Ignore imaginary roots in sympy
Solution 1
As Krastonov had mentioned mpmath provided an easier method:
y = polyroots([int(row["scaleA"]), int(row["scaleB"]), int(row["scaleC"]), int(row["scaleD"])-value])
for root in y:
if "j" not in str(root):
value = root
Solution 2
If you set x
to be real, SymPy will only give you the real solutions
x = Symbol('x', real=True)
solve(..., x)
Solution 3
solve()
doesn’t have a consistent output for various types of solutions, please use solveset(Eq,x,domain=S.Reals)
:
from sympy import ImageSet, S
x = Symbol('x')
y = solveset(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2+int(row["scaleC"])*x + int(row["scaleD"]), x, domain=S.Reals)
http://docs.sympy.org/latest/modules/solvers/solveset.html
Solution 4
This is exactly the sort of thing that real_roots
is made for and is especially applicable to your case where the coefficients are integers:
x = Symbol('x')
eq = int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"])
y = real_roots(eq, x) # gives [CRootOf(...), ...]
The value of CRootOf instances can be evaluated to whatever precision you need and should not contain any imaginary part. For example,
>>> [i.n(12) for i in real_roots(3*x**3 - 2*x**2 + 7*x - 9, x)]
[1.07951904858]
Note: As I recall, solve will send back roots that it wasn't able to confirm met the assumptions (i.e. if they weren't found to be false for the assumption then they are returned). Also, if you want more consistent output from solve, @PyRick, set the flag dict=True
.
Kreuzade
Updated on June 15, 2022Comments
-
Kreuzade almost 2 years
I'm using sympy to solve a polynomial:
x = Symbol('x') y = solve(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"]), x)
y is a list of possible solutions. However, I need to ignore the imaginary ones and only use the real solutions. Also, I would like the solution as a value not an expression. Right now it looks like:
[-2/3 - 55**(1/3)*(-1/2 - sqrt(3)*I/2)/3, -2/3 - 55**(1/3)*(-1/2 + sqrt(3)*I/2)/3, -55**(1/3)/3 - 2/3]
I need the last expression's value (-2.22756). Are there functions in sympy to simplify this?
-
asmeurer about 11 yearsA better way to filter out real roots is to check
root.is_real
. -
Alex Pacini over 7 yearsI am trying to solve an equation using this method but it still returns imaginary solutions
-
asmeurer over 7 yearsAre you using solve? Note that the new
solveset
ignores assumptions set on Symbols (usesolveset(domain=S.Reals)
to solve in the real domain. If you are using solve, this is a bug which should be reported. -
Alex Pacini over 7 yearsI tried both: solve, with the assumptions, solveset and solveset with domain=S.Reals. Solve was simply skipping the assumptions on the variable to be solved (as solveset) arriving to the solutions in 30 seconds, while solveset with a domain was not able to arrive to the solution in 20 minutes. Could it be a duplicate? (github.com/sympy/sympy/issues/9973)