How can we separate equation smaller separated equation in sympy - python

Hello I am new to Python and Sympy. i have created small project That working like sympy gamma but i want to separate my equation in to smaller sub parts.is it possible using sympy function.
I am Attaching one image for more information about my Problem.
It is my views.py file
def index(request):
if request.method == "POST":
x = symbols('x')
init_printing(use_unicode=True)
# transformations = (standard_transformations + (implicit_multiplication_application,))
transformations = standard_transformations + (implicit_multiplication_application,) + (convert_xor,) #covert_xor used
equ = latex(request.POST['equ'])
eq = parse_expr(request.POST['equ'], transformations=transformations) #request.POST['equ'].replace("^", "**")
sympifyy = latex(sympify(eq, evaluate=True))
sympifyy1 = sympify(eq)
sympifyy2 = latex(simplify(eq))
derivative = latex(sp.diff(eq,x))
integration = latex(sp.integrate(eq, x))
# integration = integrate(eq, x)
# pretty(latex(Integral(eq, x)),use_unicode=False)
# print(pretty(Integral(sqrt(1/x), x), use_unicode=False))
rootss = solve(eq)
limits = limit(eq, x, 0)
seriess = latex(series(eq, x, 0, 10))
data = {
'Sympify' : sympifyy,
# 'Equation formate' : aa formate |exsin(x) + excos(x) dx mate use karva nu in future
'Equation formate' :equ,
'Sympify1' : sympifyy1,
'Sympify2' : sympifyy2,
'Derivative' : derivative ,
'Integration' : integration, # "\int "+ latex(eq)+": "+
'Roots' : rootss,
'Limit' : limits,
'Series' : seriess
}
return render(request, 'index.html', {'data':data})
return render(request, 'index.html')
This is my Ouyput:What to do? Any suggestions please.

If you want to split a sum into terms with/without a given symbol you can do
>>> i, d = (x + x*y + 2).as_independent(x, as_Add=True); (i, d)
(2, x*y + x)
To split an expression into terms:
>>> Add.make_args(d)
(x, x*y)
To split a product into factors with/without a given symbol you can do
>>> (2*x*y).as_independent(x)
(2*y, x)
See also "See Also" references in the docstrings of these methods.

Related

Cube roots of a complex number in python

How do I calculate the cube roots of a complex number in python? Currently, the only way I've found to do it is according to this answer, but I'm trying to remove any and all import statements from my code.
Current method, reproduced here:
import math
def cuberoot( z ):
z = complex(z)
x = z.real
y = z.imag
mag = abs(z)
arg = math.atan2(y,x)
resMag = mag**(1./3)
resArg = [ (arg+2*math.pi*n)/3. for n in range(1,4) ]
return [ resMag*(math.cos(a) + math.sin(a)*1j) for a in resArg ]
Compute one and multiply with the cube roots of 1?
cuberoots_of_1 = 1, complex(-.5, .75**.5), complex(-.5, -.75**.5)
def cuberoot(z):
cuberoot = complex(z)**(1/3)
return [cuberoot * cr1 for cr1 in cuberoots_of_1]
Test:
>>> cuberoot(-2 + 3j)
[(1.1532283040274223+1.0106429470939737j),
(-1.4518566183526649+0.49340353410400484j),
(0.2986283143252425-1.5040464811979786j)]
>>> for z in cuberoot(-2 + 3j):
print(z**3, abs(z**3 - (-2 + 3j)))
(-1.999999999999999+3j) 1.1102230246251565e-15
(-1.999999999999999+3j) 1.1102230246251565e-15
(-1.9999999999999982+2.9999999999999996j) 1.831026719408895e-15
Doing the same with yours is less accurate:
(-1.999999999999996+3.000000000000002j) 4.572178254219406e-15
(-1.9999999999999933+3.000000000000004j) 7.768388458966724e-15
(-1.9999999999999956+3.0000000000000013j) 4.636427468134552e-15
Here is an approach using rotation by a factor of i^(2/3).
f = 1j**(2/3)
def cube_roots(z):
r = z**(1/3)
return [r, -r*f, r*f**2]
Python's built-in complex can handle finding one root out of the box:
def cube_root(v):
if not isinstance(v, complex):
v = complex(v, 0)
return v ** (1.0 / 3.0)
Examples:
cube_root(-3)
(0.7211247851537043+1.2490247664834064j)
cube_root(complex(1, -2))
(1.2196165079717578-0.47171126778938893j)
The function you reproduced above is one way to get all three roots.

Differential equation change of variables with sympy

I have an ordinary differential equation like this:
DiffEq = Eq(-ℏ*ℏ*diff(Ψ,x,2)/(2*m) + m*w*w*(x*x)*Ψ/2 - E*Ψ , 0)
I want to perform a variable change :
sp.Eq(u , x*sqrt(m*w/ℏ))
sp.Eq(Ψ, H*exp(-u*u/2))
How can I do this with sympy?
Use the following function:
def variable_change(ODE,dependent_var,
independent_var,
new_dependent_var = None,
new_independent_var= None,
dependent_var_relation = None,
independent_var_relation = None,
order = 2):
if new_dependent_var == None:
new_dependent_var = dependent_var
if new_independent_var == None:
new_independent_var = independent_var
# dependent variable change
if new_independent_var != independent_var:
for i in range(order, -1, -1):
# remplace derivate
a = D(dependent_var , independent_var, i )
ξ = Function("ξ")(independent_var)
b = D( dependent_var.subs(independent_var, ξ), independent_var ,i)
rel = solve(independent_var_relation, new_independent_var)[0]
for j in range(order, 0, -1):
b = b.subs( D(ξ,independent_var,j), D(rel,independent_var,j))
b = b.subs(ξ, new_independent_var)
rel = solve(independent_var_relation, independent_var)[0]
b = b.subs(independent_var, rel)
ODE = ODE.subs(a,b)
ODE = ODE.subs(independent_var, rel)
# change of variables of indpendent variable
if new_dependent_var != dependent_var:
ODE = (ODE.subs(dependent_var.subs(independent_var,new_independent_var) , (solve(dependent_var_relation, dependent_var)[0])))
ODE = ODE.doit().expand()
return ODE.simplify()
For the example posted:
from sympy import *
from sympy import diff as D
E, ℏ ,w,m,x,u = symbols("E, ℏ , w,m,x,u")
Ψ ,H = map(Function, ["Ψ ","H"])
Ψ ,H = Ψ(x), H(u)
DiffEq = Eq(-ℏ*ℏ*D(Ψ,x,2)/(2*m) + m*w*w*(x*x)*Ψ/2 - E*Ψ,0)
display(DiffEq)
display(Eq(u , x*sqrt(m*w/ℏ)))
display(Eq(Ψ, H*exp(-u*u/2)))
newODE = variable_change(ODE = DiffEq,
independent_var = x,
new_independent_var= u,
independent_var_relation = Eq(u , x*sqrt(m*w/ℏ)),
dependent_var = Ψ,
new_dependent_var = H,
dependent_var_relation = Eq(Ψ, H*exp(-u*u/2)),
order = 2)
display(newODE)
Under this substitution the differential equation outputted is then:
Eq((-E*H + u*w*ℏ*D(H, u) + w*ℏ*H/2 - w*ℏ*D(H, (u, 2))/2)*exp(-u**2/2), 0)
If anyone is wondering how they could do it as well on CoCalc notebooks/anywhere where you can mix Sage and Python, here I defined basically the same variables and functions as OP did on his accepted answer, and then after substitution the result is converted back to Sage:
# Sage objects
var("E w m x u")
var("h_bar", latex_name = r'\hbar')
Ψ = function("Ψ")(x)
H = function('H')(u)
DiffEq = (-h_bar*h_bar*Ψ.diff(x, 2)/(2*m) + m*w*w*(x*x)*Ψ/2 - E*Ψ == 0)
display(DiffEq)
display(u == x*sqrt(m*w/h_bar))
display(Ψ == H*exp(-u*u/2))
# Function is purely sympy
newODE = variable_change(
ODE = DiffEq._sympy_(),
independent_var = x._sympy_(),
new_independent_var = u._sympy_(),
independent_var_relation = (u == x*sqrt(m*w/h_bar))._sympy_(),
dependent_var = Ψ._sympy_(),
new_dependent_var = H._sympy_(),
dependent_var_relation = (Ψ == H*exp(-u*u/2))._sympy_(),
order = 2
)
display(newODE._sage_())
Note that the only difference is that here things are converted to SymPy when using as arguments inside OP's function (it'll probably break if you don't!). After you call _sympy_() only once on a variable or expression, every sympy object gets a _sage_() method to convert back.
The result given was:
# Sage object again
1/2*(2*h_bar*u*w*diff(H(u), u) + h_bar*w*H(u) - h_bar*w*diff(H(u), u, u) - 2*E*H(u))*e^(-1/2*u^2) == 0
Which is just OP's result, but Sage handles operands a little bit differently.
Note: in order to avoid overriding stuff on Sage after importing everything from SymPy, you may want to import only diff as D, Function and solve from the main library. You might also want to rename sympy's solve to something else to avoid overriding Sage's own sage.symbolic.relation.solve.

Taylor series sympy expression of a python function

I have a very complicated non-linear function f. I want to get taylor series till degree n in a form of sympy expression for the function f at value x.
f is a regular python function not a sympy expression. Output of get_polynomial should be a sympy expression.
Is there any function that will get taylor-series of a function?
from math import sin, cos, log, e
def f(x):
# a very complicated function
y = sin(x) + cos(x) + log(abs(x)+2)**2/e**2 + sin(cos(x/2)**2) + 1
return y
def get_polynomial(function, x, degree):
# .......
# using Taylor Series
# .......
return sympy_expression_for_function_at_value_x
Output:
get_polynomial(sin, 0, 3) ---> 0 + x + 0*x**2 + (1/6)*x**3
get_polynomial(lambda x: e**x, 0, 1) --> 1 + x
In a similar manner I wanna calculate get_polynomial(f, 0, 3)
The following code is close to what you're looking for. What this does it to parse the code the of the function you wish you expand into a Taylor series, convert it into a symbolic representation using Sympy and then compute the Taylor expansion.
One limitation is that you need to have an explicit function definition so you can't use lambda expressions. This can be solved with further work. Otherwise the code does what you ask for. Note that when you define a function, it has to contain a line of the form y = ... for this code to work
from inspect import *
import sympy
def f(x):
# a very complicated function
y = sin(x) + cos(x) + log(abs(x)+2)**2/e**2 + sin(cos(x/2)**2) + 1
return y
def my_sin(x):
y = sin(x)
return y
def my_exp(x):
y = e**x
return y
x = sympy.Symbol('x')
def get_polynomial(function, x0, degree):
# parse function definition code
lines_list = getsource(function).split("\n")
for line in lines_list:
if '=' in line:
func_def = line
elements = func_def.split('=')
line = ' '.join(elements[1:])
sympy_function = sympy.sympify(line)
# compute taylor expansion symbolically
i = 0
taylor_exp = sympy.Integer(0)
while i <= degree:
taylor_exp = taylor_exp + (sympy.diff(sympy_function,x,i).subs(x,x0))/(sympy.factorial(i))*(x-x0)**i
i += 1
return taylor_exp
print (get_polynomial(my_sin,0,5))
print (get_polynomial(my_exp,0,5))
print (get_polynomial(f,0,5))

python: changing symbol variable and assign numerical value

In order to calculate derivatives and other expressions I used the sympy package and said that T = sy.Symbol('T') now that I have calculated the right expression:
E= -T**2*F_deriv_T(T,rho)
where
def F_deriv_rho(T,rho):
ret = 0
for n in range(5):
for m in range(4):
inner= c[n,m]*g_rho_deriv_rho_np*g_T_np
ret += inner
return ret
that looks like this:
F_deriv_rho: [0.0 7.76971e-5*T 0.0001553942*T**2*rho
T*(-5.14488e-5*log(rho) - 5.14488e-5)*log(T) + T*(1.22574e-5*log(rho)+1.22574e-5)*log(T) + T*(1.89488e-5*log(rho) + 1.89488e-5)*log(T) + T(2.29441e-5*log(rho) + 2.29441e-5)*log(T) + T*(7.49956e-5*log(rho) + 7.49956e-5)*log(T)
T**2*(-0.0001028976*rho*log(rho) - 5.14488e-5*rho)*log(T) + T**2*(2.45148e-5*rho*log(rho) + 1.22574e-5*rho)*log(T) + T**2*(3.78976e-5*rho*log(rho) + 1.89488e-5*rho)*log(T) + T**2*(4.58882e-5*rho*log(rho) + 2.29441e-5*rho)*log(T) + T**2*(0.0001499912*rho*log(rho) + 7.49956e 5*rho)*log(T)]
with python I would like to change T (and rho) as a symbol to a value. How could I do that?
So, I would like to create 10 numbers like T_def = np.arange(2000, 10000, 800)and exchange all my sy.symbol(T) by iterating through the 10 values I created in the array.
Thanks for your help
I have found the solution according to this post:
How to substitute multiple symbols in an expression in sympy?
by usings "subs":
>>> from sympy import Symbol
>>> x, y = Symbol('x y')
>>> f = x + y
>>> f.subs({x:10, y: 20})
>>> f
30
There's more for this kinda thing here: http://docs.sympy.org/latest/tutorial/basic_operations.html
EDIT: A faster way would be by using "lamdify" as suggested by #Bjoern Dahlgren

Parsing an equation with custom functions in Python

I have a string that is a mathematical equation, but with some custom functions. I need to find all such functions and replace them with some code.
For example, I have a string:
a+b+f1(f2(x,y),x)
I want code that will replace (say) f2(x,y) with x+y^2 and f1(x,y) with sin(x+y).
It would be ideal if nested functions were supported, like in the example. However, it would still be useful if nesting was not supported.
As I understand from similar topics this can be done using a compiler module like compiler.parse(eq). How I can work with AST object created by compiler.parse(eq) to reconstruct my string back, replacing all found functions?
I need only to perform substitution and then string will be used in other program. Evaluation is not needed.
Here is a minimal working example (+, - , *, /, ** binary and unary operations and function call implemented). The priority of operations are set with parenthesis.
A little bit more than the functionality for the example given is done:
from __future__ import print_function
import ast
def transform(eq,functions):
class EqVisitor(ast.NodeVisitor):
def visit_BinOp(self,node):
#generate("=>BinOp")
generate("(")
self.visit(node.left)
self.visit(node.op)
#generate("ici",str(node.op),node._fields,node._attributes)
#generate(dir(node.op))
self.visit(node.right)
generate(")")
#ast.NodeVisitor.generic_visit(self,node)
def visit_USub(self,node):
generate("-")
def visit_UAdd(self,node):
generate("+")
def visit_Sub(self,node):
generate("-")
def visit_Add(self,node):
generate("+")
def visit_Pow(self,node):
generate("**")
def visit_Mult(self,node):
generate("*")
def visit_Div(self,node):
generate("/")
def visit_Name(self,node):
generate(node.id)
def visit_Call(self,node):
debug("function",node.func.id)
if node.func.id in functions:
debug("defined function")
func_visit(functions[node.func.id],node.args)
return
debug("not defined function",node.func.id)
#generate(node._fields)
#generate("args")
generate(node.func.id)
generate("(")
sep = ""
for arg in node.args:
generate (sep)
self.visit(arg)
sep=","
generate(")")
def visit_Num(self,node):
generate(node.n)
def generic_visit(self, node):
debug ("\n",type(node).__name__)
debug (node._fields)
ast.NodeVisitor.generic_visit(self, node)
def func_visit(definition,concrete_args):
class FuncVisitor(EqVisitor):
def visit_arguments(self,node):
#generate("visit arguments")
#generate(node._fields)
self.arguments={}
for concrete_arg,formal_arg in zip(concrete_args,node.args):
#generate(formal_arg._fields)
self.arguments[formal_arg.id]=concrete_arg
debug(self.arguments)
def visit_Name(self,node):
debug("visit Name",node.id)
if node.id in self.arguments:
eqV.visit(self.arguments[node.id])
else:
generate(node.id)
funcV=FuncVisitor()
funcV.visit(ast.parse(definition))
eqV=EqVisitor()
result = []
def generate(s):
#following line maybe usefull for debug
debug(str(s))
result.append(str(s))
eqV.visit(ast.parse(eq,mode="eval"))
return "".join(result)
def debug(*args,**kwargs):
#print(*args,**kwargs)
pass
Usage:
functions= {
"f1":"def f1(x,y):return x+y**2",
"f2":"def f2(x,y):return sin(x+y)",
}
eq="-(a+b)+f1(f2(+x,y),z)*4/365.12-h"
print(transform(eq,functions))
Result
((-(a+b)+(((sin((+x+y))+(z**2))*4)/365.12))-h)
WARNING
The code works with Python 2.7 and as it is AST dependent is not guaranteed to work with another version of Python. The Python 3 version doesn't work.
The full substitution is quite tricky. Here is my attempt to do it. Here we can successfully inline expressions,
but not in all scenarios. This code works on AST only, made by ast module. And uses codegen to stringify it back to code. The stringifying of ast and modifying ast in general is covered in other SO Q/A: "Parse a .py file, read the AST, modify it, then write back the modified source code".
First we define few helpers:
import ast
import codegen
import copy
def parseExpr(expr):
# Strip:
# Module(body=[Expr(value=
return ast.parse(expr).body[0].value
def toSource(expr):
return codegen.to_source(expr)
After that we define a substitution function using NodeTransformer.
For example:
substitute(parseExpr("a + b"), { "a": parseExpr("1") }) # 1 + b
The simulatenous substitution of multiple variables is needed to properly avoid nasty situations.
For example substituting both a and b for a + b in a + b.
The result should be (a + b) + (a + b), but if we substitute first a for a + b, we'll get (a + b) + b, and then substitute b, we'll get (a + (a + b)) + b which is the wrong result! So simultaneous is important:
class NameTransformer(ast.NodeTransformer):
def __init__(self, names):
self.names = names
def visit_Name(self, node):
if node.id in self.names:
return self.names[node.id]
else:
return node
def substitute(expr, names):
print "substitute"
for varName, varValue in names.iteritems():
print " name " + varName + " for " + toSource(varValue)
print " in " + toSource(expr)
return NameTransformer(names).visit(expr)
Then we write similar NodeTransformer to find calls, where we can inline function definitions:
class CallTransformer(ast.NodeTransformer):
def __init__(self, fnName, varNames, fnExpr):
self.fnName = fnName
self.varNames = varNames
# substitute in new fn expr for each CallTransformer
self.fnExpr = copy.deepcopy(fnExpr)
self.modified = False
def visit_Call(self, node):
if (node.func.id == self.fnName):
if len(node.args) == len(self.varNames):
print "expand call to " + self.fnName + "(" + (", ".join(self.varNames)) + ")" + " with arguments "+ ", ".join(map(toSource, node.args))
# We substitute in args too!
old_node = node
args = map(self.visit, node.args)
names = dict(zip(self.varNames, args))
node = substitute(self.fnExpr, names)
self.modified = True
return node
else:
raise Exception("invalid arity " + toSource(node))
else:
return self.generic_visit(node)
def substituteCalls(expr, definitions, n = 3):
while True:
if (n <= 0):
break
n -= 1
modified = False
for fnName, varNames, fnExpr in definitions:
transformer = CallTransformer(fnName, varNames, fnExpr)
expr = transformer.visit(expr)
modified = modified or transformer.modified
if not modified:
break
return expr
The substituteCalls is recursive so we can inline recursive functions too. Also there is an explicit limit, because some definitions might be infinitely recursive (as fact below). There is a bit of ugly looking copying, but it is required to separate different subtrees.
And the example code:
if True:
print "f1 first, unique variable names"
ex = parseExpr("a+b+f1(f2(x, y), x)")
ex = substituteCalls(ex, [
("f1", ["u", "v"], parseExpr("sin(u + v)")),
("f2", ["i", "j"], parseExpr("i + j ^ 2"))])
print toSource(ex)
print "---"
if True:
print "f1 first"
ex = parseExpr("a+b+f1(f2(x, y), x)")
ex = substituteCalls(ex, [
("f1", ["x", "y"], parseExpr("sin(x + y)")),
("f2", ["x", "y"], parseExpr("x + y ^ 2"))])
print toSource(ex)
print "---"
if True:
print "f2 first"
ex = parseExpr("f1(f1(x, x), y)")
ex = substituteCalls(ex, [
("f1", ["x", "y"], parseExpr("x + y"))])
print toSource(ex)
print "---"
if True:
print "fact"
ex = parseExpr("fact(n)")
ex = substituteCalls(ex, [
("fact", ["n"], parseExpr("n if n == 0 else n * fact(n-1)"))])
print toSource(ex)
print "---"
Which prints out:
f1 first, unique variable names
expand call to f1(u, v) with arguments f2(x, y), x
substitute
name u for f2(x, y)
name v for x
in sin((u + v))
expand call to f2(i, j) with arguments x, y
substitute
name i for x
name j for y
in ((i + j) ^ 2)
((a + b) + sin((((x + y) ^ 2) + x)))
---
f1 first
expand call to f1(x, y) with arguments f2(x, y), x
substitute
name y for x
name x for f2(x, y)
in sin((x + y))
expand call to f2(x, y) with arguments x, y
substitute
name y for y
name x for x
in ((x + y) ^ 2)
((a + b) + sin((((x + y) ^ 2) + x)))
---
f2 first
expand call to f1(x, y) with arguments f1(x, x), y
expand call to f1(x, y) with arguments x, x
substitute
name y for x
name x for x
in (x + y)
substitute
name y for y
name x for (x + x)
in (x + x)
((x + x) + ((x + x) + x))
---
fact
expand call to fact(n) with arguments n
substitute
name n for n
in n if (n == 0) else (n * fact((n - 1)))
expand call to fact(n) with arguments (n - 1)
substitute
name n for (n - 1)
in n if (n == 0) else (n * fact((n - 1)))
expand call to fact(n) with arguments ((n - 1) - 1)
substitute
name n for ((n - 1) - 1)
in n if (n == 0) else (n * fact((n - 1)))
n if (n == 0) else (n * (n - 1) if ((n - 1) == 0) else ((n - 1) * ((n - 1) - 1) if (((n - 1) - 1) == 0) else (((n - 1) - 1) * fact((((n - 1) - 1) - 1)))))
Unfortunately codegen version in pypi is buggy. It doesn't parenthesise expressions properly, even AST says they should. I used jbremer/codegen (pip install git+git://github.com/jbremer/codegen). It adds unnecessary parenthesis too, but it's better than no at all. Thanks to #XavierCombelle for the tip.
The substitution gets trickier if you have anonymous functions, i.e lambda. Then you need to rename variables. You could try to search for lambda calculus with substitution or implementation. Yet I had bad luck to find any articles which use Python for the task.
Do you know the variables beforehand?
I recommend using SymPy!
Take for example the following:
import sympy
a,b,x,y = sympy.symbols('a b x y')
f1 = sympy.Function('f1')
f2 = sympy.Function('f2')
readString = "a+b+f1(f2(x,y),x)"
z = eval(readString)
'z' will now be a symbolic term representing the mathematical formula. You can print it out. You can then use subs to replace symbolic terms or functions. You can either represent sine symbolically again (like f1 and f2) or you can possibly use the sin() in sympy.mpmath.
Depending on your needs, this approach is great because you can eventually compute, evaluate or simplify this expression.
What is your long term goal? Is it to evaluate the function or simply perform substitution? In the former case you can simply try this (note that f1 and f2 could also be dynamically defined):
import math
math.sin
def f2(x, y):
return x + y ** 2
def f1(x, y):
return math.sin(x + y)
a, b = 1, 2
x, y = 3, 4
eval('a + b + f1(f2(x, y), x)')
# 2.991148690709596
If you want to replace the functions and get back the modified version, you will indeed have to resort to some sort of AST parser. Be careful though with the use of eval, as this opens up a security hole for malicious user input code.
(Using sympy as adrianX suggested, with some extra code.)
Code below converts a given string to a new string after combining given functions. It's hasty and poorly documented, but it works.
WARNING!
Contains exec eval, malicious code could probably have an effected, if input is provided by external users.
UPDATE:
Rewrote the whole code. Works in Python 2.7.
Function arguments can be separated by comma or whitespace or both.
All examples in question and comments are working.
import re
import sympy
##################################################
# Input string and functions
initial_str = 'a1+myf1(myf2(a, b),y)'
given_functions = {'myf1(x,y)': 'cross(x,y)', 'myf2(a, b)': 'value(a,b)'}
##################################################
print '\nEXECUTED/EVALUATED STUFF:\n'
processed_str = initial_str
def fixed_power_op(str_to_fix):
return str_to_fix.replace('^', '**')
def fixed_multiplication(str_to_fix):
"""
Inserts multiplication symbol wherever omitted.
"""
pattern_digit_x = r"(\d)([A-Za-z])" # 4x -> 4*x
pattern_par_digit = r"(\))(\d)" # )4 -> )*4
pattern_digit_par = r"[^a-zA-Z]?_?(\d)(\()" # 4( -> 4*(
for patt in (pattern_digit_x, pattern_par_digit, pattern_digit_par):
str_to_fix = re.sub(patt, r'\1*\2', str_to_fix)
return str_to_fix
processed_str = fixed_power_op(processed_str)
class FProcessing(object):
def __init__(self, func_key, func_body):
self.func_key = func_key
self.func_body = func_body
def sliced_func_name(self):
return re.sub(r'(.+)\(.+', r'\1', self.func_key)
def sliced_func_args(self):
return re.search(r'\((.*)\)', self.func_key).group()
def sliced_args(self):
"""
Returns arguments found for given function. Arguments can be separated by comma or whitespace.
:returns (list)
"""
if ',' in self.sliced_func_args():
arg_separator = ','
else:
arg_separator = ' '
return self.sliced_func_args().replace('(', '').replace(')', '').split(arg_separator)
def num_of_sliced_args(self):
"""
Returns number of arguments found for given function.
"""
return len(self.sliced_args())
def functions_in_function_body(self):
"""
Detects functions in function body.
e.g. f1(x,y): sin(x+y**2), will result in "sin"
:returns (set)
"""
return set(re.findall(r'([a-zA-Z]+_?\w*)\(', self.func_body))
def symbols_in_func_body(self):
"""
Detects non argument symbols in function body.
"""
symbols_in_body = set(re.findall(r'[a-zA-Z]+_\w*', self.func_body))
return symbols_in_body - self.functions_in_function_body()
# --------------------------------------------------------------------------------------
# SYMBOL DETECTION (x, y, z, mz,..)
# Prohibited symbols
prohibited_symbol_names = set()
# Custom function names are prohibited symbol names.
for key in given_functions.keys():
prohibited_symbol_names |= {FProcessing(func_key=key, func_body=None).sliced_func_name()}
def symbols_in_str(provided_str):
"""
Returns a set of symbol names that are contained in provided string.
Allowed symbols start with a letter followed by 0 or more letters,
and then 0 or more numbers (eg. x, x1, Na, Xaa_sd, xa123)
"""
symbol_pattern = re.compile(r'[A-Za-z]+\d*')
symbol_name_set = re.findall(symbol_pattern, provided_str)
# Filters out prohibited.
symbol_name_set = {i for i in symbol_name_set if (i not in prohibited_symbol_names)}
return symbol_name_set
# ----------------------------------------------------------------
# EXEC SYMBOLS
symbols_in_given_str = symbols_in_str(initial_str)
# e.g. " x, y, sd = sympy.symbols('x y sd') "
symbol_string_to_exec = ', '.join(symbols_in_given_str)
symbol_string_to_exec += ' = '
symbol_string_to_exec += "sympy.symbols('%s')" % ' '.join(symbols_in_given_str)
exec symbol_string_to_exec
# -----------------------------------------------------------------------------------------
# FUNCTIONS
# Detects secondary functions (functions contained in body of given_functions dict)
sec_functions = set()
for key, val in given_functions.items():
sec_functions |= FProcessing(func_key=key, func_body=val).functions_in_function_body()
def secondary_function_as_exec_str(func_key):
"""
Used for functions that are contained in the function body of given_functions.
E.g. given_functions = {f1(x): sin(4+x)}
"my_f1 = sympy.Function('sin')(x)"
:param func_key: (str)
:return: (str)
"""
returned_str = "%s = sympy.Function('%s')" % (func_key, func_key)
print returned_str
return returned_str
def given_function_as_sympy_class_as_str(func_key, func_body):
"""
Converts given_function to sympy class and executes it.
E.g. class f1(sympy.Function):
nargs = (1, 2)
#classmethod
def eval(cls, x, y):
return cross(x+y**2)
:param func_key: (str)
:return: (None)
"""
func_proc_instance = FProcessing(func_key=func_key, func_body=func_body)
returned_str = 'class %s(sympy.Function): ' % func_proc_instance.sliced_func_name()
returned_str += '\n\tnargs = %s' % func_proc_instance.num_of_sliced_args()
returned_str += '\n\t#classmethod'
returned_str += '\n\tdef eval(cls, %s):' % ','.join(func_proc_instance.sliced_args())
returned_str = returned_str.replace("'", '')
returned_str += '\n\t\treturn %s' % func_body
returned_str = fixed_power_op(returned_str)
print '\n', returned_str
return returned_str
# Executes functions in given_functions' body
for name in sec_functions:
exec secondary_function_as_exec_str(func_key=name)
# Executes given_functions
for key, val in given_functions.items():
exec given_function_as_sympy_class_as_str(func_key=key, func_body=val)
final_result = eval(initial_str)
# PRINTING
print '\n' + ('-'*40)
print '\nRESULTS'
print '\nInitial string: \n%s' % initial_str
print '\nGiven functions:'
for key, val in given_functions.iteritems():
print '%s: ' % key, val
print '\nResult: \n%s' % final_result
I think you want to use something like PyBison which is a parser generator.
See an example that contains the basic code you need here:
http://freenet.mcnabhosting.com/python/pybison/calc.py
You need to add a token type for functions, and a rule for functions, and then what happens with that function if it is encountered.
If you need other information about parsing and so on, try to read some basic tutorials on Lex and (Yacc or Bison).

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