Is there a way to replace a variable with a value in a sympy function without automatically evaluating it?

Question

I'm trying to write a simple script to do some math, and print out some specific intermediate steps in LaTeX. I've found there's no real simple way to do this using sympy, so I've started manually writing functions to print out each step I want to see.

I need to take a sympy function and format it so that every variable is replaced by it's associated value, I've been accessing the values through a dictionary.

basically,

import sympy as sym
x,a = sym.symbols("x a")
var_Values = {'x':3, 'a':2}
f=(x+1)/a

print(some_Function(f, var_Values))

so that the print statement reads \frac{3+1}{2}.

I've already tried two methods, using f.subs() to replace the variable with the value, which prints out 2 in this case, since it evaluates the expression's value.

I've also tried this textual method:

def some_Function(f, var_Values):
    expression = sym.latex(f)
    for variable in f.free_symbols:
        expression = expression.replace(variable.name, var_Values.get(variable.name)
    return expression

which is even worse, as it returns \fr2c{3+1}{2}, which turns more than what I wanted into numbers. It could be possible to get around this by using variable names that don't appear in the LaTeX commands I"m using, but that approach is impossible for me, as I don't choose my variable names.

See also [Force SymPy to keep the order of terms](https://stackoverflow.com/questions/36314035/force-sympy-to-keep-the-order-of-terms) — JohanC, Dec 01 '21 at 17:04

Chris du Plessis · Answer 1 · 2021-11-30T19:31:25.210

SymPy is not great when it comes to leaving an expression unchanged because it inherently tries to simplify anything to make any future computations faster. subs and replace try to simplify the expression afterwards.

Here is the best I can think of:

import sympy as sym

x, a = sym.symbols("x a")
var_values = {x: 3, a: 2}
f = (x + 1) / a


def some_function(func: sym.Expr, var_dict: dict) -> sym.Expr:
    new_dict = {key: sym.Symbol(str(val)) for key, val in var_dict.items()}
    result = func.subs(new_dict)
    return result


print(some_function(f, var_values))

This produces (3 + 1)/2. It should work for most cases. Sadly, this does not work in general as SymPy since with addition, it will sort the terms on its own. That means (x + y + 1) / a; var_values = {x: 3, y: 1, a: 2} produces (1 + 3 + 1)/2 which is not right. I have opened up an issue about this.

The reason why the second method does not produce valid latex is because expression is a string and has nothing to do with SymPy variables. Running in debug mode or in interactive mode, you'll see that it is "x + \frac{1}{a}". When you replace "a" for 2, "\frac" becomes "\fr2c". It is best to keep in mind what variable type each object is and to use SymPy Symbols instead of strings when replacing variables in an expression.

I saw my function as (x+1)/y, but seing it as x+(1/y) doesn't change anything to the problem, I'll make it clearer. — Therk, Nov 30 '21 at 18:07
Would something other than sympy be better to do this? I don — Therk, Nov 30 '21 at 18:29
@Therk I'm sure there's a way to do it in SymPy that works in general while keeping the division sign instead of `**(-1)`. I just can't think of a way to do it. — Chris du Plessis, Nov 30 '21 at 19:48

score 0 · Answer 2 · answered Dec 01 '21 at 13:00

0

What about

>>> with evaluate(False):
...     u = f.subs(val_Values)

That gives the (1 + 3)/2 unevaluated result. If you want the order you wrote terms to be respected then you have to do 2 things: create the expression and do the replacement in an unevaluated context and 2) use a printer that doesn't re-order the args:

>>> with evaluate(False):
...     a=(1 + x).subs(var_Values)
...     b=(x + 1).subs(var_Values)
>>> a,b
(1 + 3, 1 + 3)
>>> p=StrPrinter(dict(order='none'))
>>> p.doprint(a),p.doprint(b)
(1 + 3, 3 + 1)

answered Dec 01 '21 at 13:00

smichr

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this is getting pretty close, however this method seems to only work with additions and subtractions, using multiplications or divisions gives back a long error mesage about how absolute values can't be ints – Therk Dec 06 '21 at 05:24
Can you give an example? And note that you will never be able to put a literal 0 in an Add or a 1 in a Mul unless it is faked as a `Symbol('0')` or `Symbol('1')` or something like `Mul(Pow(1,Symbol(""),evaluate=0),x,evaluate=0)`. – smichr Dec 06 '21 at 08:03

Is there a way to replace a variable with a value in a sympy function without automatically evaluating it?

2 Answers2