sympy

问题 I have the following code for a system of three nonlinear equations with three unknowns: import sympy as sp from sympy import symbols, cos, sin v0, a0, f0 = symbols('v0 a0 f0') v1, a1, f1 = symbols('v1 a1 f1') w, t = symbols('w t') g1 = v0 + a0 * w * cos(w*t + f0) - v1 - a1 * w * cos(f1) g2 = v0**2 + a0**2*w**2 -v1**2 - a1**2*w**2 g3 = a0 * sin(w*t + f0) - a1*sin(f1) sp.solvers.solve((g1,g2,g3), (a1,v1,f1)) The system of equations looks very complicated but actually it is easily solved with

问题 I'm experimenting with the sympy api for combinations. First, combinations without replacement ... from sympy.functions.combinatorial.numbers import nC from sympy.utilities.iterables import combinations nC('abc', 2) # >>> 3 list(combinations('abc', 2)) # >>> [('a', 'b'), ('a', 'c'), ('b', 'c')] I would now like to list the combinations with replacement nC('abc', 2, replacement=True) # >>> 6 However, the combinations() method doesn't seem to support a ' replacements ' argument? Init signature:

问题 I am trying to determine the equivalence of two simple functions passed to python via json like so: PHP: $data = array("2*x", "x*2"); $result = shell_exec('python /path/check.py ' . escapeshellarg(json_encode($data))); Python: import sys, json from sympy import * try: data = json.loads(sys.argv[1]) except: sys.exit(1) x = Symbol('x') response = data[0] answer = data[1] result = response==answer print json.dumps(result) My assumption is that result returns false because the response and answer

问题 I have a system of 3 equations, and I'd like to find a solution for the line of intersection, or the nullcline, of dx=dy . from sympy import * x, y, z = symbols('x, y, z') dx = x - x ** 3 / 3 - z + y dy = -y ** 2 * 0.1 + z dz = 0 xy_nullcline = nonlinsolve([dx, dy], [x, y, z]) print(xy_nullcline) # { # (x, -3.16227766016838*sqrt(z), z), # (x, 3.16227766016838*sqrt(z), z) # } In the image below, axes are x, y, z and: orange curve is the dx nullcline, (x,y,z), where dx=0 , a cubic polynomial

问题 I am trying to implement a exponential regression function. sp stands for sympy. I use numpy and sympy. Firstly, in func_exp I tried to use np.exp but it generated an error (attribute error), so I decided to use sympy instead. Well, this is the code import numpy as np from numpy.linalg import matrix_rank import scipy import scipy.integrate import random import matplotlib.pyplot as plt import matplotlib as mpl from mpl_toolkits.mplot3d import Axes3D from sympy import integrate import sympy as

问题 Suppose I have some expression from sympy import * a,b,c,x,y = symbols('a c b x y') eq=a*x + b*x*y + c*y**2 that needs to be split into an array containing the monomials. The current solution I have is parts = str(eq).split(' + ') Then, I use the eval function on each element of the array parts to be interpreted as an expression. What could I do to split the multivariate polynomial into the monomial parts without first converting the expression to a string? 回答1: You can explore a sympy

问题 I want to extend the Symbols class in SymPy so that I can add a Boolean attribute. I’m able to accomplish this for a single symbol (see my question here, and also someone else’s question here). And the code to accomplish this is repeated below: from sympy.core.symbol import Symbol class State(Symbol): def __init__(self, name, boolean_attr): self.boolean_attr = boolean_attr super(State, self).__init__(name) However, the problem with this solution is that when I am defining a polynomial, or

问题 I want to convert an expression containing symbolic variables to a numeric one so that the expression may be subsequently used in an integration method 'quad'. import numpy import math as m import scipy import sympy #define constants gammaee = 5.55e-6 MJpsi = 3.096916 alphaem = 1/137 lambdasq = 0.09 Ca = 3 qOsq = 2 def qbarsq(qsq): return (qsq+MJpsi**2)/4 def xx(qbarsq, w): return 4*qbarsq/(4*qbarsq-MJpsi**2+w**2) from sympy import * x,NN,a,b,ktsq,qbarsq,w = symbols('x NN a b ktsq qbarsq w')

问题 While using sympy (current version) to solve an polynomial equation (polynom would be d² in this case): from sympy import solve_poly_system solve_poly_system(4*d**2*sin(a)**2*sin(b)/cos(b)**2 - d*cos(a) + 4, d**2*sin(a)**2*sin(b)/cos(b)**2 - d*cos(a) + 8, 3*d**2*sin(a)**2*sin(b)/cos(b)**2 - d*cos(a) + 3 ,d ,a, b) I receive the following error: PolynomialError: cos(a) contains an element of the generators set What is the exact meaning of this error message. And why does it specifically points

问题 Why doesn't -(-1)**(1/3) + (-1)**(2/3) reduce to -1? wolfram alpha knows it's -1 but sympy gamma only does a float approximation re(_) + I*im(_) produces a NegativeOne object, but none of the other simplification functions I tried did anything to it. 回答1: I'm assuming you really mean -(-1)**Rational(1, 3) + (-1)**Rational(2, 3) , as literally -(-1)**(1/3) + (-1)**(2/3) is all Python (no SymPy), and evaluates numerically. Most SymPy objects do not do any kind of nontrivial simplification