I am writing a program that requires the user to enter an expression. This expression is entered as a string and converted to a Sympy expression using parse_expr. I then need to take the partial derivative of that expression that the user entered. However, diff is returning 0 with every expression I am testing.
For example if the user enters a*exp(-b*(x-c)**(2)), using the following code, diff returns 0 when it should (as far as I know about diff) return 2*a*b*(c - x)*exp(-b*(x - c)**2) when taking the partial derivative with respect to x:
a, b, c, x = symbols('a b c x', real=True)
str_expr = "a*exp(-b*(x-c)**(2))"
parsed_expr = parse_expr(str_expr)
result = diff(parsed_expr, x)
print(result) # prints 0
What am I doing wrong?
Bottom line: use parse_expr(str_expr,locals()).
Add global_dict=<dict of allowed entities to use>, too, if the expression may use any entities not imported into the local namespace and not accessible with the default from sympy import *.
According to Calculus — SymPy Tutorial - SymPy 1.0.1.dev documentation, you type the symbolic expression into the diff() argument as-is. Due to the fact that the letters are Symbol objects (with overridden operators), Python is tricked into constructing the SymPy object corresponding to the expression as it evaluates the argument!
Thus, if you have it as a string, you eval it to trigger the same behaviour:
<...>
>>> s="a*exp(-b*(x-c)**(2))"
>>> diff(eval(s), x)
−ab(−2c+2x)e−b(−c+x)2
But eval is a security hazard if used with untrusted input because it accepts arbitrary Python code.
This is where replacements like parse_expr come into play. However, due to the way expressions are parsed, described above, it needs access to the external entities used in the expression - like the Symbol objects for variables and function objects for the named functions used - through the local_dict and global_dict arguments.
Otherwise, it creates the Symbol objects on the fly. Which means, the Symbol object it has created for x in the expression is different from the variable x! No wonder that the derivative over it is 0!
<...>
>>> ps=parse_expr(s)
>>> ps.free_symbols
{a,b,c,x}
>>> x in _
False
>>> diff(ps,x)
0
>>> ps=parse_expr(s,locals())
>>> x in ps.free_symbols
True
>>> diff(ps,x)
-ab(−2c+2x)e−b(−c+x)2
Work is ongoing to make sympify safer than eval. Better to use something like the following:
from sympy import *
var ('a b c x')
str_expr = "a*exp(-b*(x-c)**(2))"
parsed_expr = sympify(str_expr)
result = diff(parsed_expr, x)
print(result)
Result:
-a*b*(-2*c + 2*x)*exp(-b*(-c + x)**2)
Replace a, b, c, x = symbols('a b c x', real=True) with:
a = Symbol('a')
b = Symbol('b')
c = Symbol('c')
x = Symbol('x')
Symbols with different assumptions compare unequal:
>>> Symbol('x') == Symbol('x', real=True)
False
When you use sympify or parse_expr, it parses unknown variables as symbols without assumptions. In your case, this creates Symbol('x'), which is considered distinct from the Symbol('x', real=True) you already created.
The solution is to either remove the assumptions, or include the locals() dictionary when you parse, so that it recognizes the name x as being the Symbol('x', real=True) that you already defined, like
parse_expr(str_expr,locals())
or
sympify(str_expr, locals())
Related
I am parsing a custom format like "{a} + {b}" into a sympy expression. I have that working successfully. Now, is there a way to convert that sympy expression back to the original string, assuming I have a dictionary mapping the free variable names and the corresponding string-in-braces representation?
Assuming the free variables from the above are "a" and "b", I want to do something like
str(expr.subs({'a': '{a}', 'b': '{b}'}))
but sympy doesn't seem to allow substituting in arbitrary strings like that.
You could subclass a codegenerator (e.g. StrPrinter), and overwrite the function that outputs the free variables. You can copy the original function from the sympy source and make some modifications.
Here is an example:
import sympy as sp
from sympy.printing import StrPrinter
class CustomStrPrinter(StrPrinter):
def _print_Symbol(self, expr):
return f'{{{expr.name}}}'
a, b = sp.symbols('a b')
expr = a + b
custom_strPrinter = CustomStrPrinter().doprint
print(custom_strPrinter(expr)) # {a} + {b}
Given the following math function in form of a Python function:
import math
def f(x):
a = x - math.log(x)
b = x + math.log(x)
return a / x + b / math.log(x)
Is there any way that I can convert this function into a string like
expr = '(x - math.log(x)) / x + (x + math.log(x)) / math.log(x)'
so that when I want to call the function, I can simply use it by
func = lambda x: eval(expr)
print(func(3))
# 4.364513583657809
Note that I want to keep a and b in the original function. In reality, I have a lot more intermediate variables. Also, I am aware sympy could do similar tasks, but I would like to know if it is possible to convert the function to string, as it would be much more efficient to store.
Any suggestions?
Your function is already a string the moment you write it to a file!
If the function is valid Python, you can then just import it
from myfile import expr
print(expr(3)) # 4.364513583657809
WARNING Do not ever do this
If you want some incredibly evil logic for some reason, you can save your function directly with inspect.getsource(f) and then do something like this
>>> fn_body = """def f(x):
... a = x - math.log(x)
... b = x + math.log(x)
... return a / x + b / math.log(x)
... """
>>> eval(f'lambda {fn_body.split("(")[1].split(")")[0]}, _={exec(fn_body)}: {fn_body.split(" ", 1)[-1].split(")")[0]})')(3)
4.364513583657809
This works by finding the parts needed to call the function, evaluating the source as one of the args (to smuggle it into your namespace), and then building an anonymous function to call it
Further Caveats
not remotely maintainable
extremely fragile
will clobber or conflict with an existing function with the same name depending on use
you will still need to import math or whatever other libraries
won't work with default args without more pain
calling eval() first (before creating the lambda) will allow you to use inspect to get the signature (.signature()) and you can combine it with re and/or ast for a much robust parser, but a 1-liner seemed more exciting
manages to use both eval() and exec() for an extra helping of evil
You're probably looking for a symbolic equation solver!
Sympy's lambdify feature can do this for you!
>>> fn = sympy.lambdify("x", '(x - log(x)) / x + (x + log(x)) / log(x)')
>>> fn(x=3)
4.364513583657809
Caution: this also uses eval internally as #Joshua Voskamp warns about in a comment
I am trying to convert latex expression to sympy form and then solve it.
When I feed the output of the parser(or converter actually?) to solve method, it finds no solution. However, if I manually enter the parser generated expression, it finds the roots successfully. What is wrong with parse_latex ( most probably ) or solve method?
Thanks in advance. Here is the code sample you can try:
from sympy import*
from sympy.parsing.latex import*
x = Symbol("x", real=True)
sym_eqn = parse_latex("|x-2|-1")
print sym_eqn # Abs(x - 2) - 1
print type(sym_eqn) # <class 'sympy.core.add.Add'>
print type(Abs(x - 2) - 1) # <class 'sympy.core.add.Add'>
print solve(Abs(x-2)-1) # [1,3], which is ok
#print solve(sym_eqn) # NotImplementedError: solving Abs(x - 2) when the argument is not real or imaginary.
print solve(sym_eqn,x) # []
The root issue here is whether or not your symbol 'x' has an attribute "real" set to True, or not.
Consider the following two symbols:
a = Symbol('x',real=True)
b = Symbol('x')
a and b are not of the same type and in fact a==b is False.
What happens when you execute
sym_eqn = parse_latex("|x-2|-1")
is that that sym_eqn is now an expression that contains a Symbol that does not have the attribute real set to True which is required to run solve on it.
Having understood this, the question is now how to get parse_latex to return an expression that would contain a Symbol that is real?
The only way I found is to write a function that recursively traverses the expression's tree and rebuilds a copy of it such that the result is the same, except all Symbols are now real.
def rewrite_expr_real(expr):
res_list = []
if isinstance(expr,Symbol):
return Symbol(str(expr),real=True)
if not expr.args:
return expr
for a in expr.args:
res_list.append(rewrite_expr_real(a))
return expr.func(*tuple(res_list))
Now,
if you rewrite your code as follows:
sym_eqn = rewrite_expr_real(parse_latex("|x-2|-1"))
The rest of your code will work as you expect it.
Yakov's answer is on point, but I'd like to offer a code snippet that converts all symbols to "real" in a single substitution.
from sympy import symbols
expr = expr.subs((str(symbol), symbols(str(symbol), real=True))
for symbol in expr.free_symbols)
The function expm1 is not parsed properly in the following example:
from sympy.parsing.sympy_parser import parse_expr
print parse_expr('expm1(x)').diff('x')
gives
Derivative(expm1(x), x)
How can I get sympy identifying expm1 as symbolic function, so that I get the same result as
print parse_expr('exp(x) - 1').diff('x')
which gives exp(x)?
Since there is no built-in expm1 in SymPy, the parser does not know anything about this notation. The parameter local_dict of parse_expr can be used to explain the meaning of unfamiliar functions and symbols to SymPy.
expm1 = lambda x: exp(x)-1
parse_expr('expm1(x)', local_dict={"expm1": expm1})
This returns exp(x) - 1.
For expm1 to remain a single function with known derivative, rather than exp(x)-1, define it as a SymPy function (see tutorial for more such examples).
class expm1(Function):
def fdiff(self, argindex=1):
return exp(self.args[0])
A confirmation that this works:
e = parse_expr('expm1(x)', local_dict={"expm1": expm1})
print(e) # expm1(x)
print(e.diff(x)) # exp(x)
f = lambdify(x, e)
print(f(1)) # 1.718281828459045
print(f(1e-20)) # 1e-20, unlike exp(x)-1 which would evaluate to 0
So, my code is like this:
def func(s,x):
return eval(s.replace('x',x)
#Example:
>> func('x**2 + 3*x',1)
4
The first argument of the function func must be a string because the function eval accepts only string or code objects. However, I'd like to use this function in a kind of calculator, where the user types for example 2 + sin(2*pi-0.15) + func(1.8*x-32,273) and gets the answer of the expression, and it's annoying always to have to write the quotes before in the expression inside func().
Is there a way to make python understands the s argument is always a string, even when it's not between quotes?
No, it is not possible. You can't intercept the Python interpreter before it parses and evaluates 1.8*x-32.
Using eval as a glorified calculator is a highly questionable idea. The user could pass in all kinds of malicious Python code. If you're going to do it, you should provide as minimal an environment as possible for the code to run in. Pass in your own globals dict containing only the variables the user is allowed to reference.
return eval(s, {'x': x})
Besides being safer, this is also a better way to substitute x into the expression.
You could have it handle both cases:
def func(s, x=0):
if isinstance(s, basestring):
# x is in the scope, so you don't need to replace the string
return eval(s)
else:
return s
And the output:
>>> from math import *
>>> func('2 + sin(2*pi-0.15) + func(1.8*x-32,273)')
-30.1494381324736
>>> func('x**2 + 3*x', 1)
4
Caution: eval can do more than just add numbers. I can type __import__('os').system('rm /your/homework.doc') and your calculator will delete your homework.
In a word: no, if I understand you.
In a few more, you can sort of get around the problem by making x be a special object. This is how the Python math library SymPy works. For example:
>>> from sympy import Symbol
>>> x = Symbol('x')
>>> x**2+3*x
x**2 + 3*x
>>> (x**2+3*x).subs(x,1)
4
There's even a handy function to turn strings into sympy objects:
>>> from sympy import sympify, pi
>>> sympify("x**2 - sin(x)")
x**2 - sin(x)
>>> _.subs(x, pi)
pi**2
All the warnings about untrusted user input hold. [I'm too lazy to check whether or not eval or exec is used on the sympify code path, and as they say, every weapon is loaded, even the unloaded ones.]
You can write an interpreter:
import code
def readfunc(prompt):
raw = input(prompt)
if raw.count(',')!=1:
print('Bad expression: {}'.format(raw))
return ''
s, x = raw.split(',')
return '''x={}; {}'''.format(x, s)
code.interact('Calc 0.1', readfunc)