Sympy - Rename part of an expressionSymPy: Custom print format for a custom Function expressionHow to merge two dictionaries in a single expression?Rename multiple files in a directory in PythonRenaming columns in pandasSecond derivative of the Hankel function unsing sympyExtract affine part of SymPy expressionSympy gives numerical numpy output in dtype object formatsympy autowrap (cython): limit of # of arguments, arguments in array form?sympy - symbolic sum over symbolic number of elementsReplacing part of the sympy expressionExtract the constant part of a SymPy expression

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Sympy - Rename part of an expression


SymPy: Custom print format for a custom Function expressionHow to merge two dictionaries in a single expression?Rename multiple files in a directory in PythonRenaming columns in pandasSecond derivative of the Hankel function unsing sympyExtract affine part of SymPy expressionSympy gives numerical numpy output in dtype object formatsympy autowrap (cython): limit of # of arguments, arguments in array form?sympy - symbolic sum over symbolic number of elementsReplacing part of the sympy expressionExtract the constant part of a SymPy expression






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








8















Say I have defined the following expression:



from sympy import *
N, D, i, j, d = symbols("N D i j d", integer=True)
beta, gamma = symbols(r'beta gamma')
X = IndexedBase("X", shape=(N, D))

# r(i, j) = euclidian distance between X[i] and X[j]
r = lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))
expr = r(i, j)**2 + r(i, j)


The expr variable now displays like this:





While this is fine for this minimal example, it gets quite messy in larger expressions. This really hinders my ability to see what happens later on when I compute sums over all r(i,j), derivatives etc.



My question: Is there a way to tell SymPy about r(i, j), such that it can be displayed as something like this:





while still behaving as before in subsequent expressions?



I know I could make r a Function, which would display as wanted, but then it would not work properly in a subsequent calculation (e.g. derivatives would be abstract and not evaluated).



Any help would be much appreciated!










share|improve this question
























  • I'm not too familiar with Sympy, is this helpful at all: (docs.sympy.org/latest/modules/rewriting.html)?

    – Sarcoma
    Mar 8 at 7:22











  • @Sarcoma Didn't know about that, nice! With cse it's able to auto-replace the common subexpressions, which is nice. But it's still not a full solution, as the resulting expressions "forget" their origin, and e.g. derivatives don't work as they should. I tried taking the derivative first, then cse, but then it tends to find other patterns than what I would expect it to.

    – Bendik
    Mar 8 at 7:47











  • What do you mean derivatives would be abstract and not evaluated?

    – ALFA
    Mar 8 at 11:44











  • @ALFA I mean that diff(r(i,j), X[k,l]) evaluates to zero because there is no known dependency on X within the function r (if I define r as a Function symbol). The only derivatives that make sense for SymPy is diff(r(i,j), i, 1) (or same with j), which would not be able to evaluate to anything (it's just the derivative of some function), plus not being very relevant in this case.

    – Bendik
    Mar 8 at 11:51












  • I did manage to get something working following the examples in docs.sympy.org/latest/modules/… on some built in functions. It didn't seem to work nicely with my custom class. I'll play around with it a bit more.

    – Sarcoma
    Mar 8 at 12:36

















8















Say I have defined the following expression:



from sympy import *
N, D, i, j, d = symbols("N D i j d", integer=True)
beta, gamma = symbols(r'beta gamma')
X = IndexedBase("X", shape=(N, D))

# r(i, j) = euclidian distance between X[i] and X[j]
r = lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))
expr = r(i, j)**2 + r(i, j)


The expr variable now displays like this:





While this is fine for this minimal example, it gets quite messy in larger expressions. This really hinders my ability to see what happens later on when I compute sums over all r(i,j), derivatives etc.



My question: Is there a way to tell SymPy about r(i, j), such that it can be displayed as something like this:





while still behaving as before in subsequent expressions?



I know I could make r a Function, which would display as wanted, but then it would not work properly in a subsequent calculation (e.g. derivatives would be abstract and not evaluated).



Any help would be much appreciated!










share|improve this question
























  • I'm not too familiar with Sympy, is this helpful at all: (docs.sympy.org/latest/modules/rewriting.html)?

    – Sarcoma
    Mar 8 at 7:22











  • @Sarcoma Didn't know about that, nice! With cse it's able to auto-replace the common subexpressions, which is nice. But it's still not a full solution, as the resulting expressions "forget" their origin, and e.g. derivatives don't work as they should. I tried taking the derivative first, then cse, but then it tends to find other patterns than what I would expect it to.

    – Bendik
    Mar 8 at 7:47











  • What do you mean derivatives would be abstract and not evaluated?

    – ALFA
    Mar 8 at 11:44











  • @ALFA I mean that diff(r(i,j), X[k,l]) evaluates to zero because there is no known dependency on X within the function r (if I define r as a Function symbol). The only derivatives that make sense for SymPy is diff(r(i,j), i, 1) (or same with j), which would not be able to evaluate to anything (it's just the derivative of some function), plus not being very relevant in this case.

    – Bendik
    Mar 8 at 11:51












  • I did manage to get something working following the examples in docs.sympy.org/latest/modules/… on some built in functions. It didn't seem to work nicely with my custom class. I'll play around with it a bit more.

    – Sarcoma
    Mar 8 at 12:36













8












8








8


4






Say I have defined the following expression:



from sympy import *
N, D, i, j, d = symbols("N D i j d", integer=True)
beta, gamma = symbols(r'beta gamma')
X = IndexedBase("X", shape=(N, D))

# r(i, j) = euclidian distance between X[i] and X[j]
r = lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))
expr = r(i, j)**2 + r(i, j)


The expr variable now displays like this:





While this is fine for this minimal example, it gets quite messy in larger expressions. This really hinders my ability to see what happens later on when I compute sums over all r(i,j), derivatives etc.



My question: Is there a way to tell SymPy about r(i, j), such that it can be displayed as something like this:





while still behaving as before in subsequent expressions?



I know I could make r a Function, which would display as wanted, but then it would not work properly in a subsequent calculation (e.g. derivatives would be abstract and not evaluated).



Any help would be much appreciated!










share|improve this question
















Say I have defined the following expression:



from sympy import *
N, D, i, j, d = symbols("N D i j d", integer=True)
beta, gamma = symbols(r'beta gamma')
X = IndexedBase("X", shape=(N, D))

# r(i, j) = euclidian distance between X[i] and X[j]
r = lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))
expr = r(i, j)**2 + r(i, j)


The expr variable now displays like this:





While this is fine for this minimal example, it gets quite messy in larger expressions. This really hinders my ability to see what happens later on when I compute sums over all r(i,j), derivatives etc.



My question: Is there a way to tell SymPy about r(i, j), such that it can be displayed as something like this:





while still behaving as before in subsequent expressions?



I know I could make r a Function, which would display as wanted, but then it would not work properly in a subsequent calculation (e.g. derivatives would be abstract and not evaluated).



Any help would be much appreciated!







python sympy






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Feb 28 at 11:29







Bendik

















asked Feb 28 at 9:21









BendikBendik

3811315




3811315












  • I'm not too familiar with Sympy, is this helpful at all: (docs.sympy.org/latest/modules/rewriting.html)?

    – Sarcoma
    Mar 8 at 7:22











  • @Sarcoma Didn't know about that, nice! With cse it's able to auto-replace the common subexpressions, which is nice. But it's still not a full solution, as the resulting expressions "forget" their origin, and e.g. derivatives don't work as they should. I tried taking the derivative first, then cse, but then it tends to find other patterns than what I would expect it to.

    – Bendik
    Mar 8 at 7:47











  • What do you mean derivatives would be abstract and not evaluated?

    – ALFA
    Mar 8 at 11:44











  • @ALFA I mean that diff(r(i,j), X[k,l]) evaluates to zero because there is no known dependency on X within the function r (if I define r as a Function symbol). The only derivatives that make sense for SymPy is diff(r(i,j), i, 1) (or same with j), which would not be able to evaluate to anything (it's just the derivative of some function), plus not being very relevant in this case.

    – Bendik
    Mar 8 at 11:51












  • I did manage to get something working following the examples in docs.sympy.org/latest/modules/… on some built in functions. It didn't seem to work nicely with my custom class. I'll play around with it a bit more.

    – Sarcoma
    Mar 8 at 12:36

















  • I'm not too familiar with Sympy, is this helpful at all: (docs.sympy.org/latest/modules/rewriting.html)?

    – Sarcoma
    Mar 8 at 7:22











  • @Sarcoma Didn't know about that, nice! With cse it's able to auto-replace the common subexpressions, which is nice. But it's still not a full solution, as the resulting expressions "forget" their origin, and e.g. derivatives don't work as they should. I tried taking the derivative first, then cse, but then it tends to find other patterns than what I would expect it to.

    – Bendik
    Mar 8 at 7:47











  • What do you mean derivatives would be abstract and not evaluated?

    – ALFA
    Mar 8 at 11:44











  • @ALFA I mean that diff(r(i,j), X[k,l]) evaluates to zero because there is no known dependency on X within the function r (if I define r as a Function symbol). The only derivatives that make sense for SymPy is diff(r(i,j), i, 1) (or same with j), which would not be able to evaluate to anything (it's just the derivative of some function), plus not being very relevant in this case.

    – Bendik
    Mar 8 at 11:51












  • I did manage to get something working following the examples in docs.sympy.org/latest/modules/… on some built in functions. It didn't seem to work nicely with my custom class. I'll play around with it a bit more.

    – Sarcoma
    Mar 8 at 12:36
















I'm not too familiar with Sympy, is this helpful at all: (docs.sympy.org/latest/modules/rewriting.html)?

– Sarcoma
Mar 8 at 7:22





I'm not too familiar with Sympy, is this helpful at all: (docs.sympy.org/latest/modules/rewriting.html)?

– Sarcoma
Mar 8 at 7:22













@Sarcoma Didn't know about that, nice! With cse it's able to auto-replace the common subexpressions, which is nice. But it's still not a full solution, as the resulting expressions "forget" their origin, and e.g. derivatives don't work as they should. I tried taking the derivative first, then cse, but then it tends to find other patterns than what I would expect it to.

– Bendik
Mar 8 at 7:47





@Sarcoma Didn't know about that, nice! With cse it's able to auto-replace the common subexpressions, which is nice. But it's still not a full solution, as the resulting expressions "forget" their origin, and e.g. derivatives don't work as they should. I tried taking the derivative first, then cse, but then it tends to find other patterns than what I would expect it to.

– Bendik
Mar 8 at 7:47













What do you mean derivatives would be abstract and not evaluated?

– ALFA
Mar 8 at 11:44





What do you mean derivatives would be abstract and not evaluated?

– ALFA
Mar 8 at 11:44













@ALFA I mean that diff(r(i,j), X[k,l]) evaluates to zero because there is no known dependency on X within the function r (if I define r as a Function symbol). The only derivatives that make sense for SymPy is diff(r(i,j), i, 1) (or same with j), which would not be able to evaluate to anything (it's just the derivative of some function), plus not being very relevant in this case.

– Bendik
Mar 8 at 11:51






@ALFA I mean that diff(r(i,j), X[k,l]) evaluates to zero because there is no known dependency on X within the function r (if I define r as a Function symbol). The only derivatives that make sense for SymPy is diff(r(i,j), i, 1) (or same with j), which would not be able to evaluate to anything (it's just the derivative of some function), plus not being very relevant in this case.

– Bendik
Mar 8 at 11:51














I did manage to get something working following the examples in docs.sympy.org/latest/modules/… on some built in functions. It didn't seem to work nicely with my custom class. I'll play around with it a bit more.

– Sarcoma
Mar 8 at 12:36





I did manage to get something working following the examples in docs.sympy.org/latest/modules/… on some built in functions. It didn't seem to work nicely with my custom class. I'll play around with it a bit more.

– Sarcoma
Mar 8 at 12:36












2 Answers
2






active

oldest

votes


















2





+100









You can make a custom Function subclass that doesn't evaluate by default:



class r(Function):
@classmethod
def eval(cls, i, j):
return

def doit(self, **kwargs):
i, j = self.args
return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


eval tells it when to evaluate. Since it always returns None, it never evaluates. It also tells SymPy the function has two arguments. You can also have it return explicit values in some cases, if you like. For instance, you might want it to evaluate if i and j are explicit numbers.



@classmethod
def eval(cls, i, j):
if i.is_Number and j.is_Number:
return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


With this you can use it as desired, and call expr.doit() when you want it to evaluate. You can also specifically define evaluation for certain functions to avoid doit. For example, derivatives:



def _eval_derivative(self, x):
return self.doit()._eval_derivative(x)


This will make r(i, j).diff(i) evaluate immediately without having to call doit.



Other functions have similar methods you can define. See the SymPy documentation.






share|improve this answer






























    2














    I don't really know if it may helps you but what about this:



    from sympy import *
    from sympy.utilities.lambdify import lambdify, implemented_function
    N, D, i, j, d = symbols("N D i j d", integer=True)
    beta, gamma = symbols(r'beta gamma')
    X = IndexedBase("X", shape=(N, D))

    r = implemented_function('r', lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D))));
    expr = r(i, j)**2 + r(i, j)
    print(expr)
    r = lambdify((i,j), r(i,j))
    print(diff(r(i,j), X[i,j]))


    You can display your expression as you wish, then use lambdify() and makes it behave as it should. Just guessing, maybe it's useless for you as you probably prefer a way to maintain the same expression all along the code.






    share|improve this answer























    • I'd come across this, looked like it ought to do the trick. Couldn't get it working though. I saw this in the docs for lambdify: "Be aware that this is a quick workaround, not a general method to create special symbolic functions. If you want to create a symbolic function to be used by all the machinery of SymPy you should subclass the Function class." which led me to think that the more verbose class r(Function): might also work.

      – Sarcoma
      Mar 8 at 12:21











    • That's pretty close actually. It's slightly awkward to have to change between the two all the time, but still. With r_func = lambdify((i,j), r(i,j)) defined and doing diff(expr.subs(r(i,j), r_func(i,j)), X[k,l]).subs(r_func(i,j), r(i,j)) it works the way I want it to. Wrapping this whole thing in cse makes it even simpler.

      – Bendik
      Mar 8 at 12:24






    • 1





      @Sarcoma But perhaps the Function subclass is the way to go to make this "smoother", although more work upfront... I'm gonna tinker with it a bit

      – Bendik
      Mar 8 at 12:26






    • 1





      Yes I would implement this with Function subclass. I think this is the closest option to your request, doing further researches on the web l could’t find a better way

      – ALFA
      Mar 8 at 13:59











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2





    +100









    You can make a custom Function subclass that doesn't evaluate by default:



    class r(Function):
    @classmethod
    def eval(cls, i, j):
    return

    def doit(self, **kwargs):
    i, j = self.args
    return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


    eval tells it when to evaluate. Since it always returns None, it never evaluates. It also tells SymPy the function has two arguments. You can also have it return explicit values in some cases, if you like. For instance, you might want it to evaluate if i and j are explicit numbers.



    @classmethod
    def eval(cls, i, j):
    if i.is_Number and j.is_Number:
    return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


    With this you can use it as desired, and call expr.doit() when you want it to evaluate. You can also specifically define evaluation for certain functions to avoid doit. For example, derivatives:



    def _eval_derivative(self, x):
    return self.doit()._eval_derivative(x)


    This will make r(i, j).diff(i) evaluate immediately without having to call doit.



    Other functions have similar methods you can define. See the SymPy documentation.






    share|improve this answer



























      2





      +100









      You can make a custom Function subclass that doesn't evaluate by default:



      class r(Function):
      @classmethod
      def eval(cls, i, j):
      return

      def doit(self, **kwargs):
      i, j = self.args
      return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


      eval tells it when to evaluate. Since it always returns None, it never evaluates. It also tells SymPy the function has two arguments. You can also have it return explicit values in some cases, if you like. For instance, you might want it to evaluate if i and j are explicit numbers.



      @classmethod
      def eval(cls, i, j):
      if i.is_Number and j.is_Number:
      return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


      With this you can use it as desired, and call expr.doit() when you want it to evaluate. You can also specifically define evaluation for certain functions to avoid doit. For example, derivatives:



      def _eval_derivative(self, x):
      return self.doit()._eval_derivative(x)


      This will make r(i, j).diff(i) evaluate immediately without having to call doit.



      Other functions have similar methods you can define. See the SymPy documentation.






      share|improve this answer

























        2





        +100







        2





        +100



        2




        +100





        You can make a custom Function subclass that doesn't evaluate by default:



        class r(Function):
        @classmethod
        def eval(cls, i, j):
        return

        def doit(self, **kwargs):
        i, j = self.args
        return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


        eval tells it when to evaluate. Since it always returns None, it never evaluates. It also tells SymPy the function has two arguments. You can also have it return explicit values in some cases, if you like. For instance, you might want it to evaluate if i and j are explicit numbers.



        @classmethod
        def eval(cls, i, j):
        if i.is_Number and j.is_Number:
        return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


        With this you can use it as desired, and call expr.doit() when you want it to evaluate. You can also specifically define evaluation for certain functions to avoid doit. For example, derivatives:



        def _eval_derivative(self, x):
        return self.doit()._eval_derivative(x)


        This will make r(i, j).diff(i) evaluate immediately without having to call doit.



        Other functions have similar methods you can define. See the SymPy documentation.






        share|improve this answer













        You can make a custom Function subclass that doesn't evaluate by default:



        class r(Function):
        @classmethod
        def eval(cls, i, j):
        return

        def doit(self, **kwargs):
        i, j = self.args
        return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


        eval tells it when to evaluate. Since it always returns None, it never evaluates. It also tells SymPy the function has two arguments. You can also have it return explicit values in some cases, if you like. For instance, you might want it to evaluate if i and j are explicit numbers.



        @classmethod
        def eval(cls, i, j):
        if i.is_Number and j.is_Number:
        return sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D)))


        With this you can use it as desired, and call expr.doit() when you want it to evaluate. You can also specifically define evaluation for certain functions to avoid doit. For example, derivatives:



        def _eval_derivative(self, x):
        return self.doit()._eval_derivative(x)


        This will make r(i, j).diff(i) evaluate immediately without having to call doit.



        Other functions have similar methods you can define. See the SymPy documentation.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 8 at 22:44









        asmeurerasmeurer

        59.2k17114188




        59.2k17114188























            2














            I don't really know if it may helps you but what about this:



            from sympy import *
            from sympy.utilities.lambdify import lambdify, implemented_function
            N, D, i, j, d = symbols("N D i j d", integer=True)
            beta, gamma = symbols(r'beta gamma')
            X = IndexedBase("X", shape=(N, D))

            r = implemented_function('r', lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D))));
            expr = r(i, j)**2 + r(i, j)
            print(expr)
            r = lambdify((i,j), r(i,j))
            print(diff(r(i,j), X[i,j]))


            You can display your expression as you wish, then use lambdify() and makes it behave as it should. Just guessing, maybe it's useless for you as you probably prefer a way to maintain the same expression all along the code.






            share|improve this answer























            • I'd come across this, looked like it ought to do the trick. Couldn't get it working though. I saw this in the docs for lambdify: "Be aware that this is a quick workaround, not a general method to create special symbolic functions. If you want to create a symbolic function to be used by all the machinery of SymPy you should subclass the Function class." which led me to think that the more verbose class r(Function): might also work.

              – Sarcoma
              Mar 8 at 12:21











            • That's pretty close actually. It's slightly awkward to have to change between the two all the time, but still. With r_func = lambdify((i,j), r(i,j)) defined and doing diff(expr.subs(r(i,j), r_func(i,j)), X[k,l]).subs(r_func(i,j), r(i,j)) it works the way I want it to. Wrapping this whole thing in cse makes it even simpler.

              – Bendik
              Mar 8 at 12:24






            • 1





              @Sarcoma But perhaps the Function subclass is the way to go to make this "smoother", although more work upfront... I'm gonna tinker with it a bit

              – Bendik
              Mar 8 at 12:26






            • 1





              Yes I would implement this with Function subclass. I think this is the closest option to your request, doing further researches on the web l could’t find a better way

              – ALFA
              Mar 8 at 13:59















            2














            I don't really know if it may helps you but what about this:



            from sympy import *
            from sympy.utilities.lambdify import lambdify, implemented_function
            N, D, i, j, d = symbols("N D i j d", integer=True)
            beta, gamma = symbols(r'beta gamma')
            X = IndexedBase("X", shape=(N, D))

            r = implemented_function('r', lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D))));
            expr = r(i, j)**2 + r(i, j)
            print(expr)
            r = lambdify((i,j), r(i,j))
            print(diff(r(i,j), X[i,j]))


            You can display your expression as you wish, then use lambdify() and makes it behave as it should. Just guessing, maybe it's useless for you as you probably prefer a way to maintain the same expression all along the code.






            share|improve this answer























            • I'd come across this, looked like it ought to do the trick. Couldn't get it working though. I saw this in the docs for lambdify: "Be aware that this is a quick workaround, not a general method to create special symbolic functions. If you want to create a symbolic function to be used by all the machinery of SymPy you should subclass the Function class." which led me to think that the more verbose class r(Function): might also work.

              – Sarcoma
              Mar 8 at 12:21











            • That's pretty close actually. It's slightly awkward to have to change between the two all the time, but still. With r_func = lambdify((i,j), r(i,j)) defined and doing diff(expr.subs(r(i,j), r_func(i,j)), X[k,l]).subs(r_func(i,j), r(i,j)) it works the way I want it to. Wrapping this whole thing in cse makes it even simpler.

              – Bendik
              Mar 8 at 12:24






            • 1





              @Sarcoma But perhaps the Function subclass is the way to go to make this "smoother", although more work upfront... I'm gonna tinker with it a bit

              – Bendik
              Mar 8 at 12:26






            • 1





              Yes I would implement this with Function subclass. I think this is the closest option to your request, doing further researches on the web l could’t find a better way

              – ALFA
              Mar 8 at 13:59













            2












            2








            2







            I don't really know if it may helps you but what about this:



            from sympy import *
            from sympy.utilities.lambdify import lambdify, implemented_function
            N, D, i, j, d = symbols("N D i j d", integer=True)
            beta, gamma = symbols(r'beta gamma')
            X = IndexedBase("X", shape=(N, D))

            r = implemented_function('r', lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D))));
            expr = r(i, j)**2 + r(i, j)
            print(expr)
            r = lambdify((i,j), r(i,j))
            print(diff(r(i,j), X[i,j]))


            You can display your expression as you wish, then use lambdify() and makes it behave as it should. Just guessing, maybe it's useless for you as you probably prefer a way to maintain the same expression all along the code.






            share|improve this answer













            I don't really know if it may helps you but what about this:



            from sympy import *
            from sympy.utilities.lambdify import lambdify, implemented_function
            N, D, i, j, d = symbols("N D i j d", integer=True)
            beta, gamma = symbols(r'beta gamma')
            X = IndexedBase("X", shape=(N, D))

            r = implemented_function('r', lambda i, j: sqrt(Sum((X[i, d] - X[j, d])**2, (d, 1, D))));
            expr = r(i, j)**2 + r(i, j)
            print(expr)
            r = lambdify((i,j), r(i,j))
            print(diff(r(i,j), X[i,j]))


            You can display your expression as you wish, then use lambdify() and makes it behave as it should. Just guessing, maybe it's useless for you as you probably prefer a way to maintain the same expression all along the code.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 8 at 12:06









            ALFAALFA

            1,058516




            1,058516












            • I'd come across this, looked like it ought to do the trick. Couldn't get it working though. I saw this in the docs for lambdify: "Be aware that this is a quick workaround, not a general method to create special symbolic functions. If you want to create a symbolic function to be used by all the machinery of SymPy you should subclass the Function class." which led me to think that the more verbose class r(Function): might also work.

              – Sarcoma
              Mar 8 at 12:21











            • That's pretty close actually. It's slightly awkward to have to change between the two all the time, but still. With r_func = lambdify((i,j), r(i,j)) defined and doing diff(expr.subs(r(i,j), r_func(i,j)), X[k,l]).subs(r_func(i,j), r(i,j)) it works the way I want it to. Wrapping this whole thing in cse makes it even simpler.

              – Bendik
              Mar 8 at 12:24






            • 1





              @Sarcoma But perhaps the Function subclass is the way to go to make this "smoother", although more work upfront... I'm gonna tinker with it a bit

              – Bendik
              Mar 8 at 12:26






            • 1





              Yes I would implement this with Function subclass. I think this is the closest option to your request, doing further researches on the web l could’t find a better way

              – ALFA
              Mar 8 at 13:59

















            • I'd come across this, looked like it ought to do the trick. Couldn't get it working though. I saw this in the docs for lambdify: "Be aware that this is a quick workaround, not a general method to create special symbolic functions. If you want to create a symbolic function to be used by all the machinery of SymPy you should subclass the Function class." which led me to think that the more verbose class r(Function): might also work.

              – Sarcoma
              Mar 8 at 12:21











            • That's pretty close actually. It's slightly awkward to have to change between the two all the time, but still. With r_func = lambdify((i,j), r(i,j)) defined and doing diff(expr.subs(r(i,j), r_func(i,j)), X[k,l]).subs(r_func(i,j), r(i,j)) it works the way I want it to. Wrapping this whole thing in cse makes it even simpler.

              – Bendik
              Mar 8 at 12:24






            • 1





              @Sarcoma But perhaps the Function subclass is the way to go to make this "smoother", although more work upfront... I'm gonna tinker with it a bit

              – Bendik
              Mar 8 at 12:26






            • 1





              Yes I would implement this with Function subclass. I think this is the closest option to your request, doing further researches on the web l could’t find a better way

              – ALFA
              Mar 8 at 13:59
















            I'd come across this, looked like it ought to do the trick. Couldn't get it working though. I saw this in the docs for lambdify: "Be aware that this is a quick workaround, not a general method to create special symbolic functions. If you want to create a symbolic function to be used by all the machinery of SymPy you should subclass the Function class." which led me to think that the more verbose class r(Function): might also work.

            – Sarcoma
            Mar 8 at 12:21





            I'd come across this, looked like it ought to do the trick. Couldn't get it working though. I saw this in the docs for lambdify: "Be aware that this is a quick workaround, not a general method to create special symbolic functions. If you want to create a symbolic function to be used by all the machinery of SymPy you should subclass the Function class." which led me to think that the more verbose class r(Function): might also work.

            – Sarcoma
            Mar 8 at 12:21













            That's pretty close actually. It's slightly awkward to have to change between the two all the time, but still. With r_func = lambdify((i,j), r(i,j)) defined and doing diff(expr.subs(r(i,j), r_func(i,j)), X[k,l]).subs(r_func(i,j), r(i,j)) it works the way I want it to. Wrapping this whole thing in cse makes it even simpler.

            – Bendik
            Mar 8 at 12:24





            That's pretty close actually. It's slightly awkward to have to change between the two all the time, but still. With r_func = lambdify((i,j), r(i,j)) defined and doing diff(expr.subs(r(i,j), r_func(i,j)), X[k,l]).subs(r_func(i,j), r(i,j)) it works the way I want it to. Wrapping this whole thing in cse makes it even simpler.

            – Bendik
            Mar 8 at 12:24




            1




            1





            @Sarcoma But perhaps the Function subclass is the way to go to make this "smoother", although more work upfront... I'm gonna tinker with it a bit

            – Bendik
            Mar 8 at 12:26





            @Sarcoma But perhaps the Function subclass is the way to go to make this "smoother", although more work upfront... I'm gonna tinker with it a bit

            – Bendik
            Mar 8 at 12:26




            1




            1





            Yes I would implement this with Function subclass. I think this is the closest option to your request, doing further researches on the web l could’t find a better way

            – ALFA
            Mar 8 at 13:59





            Yes I would implement this with Function subclass. I think this is the closest option to your request, doing further researches on the web l could’t find a better way

            – ALFA
            Mar 8 at 13:59

















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