The easiest way to do this is to use the symbol_names flag that was
recently added to latex(). I think this may have been added since
0.7.1, so you'll need to use a development version to get it (see
https://github.com/sympy/sympy/wiki/getting-the-bleeding-edge). With
that, you can get:
In [462]: latex(phidd**2, symbol_names={phidd:r'\ddot{\varphi}'})
Out[462]: \ddot{\varphi}^{2}
If you don't want to type that each time, you can define the dict
before hand and do latex(expr, symbol_names=translation_dict), or if
you just want to make things fast for an interactive demo, you can
redefine latex at the beginning as
def latex(*args, **kwargs): # or def my_latex if you don't want to override it
kwargs['symbol_names'] = {phidd:r'\ddot{\varphi}'}
return sympy.latex(*args, **kwargs)
which will make things a little more confusing, but will allow for
faster typing (so it depends on what your goal is).
If you're using automatic latex in the IPython notebook or qtconsole,
you'll have to modify the sympy profile.
Aaron Meurer
On Mon, Feb 20, 2012 at 7:30 AM, Carsten Knoll <[email protected]> wrote:
> Hi,
>
> I am writing a script in which I generate the an expression. This expression
> I want to use 1) with preview and 2) with lambdify.
>
>
> Lets say
>
> expr = phidd**2
>
> And I want preview to use the latex code '\ddot{\varphi}' for rendering of
> phidd. Therfore I defined:
>
> phidd = sp.Symbol(r'\ddot{\varphi}').
>
>
> However, this causes problems when I call
>
> sp.lambdify(phidd, expr)
>
>
> File "<string>", line 1
> lambda \ddot{\varphi}: (\ddot{\varphi}**2)
> ^
> SyntaxError: unexpected character after line continuation character
>
> A manual workarround would be to substitute the critical symbol with some
> uncritical one.
>
> However Im wondering if there is another (cleaner) way.
>
> I would like to use the code in a tutorial session, therefore I would like
> minimize the ballast.
>
>
> Best regards,
> Carsten
>
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