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
.
evalf() function and subs() function from sympy is used to evaluate algebraic expressions. Example 1: In this example, we import symbols from sympy package. An expression is created and evalf() function is used to evaluate the expression.
With the help of sympy. sqrt() method, we can find the square root of any number by using sympy. sqrt() method. Return : Return square root of any number.
With the help of sympy. log() function, we can simplify the principal branch of the natural logarithm. Logarithms are taken with the natural base, e. To get a logarithm of a different base b, use log(x, y), which is essentially short-hand for log(x) / log(y).
Note that by default in SymPy the base of the natural logarithm is E (capital E ). That is, exp(x) is the same as E**x .
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.
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