I can't seem to find what I'm looking for in the SymPy docs: basically what I'd like to do is to redefine a symbol so it changes in every expression.
Something like this:
from sympy import *
x, y = symbols("x y")
expr1 = 2*x + y
expr2 = x**2 + 2*y
x.redefine("foo")
print(expr1, expr2)
x.redefine(2)
print(expr1, expr2)
Output:
2*foo + y, foo**2 + 2*y
4 + y, 4 + 2*y
Is it possible?
Sympy is designed in such a way that it cannot be redefined permanently. However, you can use the below command to substitute your symbol in an expression. Here is the documentation for it.
expr1.subs({x:"foo"}) #for changing only one symbol.
expr2.subs({x:10, y:20}) #for changing both the symbols.
SymPy expressions are immutable: they never change. Hence, there can be no global switch that implicitly modifies all preexisting expressions.
To handle substitution in multiple expressions, one can use loops, list comprehension, or apply subs to a matrix. Examples:
expressions = [2*x + y, x**2 + 2*y, x - y/2]
print([expr.subs(x, 3) for expr in expressions])
mat = Matrix(expressions)
print(mat.subs(x, 3))
This prints:
[y + 6, 2*y + 9, -y/2 + 3]
Matrix([[y + 6], [2*y + 9], [-y/2 + 3]])
Related
I have an expression of many sinusoids. Some frequencies are repeated, so I would like to collect the coefficients for identical frequencies. Seems like either collect() or factor() should so the trick but they only work on simple expressions and fail when many variables are involved. Is there a better way to collect terms? Is there a way to help guide factor() to what variable to operate on like exists for collect()? I am trying to manually simplify the equation into a particular form and I think cse() goes too far in reworking terms. Examples below:
w, x, y, z = symbols('w x y z')
factor(z*sin(2*pi*x)+y*sin(2*pi*x), deep=True)
(y + z)⋅sin(2⋅π⋅x) #WORKS
factor(z*sin(2*pi*x*w)+y*sin(2*pi*x*w), deep=True)
(y + z)⋅sin(2⋅π⋅w⋅x) #WORKS
factor(z*sin(2*pi*x)+y*sin(2*pi*x)+z*sin(w)+2*y*sin(w), deep=True)
2⋅y⋅sin(w) + y⋅sin(2⋅π⋅x) + z⋅sin(w) + z⋅sin(2⋅π⋅x) # FAILS – expected Expected (2y + z)*sin(w) + (y+z)⋅sin(2⋅π⋅x)
collect(z*sin(2*pi*x)+y*sin(2*pi*x), x)
y⋅sin(2⋅π⋅x) + z⋅sin(2⋅π⋅x) # FAILS expected (y+z)⋅sin(2⋅π⋅x)
collect(z*sin(2*pi*x)+y*sin(2*pi*x)+z*sin(w)+2*y*sin(w), x)
2⋅y⋅sin(w) + y⋅sin(2⋅π⋅x) + z⋅sin(w) + z⋅sin(2⋅π⋅x) #FAILS Expected (2y + z)*sin(w) + (y+z)⋅sin(2⋅π⋅x)
It is:
>>> from sympy import *
>>> w, x, y, z = symbols('w x y z')
>>> ex= z*sin(2*pi*x)+y*sin(2*pi*x)+z*sin(w)+2*y*sin(w)
>>> ex
2*y*sin(w) + y*sin(2*pi*x) + z*sin(w) + z*sin(2*pi*x)
>>> collect(ex, [sin(w), sin(2*pi*x)])
(y + z)*sin(2*pi*x) + (2*y + z)*sin(w)
Reference: https://docs.sympy.org/latest/modules/simplify/simplify.html?highlight=collect
When I use "x" and "z" as symbols, I have no problem with this code:
from sympy import *
x, z = symbols('x z')
y = -6*x**2 + 2*x*z**0.5 + 50*x - z
solve((diff(y, x), diff(y, z)))
y.subs({x: 5, z: 25})
But when I use "q" and "a", solve does not give me any solution.
q, a = symbols('q a')
y = -6*q**2 + 2*q*a**0.5 + 50*q - a
solve((diff(y, q), diff(y, a)))
y.subs({q: 5, a: 25})
As you can see I use "subs" to check that there is no typo in the objective function.
UPDATE: I used "Symbol" to set each variable individually, but again using "q" and "a" does not work.
# This works
x = Symbol('x')
z = Symbol('z')
y = -6*x**2 + 2*x*z**0.5 + 50*x - z
solve((diff(y, x), diff(y, z)))
# This does not work
q = Symbol('q')
a = Symbol('a')
y = -6*q**2 + 2*q*a**0.5 + 50*q-a
solve((diff(y, q), diff(y, a)))
Thank you.
Got it!
It all depends on an alphabetic order of your variables.
If you substitute x for z and z for x in your first example it will also stop working.
Internally solve sends the expression to the function _solve in sympy.solvers which then tries to solve your equation and fails many times.
Finally as a last effort what it does is it tries to solve -sqrt(a) + q or x - sqrt(z) by picking symbols from it through an internal function _ok_syms, with an argument that sorts those alphabetically (even without this argument it still would, but if wrapped with reversed it magically makes your examples works in the exactly opposite way).
And so it does solve x - sqrt(z) as x: sqrt(z) and -sqrt(a) + q as a: q**2.
While in the first case it ends up with an easily solvable 50 - 10*sqrt(z), in the second case it is lost on -12*q + 2*sqrt(q**2) + 50 as it is not able to simplify sqrt(q**2).
source:
a lot of testing on:
https://github.com/sympy/sympy/blob/master/sympy/solvers/solvers.py
I am trying to use python to solve the equations: x**2*y**2 + x**2 -10*x*y + 4*y**2 + 9.0=0, due to the equations equal to (x*y-3)**2+(x-2*y)**2=0 ,so hoping to get the real solution: x = 2*sqrt（3.0/2),y = sqrt(3.0/2)&& x = -2*sqrt（3.0/2),y = -sqrt(3.0/2) Is there any way to get this solutions?
from sympy import *
x = symbols("x")
y = symbols("y")
expression = x**2*y**2 + x**2 - 10*x*y + 4*y**2 + 9
solve(expression,(x,y))
above code only get the solution: [((5*y + I*(-2*y**2 + 3))/(y**2 + 1), y),
((5*y + I*(2*y**2 - 3))/(y**2 + 1), y)],thanks for your help and advice
It looks like what you are trying to do is find where both of the terms of the expression (x*y-3)**2+(x-2*y)**2 are simultaneously zero. Instead of expanding that, ask solve for that answer:
>>> eq = (x*y-3)**2+(x-2*y)**2
>>> terms = eq.args
>>> solve(terms, x, y)
[(-sqrt(6), -sqrt(6)/2), (sqrt(6), sqrt(6)/2)]
You can use solveset_real (although the exact equation may not have any real solutions)
I am currently dealing with functions of more than one variable and need to collect like terms in an attempt to simplify an expression.
Say the expression is written as follows:
x = sympy.Symbol('x')
y = sympy.Symbol('y')
k = sympy.Symbol('k')
a = sympy.Symbol('a')
z = k*(y**2*(a + x) + (a + x)**3/3) - k((2*k*y*(a + x)*(n - 1)*(-k*(y**2*(-a + x) + (-a + x)**3/3) + k*(y**2*(a + x) + (a + x)**3/3)) + y)**2*(-a + k*(n - 1)*(y**2 + (a + x)**2)*(-k*(y**2*(-a + x)))))
zEx = z.expand()
print type(z)
print type(zEx)
EDIT: Formatting to add clarity and changed the expression z to make the problem easier to understand.
Say z contains so many terms, that sifting through them by eye. and selecting the appropriate terms, would take an unsatisfactory amount of time.
I want to collect all of the terms which are ONLY a multiple of a**1. I do not care for quadratic or higher powers of a, and I do not care for terms which do not contain a.
The type of z and zEx return the following:
print type(z)
print type(zEx)
>>>
<class 'sympy.core.add.Add'>
<class 'sympy.core.mul.Mul'>
Does anyone know how I can collect the terms which are a multiple of a , not a^0 or a^2?
tl'dr
Where z(x,y) with constants a and k described by z and zEx and their type(): How can one remove all non-a terms from z AND remove all quadratic or higher terms of a from the expression? Such that what is left is only the terms which contain a unity power of a.
In addition to the other answers given, you can also use collect as a dictionary.
print(collect(zEx,a,evaluate=False)[a])
yields the expression
k*x**2 + k*y**2
In the end it is just an one-liner. #asmeurer brought me on the right track (check the comments below this post). Here is the code; explanations can be found below:
from sympy import *
from sympy.parsing.sympy_parser import parse_expr
import sys
x, y, k, a = symbols('x y k a')
# modified string: I added a few terms
z = x*(k*a**9) + (k**1)*x**2 - k*a**8 + y*x*(k**2) + y*(x**2)*k**3 + x*(k*a**1) - k*a**3 + y*a**5
zmod = Add(*[argi for argi in z.args if argi.has(a)])
Then zmod is
a**9*k*x - a**8*k + a**5*y - a**3*k + a*k*x
So let's look at this more carefully:
z.args
is just a collection of all individual terms in your expression (please note, that also the sign is parsed which makes things easier):
(k*x**2, a**5*y, -a**3*k, -a**8*k, a*k*x, a**9*k*x, k**2*x*y, k**3*x**2*y)
In the list comprehension you then select all the terms that contain an a using the function has. All these terms can then be glued back together using Add which gives you the desired output.
EDIT
The above returns all all the expressions that contain an a. If you only want to filter out the expressions that contain a with unity power, you can use collect and Mul:
from sympy import *
from sympy.parsing.sympy_parser import parse_expr
import sys
x, y, k, a = symbols('x y k a')
z2 = x**2*(k*a**1) + (k**1)*x**2 - k*a**8 + y*x*(k**2) + y*(x**2)*k**3 + x*k*a - k*a**3 + y*a**1
zc = collect(z2, a, evaluate=False)
zmod2 = Mul(zc[a], a)
then zmod2 is
a*(k*x**2 + k*x + y)
and zmod2.expand()
a*k*x**2 + a*k*x + a*y
which is correct.
With the updated z you provide I run:
z3 = k*(y**2*(a + x) + (a + x)**3/3) - k((2*k*y*(a + x)*(n - 1)*(-k*(y**2*(-a + x) + (-a + x)**3/3) + k*(y**2*(a + x) + (a + x)**3/3)) + y)**2*(-a + k*(n - 1)*(y**2 + (a + x)**2)*(-k*(y**2*(-a + x)))))
zc3 = collect(z3.expand(), a, evaluate=False)
zmod3 = Mul(zc3[a], a)
and then obtain for zmod3.expand():
a*k*x**2 + a*k*y**2
Is this the result you were looking for?
PS: Thanks to #asmeurer for all these helpful comments!
To iterate over the terms of an expression use expr.args.
I'm unclear what a is supposed to be, but the collect function may do what you want.
I have an expression in Sympy (like
-M - n + x(n)
) and I would
like to create a formal linear function, says f, and apply it to my expression, in order to get, after simplification:
-f(M) - f(n) + f(x(n))
Is it possible to tell sympy that a property such as linearity is verified?
A very hacky way to do it would be to apply the function f to every subexpression which is in a sum.
For instance when given an expressions like the first one I gave, it would be nice to simply access the terms appearing in the sum (here it would be
[-M, -n , x(n)]
Then mapping f on the list and sum it to get what is expected.
Is there an easy way to do so, or have I necessarily to go trough the syntactic tree of the expression ?
This works:
>>> x,f = map(Function, 'xf'); n,M = symbols('n,M'); expr = -M - n + x(n)
>>> Add(*[f(a) for a in Add.make_args(expr)])
f(-M) + f(-n) + f(x(n))
If you have an expression like f(n*(M + 1)) and you expand it you will get f(n*M + n). Can you tell SymPy to apply the function to the args of f's args? Yes:
>>> expr = f(n*(M + 1))
>>> expr.expand().replace(lambda x: x.func == f,
... lambda x: Add(*[f(a) for a in Add.make_args(x.args[0])]))
f(n) + f(M*n)
If you call such a replacement linapp you can use it for any function that you want:
def linapp(expr, *f):
return expr.expand().replace(
lambda x: x.func in f,
lambda x: Add(*[x.func(a) for a in Add.make_args(x.args[0])]))
>>> print(linapp(cos(x+y) + sin(x + y), cos, sin))
sin(x) + sin(y) + cos(x) + cos(y)
(Not saying that it's a true result, just that you can do it. And if you replace a variable with something else and you want to reapply the linearization, you can:
>>> linapp(_.subs(y, z + 1), cos)
sin(x) + sin(z + 1) + cos(x) + cos(z) + cos(1)
Here's a hackey way that goes through the syntactic tree:
from sympy import *
init_session()
M,n=symbols('M n')
thing=-f(M) - f(n) + f(x(n))
def linerize_element(bro):
return bro.args[0] if len(bro.args) == 1 else bro.args[0] * bro.args[1].args[0]
print([ linerize_element(tmp) for tmp in thing.args])