Skip to main content

Prepare mathematica output to be parsed in Python


I have Mathematica output that I would like to parse in python so I can generate equivalent expressions. So suppose I have the following:


testing2 = 
 ExpandAll[
  D[(x - A)^2 + (y - B)^2 +  ( 
      v - C)^2 + (x + y - (S + v) - D)^2  - λ1*x - λ2*
     y - λ3*v - λ4*(x + y - (S + v))  , {{x, y, v}}]]


Then I solve the following system:


Solve[Thread[
  testing2 == 0 &&  -λ1*x == 0 && - λ2*y == 
    0 && -λ3*v == 0 && - λ4*(x + y - (S + v)) == 
    0], {x, y, v, x, y, 
  v, λ1, λ2, λ3, λ4}]

Which returns a list of possible solutions. One possible solution is the following:


  {x -> 0, y -> 0, 
 v -> 1/2 (C - D - S), λ1 -> -2 A - C - D - 

   S, λ2 -> -2 B - C - D - S, λ3 -> 0, λ4 -> 
  0}

Since there are a lot of possible solutions, it is quite difficult to code everything by hand. I was trying to have the output put the multiplication symbol in the appropriate place. If I can do this, then it will be easier to parse the output in python. So suppose Mathematica is outputing 2A, I would like it to output 2*A. Is this possible?



Answer



Since python has pretty close syntax as Fortran, converting the expression to FortranForm is what I usually do in this case.


testing2 = 
 ExpandAll[
  D[(x - A)^2 + (y - B)^2 + (v - C)^2 + (x + y - (S + v) - 
       D)^2 - λ1*x - λ2*y - λ3*

     v - λ4*(x + y - (S + v)), {{x, y, v}}]]
sols = {x, y, v, x, y, 
   v, λ1, λ2, λ3, λ4} /. 
  Solve[Thread[
    testing2 == 0 && -λ1*x == 0 && -λ2*y == 
      0 && -λ3*v == 0 && -λ4*(x + y - (S + v)) == 
      0], {x, y, v, x, y, 
    v, λ1, λ2, λ3, λ4}]
sols // FortranForm


This is not ideal, but a good starting point for python to work with.


If you have access to Maple. Another solution is using Maple. CodeGeneration is pretty handy in Maple, it can also generate function using numpy and scipy


with(MmaTranslator);
print(??); # input placeholder
e := FromMma("{{0, 0, 0, 0, 0, 0, -2 (A + D + S), -2 (B + D + S), -2 (C - D - S), 
  0}, {0, 0, 1/2 (C - D - S), 0, 0, 
  1/2 (C - D - S), -2 A - C - D - S, -2 B - C - D - S, 0, 0}, {0, 
  0, -S, 0, 0, -S, -2 (A + C + S), -2 (B + C + S), 0, 
  2 (C - D + S)}, {0, S, 0, 0, S, 0, -2 (A - B + S), 
  0, -2 (B + C - S), -2 (B + D - S)}, {0, 1/2 (B + C + S), 

  1/2 (B + C - S), 0, 1/2 (B + C + S), 
  1/2 (B + C - S), -2 A + B - C - S, 0, 0, -B + C - 2 D + S}, {0, 
  1/2 (B + D + S), 0, 0, 1/2 (B + D + S), 0, -2 A + B - D - S, 
  0, -B - 2 C + D + S, 0}, {0, 1/3 (2 B + C + D + S), 
  1/3 (B + 2 C - D - S), 0, 1/3 (2 B + C + D + S), 
  1/3 (B + 2 C - D - S), -(2/3) (3 A - B + C + D + S), 0, 0, 0}, {S, 
  0, 0, S, 0, 0, 0, 
  2 (A - B - S), -2 (A + C - S), -2 (A + D - S)}, {1/2 (A - B + S), 
  1/2 (-A + B + S), 0, 1/2 (A - B + S), 1/2 (-A + B + S), 0, 0, 
  0, -A - B - 2 C + S, -A - B - 2 D + S}, {1/2 (A + C + S), 0, 

  1/2 (A + C - S), 1/2 (A + C + S), 0, 1/2 (A + C - S), 0, 
  A - 2 B - C - S, 0, -A + C - 2 D + S}, {1/3 (2 A - B + C + S), 
  1/3 (-A + 2 B + C + S), 1/3 (A + B + 2 C - S), 
  1/3 (2 A - B + C + S), 1/3 (-A + 2 B + C + S), 
  1/3 (A + B + 2 C - S), 0, 0, 
  0, -(2/3) (A + B - C + 3 D - S)}, {1/2 (A + D + S), 0, 0, 
  1/2 (A + D + S), 0, 0, 0, A - 2 B - D - S, -A - 2 C + D + S, 
  0}, {1/3 (2 A - B + D + S), 1/3 (-A + 2 B + D + S), 0, 
  1/3 (2 A - B + D + S), 1/3 (-A + 2 B + D + S), 0, 0, 
  0, -(2/3) (A + B + 3 C - D - S), 0}, {1/3 (2 A + C + D + S), 0, 

  1/3 (A + 2 C - D - S), 1/3 (2 A + C + D + S), 0, 
  1/3 (A + 2 C - D - S), 0, 2/3 (A - 3 B - C - D - S), 0, 
  0}, {1/4 (3 A - B + C + D + S), 1/4 (-A + 3 B + C + D + S), 
  1/4 (A + B + 3 C - D - S), 1/4 (3 A - B + C + D + S), 
  1/4 (-A + 3 B + C + D + S), 1/4 (A + B + 3 C - D - S), 0, 0, 0, 0}}");
with(CodeGeneration);
Python(e);

Here is an example from help of Maple


# Translate a procedure involving linear algebra.

detHilbert := proc(M, n :: posint) uses LinearAlgebra;
   return Determinant( HilbertMatrix( n ) );
end proc:
Python(detHilbert);
import numpy.linalg
import scipy.linalg

def detHilbert (M, n):
    return(numpy.linalg.det(scipy.linalg.hilbert(n)))


Update: sympy method


sympy has now support translate MMA code to sympy. Since this is a pretty new method, I will demonstrate a bit here.


In [1]: from sympy.parsing import mathematica

In [2]: mathematica.parse('Sin[a]^2 27 + 54 x + 36 x^2 + 8 x^3')
Out[2]: 'sin(a)**2 27+54 x+36 x**2+8 x**3'

Noted: sympy does not handle the multiplication correctly now, but I believe this will be solved in the future. The list conversion is not correct as well.


Currently, you can do something like this:


in MMA:



In[178]:= {Sin[a]^2 27 + 54 x + 36 x^2 + 8 x^3, ArcTan[x]} // InputForm

Out[178]//InputForm=
{54*x + 36*x^2 + 8*x^3 + 27*Sin[a]^2, ArcTan[x]}

Copy the output to Python:


In [3]: mathematica.parse('54*x + 36*x^2 + 8*x^3 + 27*Sin[a]^2')
Out[3]: '54*x+36*x**2+8*x**3+27*sin(a)**2'

This result can be further converted to sympy object



In [4]: mathematica.sympify(_)
Out[4]: 8*x**3 + 36*x**2 + 54*x + 27*sin(a)**2

You may also use mathematica function in the module to merge above two functions. But I do not suggest to use the function. Because for parse function, you get the parse result in any case, but mathematica function returns a result only the result is a valid sympy expression. Here is a example of using mathematica function:


In [1]: from sympy.parsing import mathematica as M
In [2]: M.mathematica('4a+8b^2+Cos[9a]')
Out[2]: 4*a + 8*b**2 + cos(9*a)

It is very welcomed if you can improve Mathematica parser in sympy.


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more