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Python sympy.solve函数代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

本文整理汇总了Python中sympy.solve函数的典型用法代码示例。如果您正苦于以下问题:Python solve函数的具体用法?Python solve怎么用?Python solve使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。



在下文中一共展示了solve函数的20个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。

示例1: SequentialSolving

def SequentialSolving(eqns):
    sorteq = SortEquations(eqns)
    seqsol = {}
    unkn = sorted(sorteq[0].atoms(sy.Symbol)) 
    num_unkn = len(unkn)
    while num_unkn == 1 and not sorteq == []:  
        val_unkn = sy.solve(sorteq[0], unkn[0], simplify = False) 
        seqsol[unkn[0]] = sy.Rational(str(round(val_unkn[0], 4)))
    unkn = sorted(sorteq[0].atoms(sy.Symbol))    
    num_unkn = len(unkn)
    while num_unkn == 1:    
        val_unkn = sy.solve(sorteq[0], unkn[0])   
        seqsol[unkn[0]] = val_unkn[0]
        i = 0
        for eq in sorteq:
            if unkn[0] in sorteq[i]:    
                val_unkn[0] = round(val_unkn[0], 5)    
                repl = eq.subs(unkn[0], sy.Rational(str(val_unkn[0])))
                if not isinstance(repl, sy.Float):
                    sorteq[i] = repl
                else:
                    sorteq[i] = repl
            i+=1 
        
        sorteq = RemoveZeros(sorteq)
        if sorteq == []:
            num_unkn = 0
        else:
            sorteq = SortEquations(sorteq)
            unkn = sorted(sorteq[0].atoms(sy.Symbol))
            num_unkn = len(unkn)
                
    return seqsol
开发者ID:Johies,项目名称:CSC411-JJ-Coetzee,代码行数:33,代码来源:solverc.py


示例2: benefit_from_demand

def benefit_from_demand(x, p, demand):
    """Converts the demand curve to the benefit.  It assumes that the
    demand is a x=d(p) function, where the x= is implicit.

    >>> sp.var('x p')
    (x, p)
    >>> sp.simplify(benefit_from_demand(x, p, 10/p -1) -
    ...             10*sp.log(x+1))
    0
    >>> sp.simplify(benefit_from_demand(x, p, sp.Eq(x, 10/p -1)) -
    ...             10*sp.log(x+1))
    0
    >>> sp.simplify(benefit_from_demand(x, p, 100-p) -
    ...             (-x**2/2 + 100*x))
    0
    >>> benefit_from_demand(x, p, sp.Piecewise((0, p < 0),
    ...                                        (-p + 100, p <= 100),
    ...                                        (0, True)))
    -x**2/2 + 100*x
    """
    if isinstance(demand, sp.relational.Relational):
        return sp.integrate(sp.solve(demand, p)[0], (x, 0, x))
    substracting = sp.solve(demand-x, p)
    if substracting:
        toint = substracting[0]
    else:
        substracting = sp.solve(demand, p)
        if substracting:
            toint = substracting[0] - x
        else:
            return None

    return sp.integrate(toint, (x, 0, x))
开发者ID:juanre,项目名称:econopy,代码行数:33,代码来源:tools.py


示例3: getfunc

def getfunc(p1, p2, p3, p4):
    """ Get a point-returning function for a cubic
        curve over four points, with domain [0 - 3].
    """
    # knowns
    points = p1, p2, p3, p4

    # unknowns
    a, b, c, d, e, f, g, h = [Symbol(n) for n in 'abcdefgh']

    # coefficients
    xco = solve([(a * i**3 + b * i**2 + c * i + d - p.x) for (i, p) in enumerate(points)], [a, b, c, d])
    yco = solve([(e * i**3 + f * i**2 + g * i + h - p.y) for (i, p) in enumerate(points)], [e, f, g, h])

    # shorter variable names
    a, b, c, d = [xco[n] for n in (a, b, c, d)]
    e, f, g, h = [yco[n] for n in (e, f, g, h)]

    def func(t, d1=False):
        """ Return a position for given t or velocity (1st derivative) if arg. is True.
        """
        if d1:
            # first derivative
            return Point(3 * a * t**2 + 2 * b * t + c,
                         3 * e * t**2 + 2 * f * t + g)

        else:
            # actual function
            return Point(a * t**3 + b * t**2 + c * t + d,
                         e * t**3 + f * t**2 + g * t + h)

    return func
开发者ID:straup,项目名称:py-modestMMarkers,代码行数:32,代码来源:spline.py


示例4: print_assignment

def print_assignment(eq, s, s0=0, pochoir=False):
	s1 = print_myccode(s, None, pochoir)
	if(s0==0):
		s2 = print_myccode(simplify(solve(eq,s)[0]), None, pochoir)
	else:
		s2 = print_myccode(simplify(solve(eq,s)[0] - s0) + s0, None, pochoir)
	return s1 + '=' + s2
开发者ID:tj-sun,项目名称:propagator,代码行数:7,代码来源:fdlib.py


示例5: __init__

 def __init__(self):
     h, g, h0, g0, a, d = sympy.symbols('h g h0 g0 a d')
     #g1 = sympy.Max(-d, g0 + h0 - 1/(4*a))
     g1 = g0 + h0 - 1/(4*a)
     h1 = h0 - 1/(2*a)
     parabola = d - a*h*h       # =g on boundary
     slope = g1 + h - h1        # =g on line with slope=1 through z1
     h2 = sympy.solve(parabola-slope,h)[0] # First solution is above z1
     g2 = h2 - h1 + g1          # Line has slope of 1
     g3 = g1
     h3 = sympy.solve((parabola-g).subs(g, g1),h)[1] # Second is on right
     r_a = sympy.Rational(1,24) # a=1/24 always
     self.h = tuple(x.subs(a,r_a) for x in (h0, h1, h2, h3))
     self.g = tuple(x.subs(a,r_a) for x in (g0, g1, g2, g3))
     def integrate(f):
         ia = sympy.integrate(
             sympy.integrate(f,(g, g1, g1+h-h1)),
             (h, h1, h2)) # Integral of f over right triangle
         ib = sympy.integrate(
             sympy.integrate(f,(g, g1, parabola)),
             (h, h2, h3)) # Integral of f over region against parabola
         return (ia+ib).subs(a, r_a)
     i0 = integrate(1)               # Area = integral of pie slice
     E = lambda f:(integrate(f)/i0)  # Expected value wrt Lebesgue measure
     sigma = lambda f,g:E(f*g) - E(f)*E(g)
     self.d = d
     self.Eh = collect(E(h),sympy.sqrt(d-g0-h0+6))
     self.Eg = E(g)
     self.Sigmahh = sigma(h,h)
     self.Sigmahg = sigma(h,g)
     self.Sigmagg = sigma(g,g)
     return
开发者ID:fraserphysics,项目名称:metfie,代码行数:32,代码来源:eric.py


示例6: solve_eq

def solve_eq(rho_m, u_m, u_s):
    u_max, u_star, rho_max, rho_star, A, B = sympy.symbols("u_max u_star rho_max rho_star A B")

    eq1 = sympy.Eq(0, u_max * rho_max * (1 - A * rho_max - B * rho_max ** 2))
    eq2 = sympy.Eq(0, u_max * (1 - 2 * A * rho_star - 3 * B * rho_star ** 2))
    eq3 = sympy.Eq(u_star, u_max * (1 - A * rho_star - B * rho_star ** 2))

    eq4 = sympy.Eq(eq2.lhs - 3 * eq3.lhs, eq2.rhs - 3 * eq3.rhs)
    eq4.simplify()
    eq4.expand()

    rho_sol = sympy.solve(eq4, rho_star)[0]
    B_sol = sympy.solve(eq1, B)[0]

    quadA = eq2.subs([(rho_star, rho_sol), (B, B_sol)])
    quadA.simplify()

    A_sol = sympy.solve(quadA, A)[0]

    aval = A_sol.evalf(subs={u_star: u_s, u_max: u_m, rho_max: rho_m})
    bval = B_sol.evalf(subs={rho_max: rho_m, A: aval})

    rho_sol = sympy.solve(eq2, rho_star)[0]
    rho_val = rho_sol.evalf(subs={u_max: u_m, A: aval, B: bval})

    return aval, bval, rho_val
开发者ID:archphy,项目名称:numerical-mooc-1,代码行数:26,代码来源:traffic_03.py


示例7: PlotFor2Param

def PlotFor2Param(k1, k2, x, y, yVal, eq1, eq2, eqDet, eqTr, valK1m, valK3m, valK2 = 0.95, valK3 = 0.0032):
    eqk1_1Det = solve(eqDet.subs(x, eq2), k1)
    eqForK1Det = eq1 - eqk1_1Det[0]
    eqK2Det = solve(eqForK1Det.subs(km1, valK1m).subs(km3, valK3m).subs(k3, valK3), k2)
    funcK2Det = lambdify(y, eqK2Det[0])
    funcK1Det = lambdify(y, eqk1_1Det[0].subs(x, eq2).subs(k2, eqK2Det[0]).subs(km1, valK1m).subs(km3, valK3m).subs(k3, valK3))
    k1DetAll = funcK1Det(yVal)
    k2DetAll = funcK2Det(yVal)
    k1DetPos = [k1DetAll[i] for i in range(len(k1DetAll)) if k1DetAll[i] > 0 and k2DetAll[i] > 0]
    k2DetPos = [k2DetAll[i] for i in range(len(k2DetAll)) if k1DetAll[i] > 0 and k2DetAll[i] > 0]
    hopf, = plt.plot(k1DetPos, k2DetPos, linestyle = '--', color = 'r', label = 'hopf line')

    eqk1_1Tr = solve(eqTr.subs(x, eq2), k1)
    eqForK1Tr = eq1 - eqk1_1Tr[0]
    eqK2Tr = solve(eqForK1Tr.subs(km1, valK1m).subs(km3, valK3m).subs(k3, valK3), k2)
    funcK2Tr = lambdify(y, eqK2Tr[0])
    funcK1Tr = lambdify(y, eq1.subs(x, eq2).subs(k2, eqK2Tr[0]).subs(km1, valK1m).subs(km3, valK3m).subs(k3, valK3))
    k1AllTr = funcK1Tr(yVal)
    k2AllTr = funcK2Tr(yVal)
    k1PosTr = [k1AllTr[i] for i in range(len(k1AllTr)) if k1AllTr[i] > 0 and k2AllTr[i] > 0]
    print(len(k1PosTr))
    k2PosTr = [k2AllTr[i] for i in range(len(k2AllTr)) if k1AllTr[i] > 0 and k2AllTr[i] > 0]
    print(len(k2PosTr))
    sadle, = plt.plot(k1PosTr, k2PosTr, color = 'g', label = 'sadle-nodle line')

    plt.xlim([0, 2])
    plt.ylim([0,5])
    plt.legend(handles=[hopf, sadle])
    plt.xlabel('k1')
    plt.ylabel('k2')
    plt.show()
开发者ID:ChesnakovKonstantin,项目名称:Test-Example,代码行数:31,代码来源:PythonApplication4.py


示例8: test_issue_1572_1364_1368

def test_issue_1572_1364_1368():
    assert solve((sqrt(x**2 - 1) - 2)) in ([sqrt(5), -sqrt(5)],
                                           [-sqrt(5), sqrt(5)])
    assert set(solve((2**exp(y**2/x) + 2)/(x**2 + 15), y)) == set([
        -sqrt(x)*sqrt(-log(log(2)) + log(log(2) + I*pi)),
        sqrt(x)*sqrt(-log(log(2)) + log(log(2) + I*pi))])

    C1, C2 = symbols('C1 C2')
    f = Function('f')
    assert solve(C1 + C2/x**2 - exp(-f(x)), f(x)) == [log(x**2/(C1*x**2 + C2))]
    a = Symbol('a')
    E = S.Exp1
    assert solve(1 - log(a + 4*x**2), x) in (
        [-sqrt(-a + E)/2, sqrt(-a + E)/2],
        [sqrt(-a + E)/2, -sqrt(-a + E)/2]
    )
    assert solve(log(a**(-3) - x**2)/a, x) in (
        [-sqrt(-1 + a**(-3)), sqrt(-1 + a**(-3))],
        [sqrt(-1 + a**(-3)), -sqrt(-1 + a**(-3))],)
    assert solve(1 - log(a + 4*x**2), x) in (
        [-sqrt(-a + E)/2, sqrt(-a + E)/2],
        [sqrt(-a + E)/2, -sqrt(-a + E)/2],)
    assert set(solve((
        a**2 + 1) * (sin(a*x) + cos(a*x)), x)) == set([-pi/(4*a), 3*pi/(4*a)])
    assert solve(3 - (sinh(a*x) + cosh(a*x)), x) == [2*atanh(S.Half)/a]
    assert set(solve(3 - (sinh(a*x) + cosh(a*x)**2), x)) == \
        set([
            2*atanh(-1 + sqrt(2))/a,
            2*atanh(S(1)/2 + sqrt(5)/2)/a,
            2*atanh(-sqrt(2) - 1)/a,
            2*atanh(-sqrt(5)/2 + S(1)/2)/a
             ])
    assert solve(atan(x) - 1) == [tan(1)]
开发者ID:Maihj,项目名称:sympy,代码行数:33,代码来源:test_solvers.py


示例9: generate_x

    def generate_x(self, N=1000):
        '''
        this sampling method works wit all margins,
        but only frank and independent copulas can be used
        and it is limited to 2 dimensions
        '''
        # compute marginal prob of u1
        P_U1 = sy.simplify(sy.diff(self.C,self.U[0]))

        # invert marginal prob of u1
        y = sy.symbols('y')
        tmp = sy.solve(sy.Eq(P_U1,y),self.U[1])
        inv_P_U1 = sy.lambdify((self.U[0],y,self.D),tmp,'numpy')

        # invert margins
        inv_F = {}
        for m in [0,1]:
            u = sy.symbols('u')
            tmp = sy.solve(sy.Eq(self.F[m],u),self.X[m])[0]
            inv_F[m] = sy.lambdify((u,self.P[m]),tmp,'numpy')


        X = np.zeros((N,2))
        for i in range(N):
            u0, y = np.random.uniform(size=2)
            u1 = inv_P_U1(u0,y,self.C_para)
            X[i,0] = inv_F[0](u0,self.F_para[0])
            X[i,1] = inv_F[1](u1,self.F_para[1])

        return X
开发者ID:RWalecki,项目名称:copula_sandbox,代码行数:30,代码来源:copula_sandbox.py


示例10: test_minsolve_linear_system

def test_minsolve_linear_system():
    def count(dic):
        return len([x for x in dic.itervalues() if x == 0])
    assert count(solve([x + y + z, y + z + a + t], minimal=True, quick=True)) == 3
    assert count(solve([x + y + z, y + z + a + t], minimal=True, quick=False)) == 3
    assert count(solve([x + y + z, y + z + a], minimal=True, quick=True)) == 1
    assert count(solve([x + y + z, y + z + a], minimal=True, quick=False)) == 2
开发者ID:amar47shah,项目名称:sympy,代码行数:7,代码来源:test_solvers.py


示例11: test_issue_2813

def test_issue_2813():
    assert solve(x ** 2 - x - 0.1, rational=True) == [S(1) / 2 + sqrt(35) / 10, -sqrt(35) / 10 + S(1) / 2]
    # [-0.0916079783099616, 1.09160797830996]
    ans = solve(x ** 2 - x - 0.1, rational=False)
    assert len(ans) == 2 and all(a.is_Number for a in ans)
    ans = solve(x ** 2 - x - 0.1)
    assert len(ans) == 2 and all(a.is_Number for a in ans)
开发者ID:vipulnsward,项目名称:sympy,代码行数:7,代码来源:test_solvers.py


示例12: test_CRootOf___eval_Eq__

def test_CRootOf___eval_Eq__():
    f = Function('f')
    eq = x**3 + x + 3
    r = rootof(eq, 2)
    r1 = rootof(eq, 1)
    assert Eq(r, r1) is S.false
    assert Eq(r, r) is S.true
    assert Eq(r, x) is S.false
    assert Eq(r, 0) is S.false
    assert Eq(r, S.Infinity) is S.false
    assert Eq(r, I) is S.false
    assert Eq(r, f(0)) is S.false
    assert Eq(r, f(0)) is S.false
    sol = solve(eq)
    for s in sol:
        if s.is_real:
            assert Eq(r, s) is S.false
    r = rootof(eq, 0)
    for s in sol:
        if s.is_real:
            assert Eq(r, s) is S.true
    eq = x**3 + x + 1
    sol = solve(eq)
    assert [Eq(rootof(eq, i), j) for i in range(3) for j in sol] == [
        False, False, True, False, True, False, True, False, False]
    assert Eq(rootof(eq, 0), 1 + S.ImaginaryUnit) == False
开发者ID:cmarqu,项目名称:sympy,代码行数:26,代码来源:test_rootoftools.py


示例13: test_solve_inequalities

def test_solve_inequalities():
    system = [Lt(x ** 2 - 2, 0), Gt(x ** 2 - 1, 0)]

    assert solve(system) == And(
        Or(And(Lt(-sqrt(2), re(x)), Lt(re(x), -1)), And(Lt(1, re(x)), Lt(re(x), sqrt(2)))), Eq(im(x), 0)
    )
    assert solve(system, assume=Q.real(x)) == Or(And(Lt(-sqrt(2), x), Lt(x, -1)), And(Lt(1, x), Lt(x, sqrt(2))))
开发者ID:vipulnsward,项目名称:sympy,代码行数:7,代码来源:test_solvers.py


示例14: test_RootOf___eval_Eq__

def test_RootOf___eval_Eq__():
    f = Function('f')
    r = RootOf(x**3 + x + 3, 2)
    r1 = RootOf(x**3 + x + 3, 1)
    assert Eq(r, r1) is S.false
    assert Eq(r, r) is S.true
    assert Eq(r, x) is S.false
    assert Eq(r, 0) is S.false
    assert Eq(r, S.Infinity) is S.false
    assert Eq(r, I) is S.false
    assert Eq(r, f(0)) is S.false
    assert Eq(r, f(0)) is S.false
    sol = solve(r.expr)
    for s in sol:
        if s.is_real:
            assert Eq(r, s) is S.false
    r = RootOf(r.expr, 0)
    for s in sol:
        if s.is_real:
            assert Eq(r, s) is S.true
    eq = (x**3 + x + 1)
    assert [Eq(RootOf(eq, i), j) for i in range(3) for j in solve(eq)] == [
        False, False, True, False, True, False, True, False, False
    ]
    assert Eq(RootOf(eq, 0), 1 + S.ImaginaryUnit) == False
开发者ID:artcompiler,项目名称:artcompiler.github.com,代码行数:25,代码来源:test_rootoftools.py


示例15: __init__

    def __init__(self, width, height, x_radius=None,
                 curve=CIRCULAR):
        self.width = width
        self.height = height
        self.x_radius = x_radius
        self.curve = curve

        self.x, self.y, dx, dy = sympy.symbols("x y dx dy", real=True)
        a, b = sympy.symbols("a b", positive=True, real=True)

        self.ellipse = ((self.x - dx) / a) ** 2 + ((self.y - dy) / b) ** 2 - 1

        self.ellipse = self.ellipse.subs([(dx, self.width),
                                          (dy, b)])

        if curve == Transition.CIRCULAR:
            self.ellipse = self.ellipse.subs([(a, b)])

        ellipse = self.ellipse.subs([(self.x, 0),
                                     (self.y, self.height)])

        if curve == Transition.CIRCULAR:
            self.x_radius = self.y_radius = max(sympy.solve(ellipse, b))
            self.angle = math.asin(self.width / self.x_radius)
            self.arc_length = self.x_radius * self.angle

        elif curve == Transition.ELLIPTICAL:
            self.y_radius = max(sympy.solve(ellipse, b))
            # TODO: Fix me
            self.angle = 0
            self.arc_length = 0


        self.ellipse = self.ellipse.subs([(a, self.x_radius),
                                          (b, self.y_radius)])
开发者ID:oampo,项目名称:mini-ramp,代码行数:35,代码来源:ramp.py


示例16: __init__

    def __init__(self):
        self._qp = {}
        function_type = random.choice([sympy.sin, sympy.cos, 'linear', 'quadratic'])

        if function_type in [sympy.sin, sympy.cos]:
            while True:
                self._qp['domain'] = domains.integer_domain(low=0, high=2, minimum_distance=1)
                m = random.choice([sympy.pi / 2, sympy.pi])
                self._qp['equation'] = k * function_type(m * x)

                area = self._qp['equation'].integrate((x, self._qp['domain'].left, self._qp['domain'].right))

                solution = sympy.solve(area - 1)
                if len(solution) == 0:
                    continue

                self._qp['k'] = solution[0]
                break

        elif function_type == 'linear':
            self._qp['domain'] = domains.integer_domain()
            m = random.randint(1, 2)
            a = random.randint(7, 12)
            self._qp['equation'] = ((m * x + k) / a).together()

            area = self._qp['equation'].integrate((x, self._qp['domain'].left, self._qp['domain'].right))
            self._qp['k'] = sympy.solve(area - 1)[0]

        elif function_type == 'quadratic':
            self._qp['domain'] = domains.integer_domain()
            self._qp['equation'] = k * (x - self._qp['domain'].left) * (x - self._qp['domain'].right)

            area = self._qp['equation'].integrate((x, self._qp['domain'].left, self._qp['domain'].right))
            self._qp['k'] = sympy.solve(area - 1)[0]
开发者ID:nebffa,项目名称:MathsExams,代码行数:34,代码来源:piecewise_prob_density_function.py


示例17: test_polysys

def test_polysys():
    assert set(solve([x**2 + 2/y - 2 , x + y - 3], [x, y])) == \
        set([(S(1), S(2)), (1 + sqrt(5), 2 - sqrt(5)),
        (1 - sqrt(5), 2 + sqrt(5))])
    assert solve([x**2 + y - 2, x**2 + y]) == []
    # the ordering should be whatever the user requested
    assert solve([x**2 + y - 3, x - y - 4], (x, y)) != solve([x**2 + y - 3, x - y - 4], (y, x))
开发者ID:StefenYin,项目名称:sympy,代码行数:7,代码来源:test_solvers.py


示例18: qExponentialSameBase_template

def qExponentialSameBase_template():
    '''Solves the same base e.g. 2^(2x+1) = 32.'''
    base = choice([2,3,5,7])
    pow_rs = randint(3,6)
    rs = int(pow(base,pow_rs))
    front_num = randint(-100,100)
    while front_num == 0:
        front_num = randint(-100,100)
    lspow = front_num*x+randint(-100,100)
    question = 'Solve ' + tostring(am.parse('%s^(%s) = %s' % (base, lspow, rs)))
    ls_samebase = tostring(am.parse('%s^(%s)' % (base, lspow)))
    rs_samebase = tostring(am.parse('%s^(%s)' % (base, pow_rs)))
    steps = []
    steps.append('Covert right side to be same base as left side. Left side' \
                 ' has a base of: ' + str(base))
    steps.append('As ' + tostring(am.parse('%s^(%s)=%s' % 
                                                 (base, pow_rs, rs))))
    steps.append('Right side is now: ' + tostring(am.parse('%s^(%s)' % 
                                                           (base, pow_rs))))
    steps.append('Therefore ' + ls_samebase + tostring(am.parse('=')) + \
                 rs_samebase)
    steps.append('Therefore solve: ' + tostring(am.parse('%s%s%s' %
                                                        (lspow,'=',pow_rs)))) 
    steps.append('Note: As bases are same the power equates to each other.')
    answer = []
    answer.append(steps)

    if len(solve(Eq(lspow, rs))) > 1:
        answer.append(tostring(am.parse('x = %s' % solve(Eq(lspow, rs))[0])))
    else:
        answer.append(tostring(am.parse('x = %s' % solve(Eq(lspow, rs))[0])))

    return question, answer
开发者ID:shaond,项目名称:mathbrain,代码行数:33,代码来源:mathbrain.py


示例19: test_swap_back

def test_swap_back():
    f, g = map(Function, 'fg')
    fx, gx = f(x), g(x)
    assert solve([fx + y - 2, fx - gx - 5], fx, y, gx) == \
                 {fx: gx + 5, y: -gx - 3}
    assert solve(fx + gx*x - 2, [fx, gx]) == {fx: 2, gx: 0}
    assert solve(fx + gx**2*x - y, [fx, gx]) == [{fx: y - gx**2*x}]
开发者ID:Enchanter12,项目名称:sympy,代码行数:7,代码来源:test_solvers.py


示例20: test_exclude

def test_exclude():
    R, C, Ri, Vout, V1, Vminus, Vplus, s = \
        symbols('R, C, Ri, Vout, V1, Vminus, Vplus, s')
    Rf = symbols('Rf', positive=True)  # to eliminate Rf = 0 soln
    eqs = [C*V1*s + Vplus*(-2*C*s - 1/R),
           Vminus*(-1/Ri - 1/Rf) + Vout/Rf,
           C*Vplus*s + V1*(-C*s - 1/R) + Vout/R,
           -Vminus + Vplus]
    assert solve(eqs, exclude=s*C*R) == [
        {
            Rf: Ri*(C*R*s + 1)**2/(C*R*s),
            Vminus: Vplus,
            V1: Vplus*(2*C*R*s + 1)/(C*R*s),
            Vout: Vplus*(C**2*R**2*s**2 + 3*C*R*s + 1)/(C*R*s)},
        {
            Vplus: 0,
            Vminus: 0,
            V1: 0,
            Vout: 0},
    ]
    assert solve(eqs, exclude=[Vplus, s, C]) == [
        {
            Rf: Ri*(V1 - Vplus)**2/(Vplus*(V1 - 2*Vplus)),
            Vminus: Vplus,
            Vout: (V1**2 - V1*Vplus - Vplus**2)/(V1 - 2*Vplus),
            R: Vplus/(C*s*(V1 - 2*Vplus))}]
开发者ID:Maihj,项目名称:sympy,代码行数:26,代码来源:test_solvers.py



注:本文中的sympy.solve函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。


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