -1, 0.2], {3, 2}] Plot[Sin[x], {x, -2, -1}, Frame -> True, Axes -> False, FrameTicks -> {{Automatic, Automatic},{Transpose[{#, Map[PaddedForm[#, {3, 2}] &, #]}] & [Range[-2, -1, 0.2]], Automatic}}]


4

The solution depends on how you want a zero integer coefficient treated. list1 = {0 v, 3 a, 4 b, -2 c, 9 d}; Variables[list1] {a, b, c, d} list1 /. _Integer :> 1 {1, a, b, c, d} list1 /. x_Integer :> Unitize[x] {0, a, b, c, d}


4

You can use a pattern to just operate on the integers. For example, list1 /. _Integer -> 1 yields {a,b,c,d}. Not sure if it's as fast as you need it. Explanation: the underscore _ is an unnamed pattern and the attached Integer says that the pattern should only match things whose Head is Integer. You can use other heads after the _, of course, to ...



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