Can SymPy Lambdify translate core functions like Add and Mul?

Question

My goal is to evaluate a basic symbolic equation such as ad(b + c) with my own custom implementaions of multiply and addition.

I'm trying to use lambdify to translate the two core SymPy functions (Add and Mul) with my own functions, but I cant get them recognised.

At this stage I'm just trying to get Add working. The code I have is below.

from sympy import *
import numpy as np
x, y = symbols('x y')
A = [1,1]
B = [2,2]

def addVectors(inA, inB):
    print("running addVectors")
    return np.add(inA, inB)

# Test vector addition
print(addVectors(A,B))

# Now using lambdify
f = lambdify([x, y], x + y, {"add":addVectors})
print(f(A, B)) # <------- expect [3,3] and addVectors to be run a second time

# but I get the same as this
print(A + B)

which yields

running addVectors
[3 3]
[1, 1, 2, 2]
[1, 1, 2, 2]

I was expecting the + operator in the expression to be evaluated using the custom addVectors function. Which would mean the results looks like this.

running addVectors
[3 3]
running addVectors
[3 3]
[1, 1, 2, 2]

I tried several different configurations of the lambdify line and these all give the same original result.

f = lambdify([x, y], x + y, {"add":addVectors})
f = lambdify([x, y], x + y, {"Add":addVectors})
f = lambdify([x, y], x + y, {"+":addVectors})
f = lambdify([x, y], Add(x,y), {"Add":addVectors})
f = lambdify([x, y], x + y)

To confirm I have the syntax correct I used an example closer to the documentation and replaced the symbolic cos function with a sin implementation.

from sympy import *
import numpy as np
x = symbols('x')
    
def mysin(x):
    print('taking the sin of', x)
    return np.sin(x)

print(mysin(1))

f = lambdify(x, cos(x), {'cos': mysin})
f(1)

which works as expected and yields

taking the sin of 1
0.8414709848078965
taking the sin of 1
0.8414709848078965

Is it even possible to implement my own Add and Mul functions using lambdify?

I suspect my trouble is Add (and Mul) are not SymPy 'functions'. The documentation refers to them as an 'expression' and that somehow means they dont get recognised for substitution in the lambdify process.

Some links that I've been reading: SymPy cos SymPy Add SymPy Lambdify

Any pointers would be appreciated. Thanks for reading this far.

EDIT: Got a more general case working

This uses a combination of the lambdify and replace functions to replace Add and Mul. This example then evaluates an expression in the form ad(b + c), which was the goal.

from sympy import *
import numpy as np
 
w, x, y, z = symbols('w x y z')
A = [3,3]
B = [2,2]
C = [1,1]
D = [4,4]
 
def addVectors(*args):
    result = args[0]
    for arg in args[1:]:
        result = np.add(result, arg)
    return result
 
def mulVectors(*args):
    result = args[0]
    for arg in args[1:]:
        result = np.multiply(result, arg)
    return result
 
expr = w*z*(x + y)
print(expr)
expr = expr.replace(Add, lambda *args: lerchphi(*args))
expr = expr.replace(Mul, lambda *args: Max(*args))
print(expr)
 
f = lambdify([w, x, y, z], expr, {"lerchphi":addVectors, "Max":mulVectors})
print(f(A, B, C, D))
 
print(mulVectors(A,D,addVectors(B,C)))

which yields

w*z*(x + y)
Max(w, z, lerchphi(x, y))
[36 36]
[36 36]

A few things to note with this solution:

1. Using the replace function you can replace a type with a function (type -> func). See the docs.
2. The function I replace the types with have to accept multiple inputs because each type in the expression may have more than two arguments (like multiply in the example above). I only found 3 functions that accept *args as an input. These were Min, Max and lerchphi.
3. SymPy simplifies Min and Max functions since Max(x, Min(x, y)) = x. That meant I couldn't use Min and Max together. So I used lerchphi and Max. These functions are arbitary as I'll be translating their implementation to a custom function in the next step. However, this means I can only replace two.
4. Final step was to translate lerchphi and Max to the custom functions.

Answer 1

With sympy, addition is an operation. Hence, I'm not sure if it's possible to achieve your goal by passing in custom modules...

However, at the heart of lambdify there is the printing module. Essentially, lambdify uses some printer to generate a string representation of the expression to be evaluated. If you look at lambdify's signature, you'll see that it's possible to pass a custom printer.

Given a printer class, the addition with + is performed by the _print_Add method. One way to achieve your goal is to modify this method of the NumPyPrinter.

from sympy.printing.lambdarepr import NumPyPrinter
import inspect

class MyNumPyPrinter(NumPyPrinter):
    def _print_Add(self, expr, **kwargs):
        str_args = [self.doprint(t) for t in expr.args]
        return "add(*[%s])" % ", ".join(str_args)

f = lambdify([x, y], x + y, printer=MyNumPyPrinter)
print(inspect.getsource(f))
# def _lambdifygenerated(x, y):
#    return add(*[x, y])

print(f(A, B))
# [3 3]

Note that I've no idea what implication this might creates. That's for you to find out...

Answer 2

Help for your first f is:

In [26]: help(f)
Help on function _lambdifygenerated:

_lambdifygenerated(x, y)
    Created with lambdify. Signature:
    
    func(x, y)
    
    Expression:
    
    x + y
    
    Source code:
    
    def _lambdifygenerated(x, y):
        return x + y
    

In the mysin case it produces

def _lambdifygenerated(x):
    return cos(x)

The sympy + produces an Add object

In [20]: expr = x+y
In [21]: expr
Out[21]: 
x + y
In [22]: type(expr)
Out[22]: sympy.core.add.Add

but I don't know if that information can be used in lambdify. I know from other SO, that lambdify does a rather superficial, lexical, translation of the sympy. It does not show much sign of having a 'deep' understanding of the target module (numpy).

The documenation for lambdify is quite long. I haven't read it all myself, yet I've answered a number of SO about it :(