Hi Aaron,
this remark really throws me off. I certainly didn't think that Round was a
shortcut for anything. Here is how I understood Sympy:
When I write
Round1 = symbols('\\(1\\)')
then I assume that the right-hand side is the form used by Sympy for
printout, especially for print(latex(...)), and the left-hand side is just
a name that I invent, meaning that Round is my way of remembering what it
is, and the name should avoid special characters that I want in the
printout, such as parentheses and minus signs, because they cause unwanted
things to happen in an expression. By the way, that is why I have a Roundm2
= symbols('\\(-2\\)')
meaning that I have used "m" to indicate (to me) that this is minus 2, but
avoiding "-" in a name.
Now, as a bit more background, (j) is an arithmetic expression involving j
and m (a constant in this context), meaning that (j) is a shortcut for
something that would fill a lot of space in a normal printout.
By the way, in the recent test code
print(latex(Round1))
print(latex(parse_expr('Round1')))
print(latex(Round1Bar))
print(latex(parse_expr('Round1Bar')))
produces
\(1\)
Round_{1}
\bar{\(1\)}
\bar{Round1}
so the problem I have is that
print(latex(symbolname))
and
print(latex(parse_expr('symbolname')))
are different. My old code with static limits uses the first case, and the
new code with variable limits uses the second.
I can see now that parse_expr('string') is not converting 'string' into the
name of a symbol, as I expected. I also see that suffix 1 is being
converted to _{1} (as you mentioned earlier) and suffix Bar is being
converted to prefix \bar. That explains where this form is coming from.
Finally, here is one more test to avoid the automatic assumption about
suffixes (like Microsoft Word's autocorrect, but I won't go into that rant
here).
Round2BarX = symbols('\\bar{\\(2\\)}')
print(latex(Round2BarX))
print(latex(parse_expr('Round2BarX')))
produces
\bar{\(2\)}
Round2BarX
On Friday, November 6, 2020 at 10:25:54 PM UTC+1 [email protected] wrote:
> I think Round as a shortcut for putting parentheses around something
> is not implemented in the LaTeX printer. I've never seen that before,
> but if it is common, we can add it. Otherwise, if you need custom
> LaTeX in your symbol names you should just use it directly in the
> symbol name.
>
> Aaron Meurer
>
> On Fri, Nov 6, 2020 at 1:15 PM Thomas Ligon <[email protected]> wrote:
> >
> > Thanks again for the very valuable help. I think I am getting very close
> to a good solution, but I must still have a bug in there.
> > First, just a small piece of background: I am using some symbols like
> > Round1 = symbols('\\(1\\)')
> > Round1Bar = symbols('\\Bar{\\(1\\)}')
> > In other words, a "1" in parentheses and the same thing with a bar over
> it, plus some using square brackets. These were defined and used in two
> papers I am referring to, published in 1854 and 1896, both by famous
> mathematicians. One version of my code used something like E_1 instead, but
> this time I chose to stick to the old symbols, even though they are not
> typical today, since they could be confused with normal parentheses.
> > Now, I have working code that has a hard-coded limit on the number of
> symbols and I need to upgrade it so that it can handle an undetermined
> (perhaps large) number of symbols. When creating expressions, I am using
> functions like this one:
> > def getRoundjBar (j):
> > strJ = str(abs(j))
> > if j < 0:
> > strJ = 'm' + strJ
> > strSym = 'Round' + strJ + 'Bar'
> > return parse_expr(strSym)
> > and here is some test code showing it in action:
> > ret1 = getRoundjBar(1)
> > print(ret1)
> > print(latex(ret1))
> > I can see that it is important for me to understand the type of the
> variable and, fortunately, the "Locals" window in my debugger shows that
> ret1 contains the value Round1Bar and has type Symbol. Good, that makes
> almost everything work well, but
> > print(latex(ret1))
> > produces
> > \bar{Round1}
> > where it should produce
> > \\bar{\(1\)}
> >
> > On Wednesday, November 4, 2020 at 9:46:03 PM UTC+1 [email protected]
> wrote:
> >>
> >> You can't use exec() to do a return statement for a function that is
> >> not also part of the string. What you want is eval(), which evaluates
> >> an expression and returns it, like
> >>
> >> return eval(string)
> >>
> >> However, you should consider just using sympify() or parse_expr() if
> >> you are evaluating strings to SymPy expressions, as they are designed
> >> for that and will give you things like automatically defining the
> >> symbols.
> >>
> >> Actually, for what you've shown, I don't see the need to use exec() or
> >> eval() at all. You can just create the symbols from the strings, like
> >> Symbol('a' + str(j)). Symbols are completely defined by name, so if
> >> two symbols have the same name, they will be equal.
> >>
> >> Aaron Meurer
> >>
> >> On Wed, Nov 4, 2020 at 2:24 AM Thomas Ligon <[email protected]>
> wrote:
> >> >
> >> > Now I see a solution: forget the function and create the expression
> exp using string manipulation and exec, something like
> >> > strExp = 'complex string'
> >> > strSym = 'exp = ' + strExp
> >> > exec(strSym)
> >> >
> >> > On Wednesday, November 4, 2020 at 9:51:35 AM UTC+1 Thomas Ligon wrote:
> >> >>
> >> >> Thanks!
> >> >> The automatic subscript is just a minor irritation, and your
> response is helpful.
> >> >> I have included a new test program that hopefully no longer
> oversimplifies the situation. Currently, I have code (2000 lines) for
> investigating a very complex series, and it runs to a maximum of 3 terms of
> the series, requiring index values up to 6. I showed the results to the
> mathematics professor who has mentored my work and he immediately says that
> it is valuable and I should publish a version that can go up to a very
> large number of terms. In order to do that, I need to produce a fully
> automated version with a variable number of terms and indices.
> >> >> So, for variable j, I want to use a_j in a formula.
> >> >> The expression exp on line 20 does that, where I use get_a(1) to
> represent aj for variable j, where j happens to be 1.
> >> >> In line 5, the return is part of a function. In line 8, I expect
> return in exec ('return...') to be a part of a function.
> >> >>
> >> >> maxIndex = 2
> >> >> import sympy
> >> >> from sympy import symbols, latex
> >> >> def get_a1():
> >> >> return a1
> >> >> def get_a2():
> >> >> strSym = 'return a2'
> >> >> exec(strSym)
> >> >> def get_a (j):
> >> >> strSym = 'return a' + str(j)
> >> >> exec(strSym)
> >> >> lam = symbols('\\lambda')
> >> >> for ind in range(1,maxIndex+1):
> >> >> strSym = 'a' + str(ind) + ' = symbols(\'a_' + str(ind) + '\')'
> >> >> exec(strSym)
> >> >> #a1 = symbols('a_1')
> >> >> #a2 = symbols('a_2')
> >> >> print(latex(get_a1()))
> >> >> #print(latex(get_a2())) # causes 'return' outside function
> (<string>, line 1)
> >> >> exp = 3*get_a(1) + 5*get_a(2)*lam**2 # causes 'return' outside
> function (<string>, line 1)
> >> >> print(latex(exp))
> >> >> print('end testSym')
> >> >>
> >> >> On Tuesday, November 3, 2020 at 11:29:00 PM UTC+1 [email protected]
> wrote:
> >> >>>
> >> >>> I'm unclear what you're trying to achieve with the functions. exec()
> >> >>> takes a string of Python code and executes it. The entire string
> must
> >> >>> be valid Python code by itself, so exec('return x') fails because a
> >> >>> bare "return x" is not valid Python. "return" must be inside a
> >> >>> function definition to be valid. But even defSym1 doesn't do
> anything
> >> >>> useful beyond just returning Sym1, so there's no point to having it
> >> >>> instead of just "Sym1" directly.
> >> >>>
> >> >>> The LaTeX version of Sym1 contains _ because SymPy automatically
> >> >>> assumes that symbol names ending in numbers are subscripted, so it
> >> >>> renders it as a Sym_1 in LaTeX. If you don't want the 1 to be
> >> >>> subscripted, you can use something like Symbol("{Sym1}").
> >> >>>
> >> >>> Aaron Meurer
> >> >>>
> >> >>> On Tue, Nov 3, 2020 at 9:42 AM Thomas Ligon <[email protected]>
> wrote:
> >> >>> >
> >> >>> > The following code is not working as I expected.
> >> >>> > Why does Sym1 contain an underline (_{1})?
> >> >>> > Why does return fail (unhandled exception) in line 7?
> >> >>> >
> >> >>> > Code:
> >> >>> > import sympy
> >> >>> > from sympy import symbols, latex
> >> >>> > def defSym1():
> >> >>> > return Sym1
> >> >>> > def defSym2():
> >> >>> > strSym = 'return Sym2'
> >> >>> > exec(strSym)
> >> >>> > Sym1 = symbols('Sym1')
> >> >>> > Sym2 = symbols('Sym2')
> >> >>> > print(latex(defSym1()))
> >> >>> > print(latex(defSym2()))
> >> >>> > print('end testSym')
> >> >>> >
> >> >>> > Output:
> >> >>> > Sym_{1}
> >> >>> > 'return' outside function (<string>, line 1)
> >> >>> >
> >> >>> >
> >> >>> >
> >> >>> >
> >> >>> > --
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> .
> >> >
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