My guess is that your debugger is trying to filter classes from the
global variables, since there typically are a lot of them and showing
them doesn't provide a lot of info. Function('f') is actually a class
(same as if you typed "class f(Function): ..."). There's no reason why
global would behave differently for one type of object or another.
Python doesn't care what the type of an object is when it does
scoping.
By the way, personally I would not do this dance with global. If you
want your variables to be defined in the global namespace, just define
them at the module level instead of inside of a function.
Aaron Meurer
On Fri, Mar 10, 2023 at 1:25 PM Thomas Ligon <thomassli...@gmail.com> wrote:
>
> I now have a test showing that I can write this either way, and also take the
> derivative either way. But, it is still irritating not to see q2F in the list
> of global variables in the debugger. Maybe there is a bug in the debugger.
> After all I can assign q2F(t) to the variable C, which wouldn't work if it
> weren't defined. Here is the code:
> global t, q1F, q2F
> t = symbols('t')
> q1F = Function('q1F')(t)
> q2F = Function('q2F')
> #A = q1F(t) # 'q1F' object is not callable
> A = q1F
> print(A) # q1F(t)
> B = diff(A, t)
> print(B) # Derivative(q1F(t), t)
> C = q2F(t)
> print(C) # q2F(t)
> D = diff(C, t)
> print(D) # Derivative(q2F(t), t)
>
> I am using Python inside of Visual Studio and looking for the global
> variables in the Locals window (with {}Globals expanded). Could someone try
> this in a different IDE?'
>
> On Friday, March 10, 2023 at 5:31:57 PM UTC+1 Thomas Ligon wrote:
>>
>> Thanks, but this just shows how poorly I formulated my question. After we
>> solved some printing problems, I am very happy with the way it is working.
>> My problem is Function. With
>> q1F = Function('q1F')(t)
>> I get: q1F has value q1F(t) and type Q1F. With
>> q1F = Function('q1F')
>> I get nothing, even though this was preceded by
>> global q1F
>> I was previously using
>> q1F = Function('q1F')
>> and I think it was creating a global q1F with type Function, but I must have
>> broken something.
>> I am defining both q1S and q1F because I need a symbol for q1 for most
>> calculations, and I also need a function for q1 in order to take derivatives.
>> if nDer == 0:
>> retVal = q1F#(t)
>> returns q1F with no derivative (nDer = number of derivatives), and
>> elif nDer == 1:
>> retVal = diff(qd(1, 0), t)
>> takes the first derivative of the above. (qd(qi=1, nDer) returns the
>> nDer-th derivative of qi=1, i.e. q1).
>>
>> On Thursday, March 9, 2023 at 8:59:19 PM UTC+1 asme...@gmail.com wrote:
>>>
>>> If you want your code to match what is printed, you should write
>>>
>>> f = Function('f')
>>>
>>> instead of
>>>
>>> f = Function('f')(t)
>>>
>>> Alternatively you could use a special printer that prints things as
>>> you want. There is already a printer in the mechanics module that
>>> ignores the (t) part when printing a function (mprint()).
>>>
>>> Aaron Meurer
>>>
>>> On Thu, Mar 9, 2023 at 8:12 AM Thomas Ligon <thomas...@gmail.com> wrote:
>>> >
>>> > I am now using Symbol and Function where the name and the value match.
>>> > That was recommended for easier reading and then required for pickle. In
>>> > order to avoid confusing these, I have a different name for Symbol and
>>> > for Function, then I convert them just before printing. My problem is
>>> > that I am not sure how to use this when I take derivatives. When I look
>>> > at q1S in the debugger I see it has type Symbol, but q1F has value q1F(t)
>>> > and type Q1F (not Function).
>>> > The code looks like this:
>>> > def createSymbols():
>>> > global t, q1S, q2S, q1F, q2F, q1dotS, q2dotS, q1ddotS, q2ddotS
>>> > t = symbols('t')
>>> > q1S, q2S = symbols('q1S, q2S')
>>> > q1dotS, q2dotS = symbols('q1dotS, q2dotS')
>>> > q1ddotS, q2ddotS = symbols('q1ddotS, q2ddotS')
>>> > q1F = Function('q1F')(t)
>>> > q2F = Function('q2F')(t)
>>> > print('end of createSymbols HillCusp')
>>> >
>>> > def symForLatex(expIn):
>>> > expOut = expIn
>>> > expOut = expOut.subs(q1S, symbols('q1'))
>>> > expOut = expOut.subs(q2S, symbols('q2'))
>>> > expOut = expOut.subs(q1F, symbols('q1'))
>>> > expOut = expOut.subs(q2F, symbols('q2'))
>>> > expOut = expOut.subs(q1dotS, symbols('{\dot{q}}_{1}'))
>>> > expOut = expOut.subs(q2dotS, symbols('{\dot{q}}_{2}'))
>>> > expOut = expOut.subs(q1ddotS, symbols('{\ddot{q}}_{1}'))
>>> > expOut = expOut.subs(q2ddotS, symbols('{\ddot{q}}_{2}'))
>>> > return expOut
>>> >
>>> > Here is the code that calculates higher-order derivatives. The
>>> > second-order derivative is defined by Newton's equations (ODEs).
>>> > if qi == 1:
>>> > if nDer == 0:
>>> > retVal = q1F#(t)
>>> > elif nDer == 1:
>>> > retVal = diff(qd(1, 0), t)
>>> > elif nDer == 2:
>>> > retVal = 2*qd(2, 1) + 3*qd(1, 0) - qd(1, 0)*(qd(1, 0)**2 + qd(2,
>>> > 0)**2)**Rational(-3,2)
>>> > else:
>>> > retVal = diff(qd(1, nDer-1), t)
>>> > # replace q1dd & q2dd
>>> > #retVal = retVal.subs(Derivative(q1F(t), (t, 2)), qd(1, 2))
>>> > #retVal = retVal.subs(Derivative(q2F(t), (t, 2)), qd(2, 2))
>>> > retVal = retVal.subs(Derivative(q1F, (t, 2)), qd(1, 2))
>>> > retVal = retVal.subs(Derivative(q2F, (t, 2)), qd(2, 2))
>>> >
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