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How can I automatically generate an expression in Sympy using the coefficients of a 5th-order expression?

How can I automatically write a 5th order expression using a for loop using sympy in Python?
For example, let the expression be: y = 2x^5 + 3x^4 + 17x^3 + x^2 - 8x +101
I have the coefficients in the 5th order expression as a list. You can think of it like [101,-8,1,17,3,2].
I want to create a 5th order expression using the elements of this list using sympy.
I would be very grateful if you could help with this.

from sympy import *
x = symbols("x")
coeff = [101,-8,1,17,3,2]
# create a list of exponents going from 0 to 5 (included)
expon = list(range(len(coeff)))
# create a list of polynomial terms using list-comprehension syntax
# and sum up the elements
expr = sum(c * x**e for c, e in zip(coeff, expon))
Edit to satisfy comment:
from sympy import *
x = symbols("x")
coeff = [101,-8,1,17,3,2]
# create a list of exponents going from 0 to 5 (included)
expon = list(range(len(coeff)))
terms = []
for c, e in zip(coeff, expon):
terms.append(c * x**e)
expr = sum(terms)

Poly can accept a list of coefficients and the generator directly. It expects them in order from high to low. If you want an Expr, not a Poly, you can append the as_expr() call:
>>> little_endian = [101,-8,1,17,3,2]
>>> Poly(reversed(little_endian), x).as_expr()
2*x**5 + 3*x**4 + 17*x**3 + x**2 - 8*x + 101

How to edit the order of sympy latex print? [duplicate]

I have the following code:
from sympy import *
init_printing()
x,y = symbols('x y')
u = Function('u')(x,y)
ux,uy,uxx,uxy,uyy = symbols("u_x u_y u_xx u_xy u_yy")
mainEvaluation = uxx - 2*sin(x)*uxy - (cos(x) ** 2) * uyy - 2*ux + (2 - cos(x) + 2*sin(x) )*uy
And when the output of print(mainExpression) is
-2*u_x + u_xx - 2*u_xy*sin(x) + u_y*(2*sin(x) - cos(x) + 2) - u_yy*cos(x)**2
The problem is: I want the original order of variables.
u_xx - 2*u_xy*sin(x) - u_yy*cos(x)**2 - 2*u_x + u_y*(2*sin(x) - cos(x) + 2)
All this is done in IPython notebook.
Is there any way to keep order?

Sadly, SymPy does not keep track of the input order (see the other question I linked in a comment on the question). You can define your own ordering function that orders expressions however you want, but there's no way to order things exactly as they were input, since that information isn't saved.

If you know what your arguments/terms are then you can manually create the Add with evaluate=False to keep them in order and print them with a printer initialized to not change the order:
x,y = symbols('x y')
u = Function('u')(x,y)
ux,uy,uxx,uxy,uyy = symbols("u_x u_y u_xx u_xy u_yy")
args = uxx , -2*sin(x)*uxy, -cos(x)**2*uyy, -2*ux, +(2-cos(x)+2*sin(x))*uy
expr = Add(*args, evaluate=False)
from sympy.printing.str import StrPrinter # or LatexPrinter from .latex)
StrPrinter(dict(order='none'))._print_Add(expr)
This outputs
u_xx - 2*u_xy*sin(x) - u_yy*cos(x)**2 - 2*u_x + u_y*(2 - cos(x) + 2*sin(x))

try read this http://docs.sympy.org/0.7.2/modules/utilities/misc.html, may be could help you
Note:
The key returned is useful for getting items into a canonical order that will be the same across platforms. It is not directly useful for sorting lists of expressions:
>>> a, b = x, 1/x
Since a has only 1 term, its value of sort_key is unaffected by order:
>>> a.sort_key() == a.sort_key('rev-lex')
True
If a and b are combined then the key will differ because there are terms that can be ordered:
>>> eq = a + b
>>> eq.sort_key() == eq.sort_key('rev-lex')
False
>>> eq.as_ordered_terms()
[x, 1/x]
>>> eq.as_ordered_terms('rev-lex')
[1/x, x]
But since the keys for each of these terms are independent of order‘s value, they don’t sort differently when they appear separately in a list:
>>> sorted(eq.args, key=default_sort_key)
[1/x, x]
>>> sorted(eq.args, key=lambda i: default_sort_key(i, order='rev-lex'))
[1/x, x]
The order of terms obtained when using these keys is the order that would be obtained if those terms were factors in a product.

For the teaching purpose, I also don't want to simpilify or change the order of terms too early.
I'm using pytexit with jupyter notebook: https://pytexit.readthedocs.io/en/latest/
from pytexit import py2tex
def showeq(str):
py2tex(str, print_formula=False);
eq1 = "angle = acos((side_A**2 + side_B**2 - side_C**2)/(2*side_A*side_B))"
show(eq1)
side_A = 14.8
side_B = 16.3
side_C = 13.2
exec(eq1)

Sympy outputs a derivative with log(e)

I'm using Sympy to calculate derivatives and some other things. I tried to calculate the derivative of "e**x + x + 1", and it returns e**x*log(e) + 1 as the result, but as far as I know the correct result should be e**x + 1. What's going on here?
Full code:
from sympy import *
from sympy.parsing.sympy_parser import parse_expr
x = symbols("x")
_fOfX = "e**x + x + 1"
sympyFunction = parse_expr(_fOfX)
dSeconda = diff(sympyFunction,x,1)
print(dSeconda)

The answer correctly includes log(e) because you never specified what "e" is. It's just a letter like "a" or "b".
The Euler number 2.71828... is represented as E in SymPy. But usually, writing exp(x) is preferable because the notation is unambiguous, and also because SymPy is going to return exp(x) anyway. Examples:
>>> fx = E**x + x + 1
>>> diff(fx, x, 1)
exp(x) + 1
or with exp notation:
>>> fx = exp(x) + x + 1
>>> diff(fx, x, 1)
exp(x) + 1
Avoid creating expressions by parsing strings, unless you really need to and know why you need it.

Create a formal linear function in Sympy

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])

Corresponding Coefficients in Python SymPy Pattern Matching

I have a function named f = 0.5/(z-3). I would like to know what would the coefficients p and q be if f was written in the following form: q/(1-p*z) but unfortunately sympy match function returns None. Am I doing something wrong? or what is the right way of doing something like this?
Here is the code:
z = symbols('z')
p, q = Wild('p'), Wild('q')
print (0.5/(z-3)).match(q/(1-p*z))
EDIT:
My expected answer is: q=-1/6 and p = 1/3
One way of course is
p, q = symbols('p q')
f = 0.5/(z-3)
print solve(f - q/(1-p*z), p, q,rational=True)
But I don't know how to do that in pattern matching, or if it's capable of doing something like this.
Thanks in Advance =)

If you start by converting to linear form,
1 / (2*z - 6) == q / (1 - p*z)
# multiply both sides
# by (2*z - 6) * (1 - p*z)
1 - p*z == q * (2*z - 6)
then
from sympy import Eq, solve, symbols, Wild
z = symbols("z")
p,q = symbols("p q", cls=Wild)
solve(Eq(1 - p*z, q*(2*z - 6)), (p,q))
gives
{p_: 1/3, q_: -1/6}
as expected.
Edit: I found a slightly different approach:
solve(Eq(f, g)) is equivalent to solve(f - g) (implicitly ==0)
We can reduce f - g like simplify(f - g), but by default it doesn't do anything because the resulting equation is more than 1.7 times longer than the original (default value for ratio argument).
If we specify a higher ratio, like simplify(f - g, ratio=5), we get
>>> simplify(1/(2*z-6) - q/(1-p*z), ratio=5)
(z*p_ + 2*q_*(z - 3) - 1)/(2*(z - 3)*(z*p_ - 1))
This is now in a form the solver will deal with:
>>> solve(_, (p,q))
{p_: 1/3, q_: -1/6}

SymPy's pattern matcher only does minimal algebraic manipulation to match things. It doesn't match in this case because there is no 1 in the denominator. It would be better to match against a/(b + c*z) and manipulate a, b, and c into the p and q. solve can show you the exact formula:
In [7]: solve(Eq(a/(b + c*z), q/(1 - p*z)), (q, p))
Out[7]:
⎧ -c a⎫
⎨p: ───, q: ─⎬
⎩ b b⎭
Finally, it's always a good idea to use exclude when constructing Wild object, like Wild('a', exclude=[z]). Otherwise you can get unexpected behavior like
In [11]: a, b = Wild('a'), Wild('b')
In [12]: S(2).match(a + b*z)
Out[12]:
⎧ 2⎫
⎨a: 0, b: ─⎬
⎩ z⎭
which is technically correct, but probably not what you want.