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Python Matrix.simplify方法代码示例

本文整理汇总了Python中sympy.Matrix.simplify方法的典型用法代码示例。如果您正苦于以下问题:Python Matrix.simplify方法的具体用法?Python Matrix.simplify怎么用?Python Matrix.simplify使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在sympy.Matrix的用法示例。


在下文中一共展示了Matrix.simplify方法的6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,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: test_symbolic_twoport

# 需要导入模块: from sympy import Matrix [as 别名]
# 或者: from sympy.Matrix import simplify [as 别名]
def test_symbolic_twoport():
    circuit.default_toolkit = symbolic
    cir = SubCircuit()

    k = symbolic.kboltzmann
    var('R1 R0 C1 w T', real=True, positive=True)
    s = 1j*w

    cir['R0'] = R(1, gnd, r=R0)
    cir['R1'] = R(1, 2, r=R1)
#    cir['C1'] = C(2, gnd, c=C1)

    ## Add an AC source to verify that the source will not affect results
#    cir['IS'] = IS(1, gnd, iac=1) 

    ## Run symbolic 2-port analysis
    twoport_ana = TwoPortAnalysis(cir, Node('1'), gnd, Node('2'), gnd,
                                  noise = True, toolkit=symbolic,
                                  noise_outquantity = 'v')
    result = twoport_ana.solve(freqs=s, complexfreq=True)
    
    ABCD = Matrix(result['twoport'].A)
    ABCD.simplify()

    assert_array_equal(ABCD, np.array([[1 + 0*R1*C1*s, R1],
                                    [(1 + 0*R0*C1*s + 0*R1*C1*s) / R0,  (R0 + R1)/R0]]))

    assert_array_equal(simplify(result['Sin'] - (4*k*T/R0 + 4*R1*k*T/R0**2)), 0)
    assert_array_equal(simplify(result['Svn']), 4*k*T*R1)
开发者ID:gadamc,项目名称:pycircuit,代码行数:31,代码来源:test_analysis_nport.py

示例2: _mat_inv_mul

# 需要导入模块: from sympy import Matrix [as 别名]
# 或者: from sympy.Matrix import simplify [as 别名]
    def _mat_inv_mul(self, A, B):
        """Internal Function

        Computes A^-1 * B symbolically w/ substitution, where B is not
        necessarily a vector, but can be a matrix.

        """

        # Note: investigate difficulty in only creating symbols for non-zero
        # entries; this could speed things up, perhaps?

        r1, c1 = A.shape
        r2, c2 = B.shape
        temp1 = Matrix(r1, c1, lambda i, j: Symbol('x' + str(j + r1 * i)))
        temp2 = Matrix(r2, c2, lambda i, j: Symbol('y' + str(j + r2 * i)))
        for i in range(len(temp1)):
            if A[i] == 0:
                temp1[i] = 0
        for i in range(len(temp2)):
            if B[i] == 0:
                temp2[i] = 0
        temp3 = []
        for i in range(c2):
            temp3.append(temp1.LUsolve(temp2.extract(range(r2), [i])))
        temp3 = Matrix([i.T for i in temp3]).T
        if Kane.simp == True:
            temp3.simplify()
        return temp3.subs(dict(zip(temp1, A))).subs(dict(zip(temp2, B)))
开发者ID:ENuge,项目名称:sympy,代码行数:30,代码来源:kane.py

示例3: test_simplify

# 需要导入模块: from sympy import Matrix [as 别名]
# 或者: from sympy.Matrix import simplify [as 别名]
def test_simplify():
    x,y,f,n = symbols('xyfn')
    M = Matrix([ [    1/x + 1/y,               (x + x*y)/ x             ],
                 [(f(x) + y*f(x))/f(x), 2 * (1/n - cos(n * pi)/n)/ pi ]
                 ])
    M.simplify()
    assert M ==  Matrix([[(x + y)/(x * y),               1 + y           ],
                         [   1 + y,       (2 - 2*cos(pi*n))/(pi*n) ]])
开发者ID:Lucaweihs,项目名称:sympy,代码行数:10,代码来源:test_matrices.py

示例4: test_simplify

# 需要导入模块: from sympy import Matrix [as 别名]
# 或者: from sympy.Matrix import simplify [as 别名]
def test_simplify():
    x,y,f,n = symbols('x y f n')
    M = Matrix([ [    1/x + 1/y,               (x + x*y)/ x           ],
                 [(f(x) + y*f(x))/f(x), 2 * (1/n - cos(n * pi)/n)/ pi ]
                 ])
    M.simplify()
    assert M ==  Matrix([[(x + y)/(x * y),                 1 + y       ],
                         [   1 + y,       2*((1 - 1*cos(pi*n))/(pi*n)) ]])
    M = Matrix([[(1 + x)**2]])
    M.simplify()
    assert M == Matrix([[(1 + x)**2]])
    M.simplify(ratio=oo)
    assert M == Matrix([[1 + 2*x + x**2]])
开发者ID:robotment,项目名称:sympy,代码行数:15,代码来源:test_matrices.py

示例5: test_linearize_pendulum_kane_nonminimal

# 需要导入模块: from sympy import Matrix [as 别名]
# 或者: from sympy.Matrix import simplify [as 别名]
def test_linearize_pendulum_kane_nonminimal():
    # Create generalized coordinates and speeds for this non-minimal realization
    # q1, q2 = N.x and N.y coordinates of pendulum
    # u1, u2 = N.x and N.y velocities of pendulum
    q1, q2 = dynamicsymbols('q1:3')
    q1d, q2d = dynamicsymbols('q1:3', level=1)
    u1, u2 = dynamicsymbols('u1:3')
    u1d, u2d = dynamicsymbols('u1:3', level=1)
    L, m, t = symbols('L, m, t')
    g = 9.8

    # Compose world frame
    N = ReferenceFrame('N')
    pN = Point('N*')
    pN.set_vel(N, 0)

    # A.x is along the pendulum
    theta1 = atan(q2/q1)
    A = N.orientnew('A', 'axis', [theta1, N.z])

    # Locate the pendulum mass
    P = pN.locatenew('P1', q1*N.x + q2*N.y)
    pP = Particle('pP', P, m)

    # Calculate the kinematic differential equations
    kde = Matrix([q1d - u1,
                  q2d - u2])
    dq_dict = solve(kde, [q1d, q2d])

    # Set velocity of point P
    P.set_vel(N, P.pos_from(pN).dt(N).subs(dq_dict))

    # Configuration constraint is length of pendulum
    f_c = Matrix([P.pos_from(pN).magnitude() - L])

    # Velocity constraint is that the velocity in the A.x direction is
    # always zero (the pendulum is never getting longer).
    f_v = Matrix([P.vel(N).express(A).dot(A.x)])
    f_v.simplify()

    # Acceleration constraints is the time derivative of the velocity constraint
    f_a = f_v.diff(t)
    f_a.simplify()

    # Input the force resultant at P
    R = m*g*N.x

    # Derive the equations of motion using the KanesMethod class.
    KM = KanesMethod(N, q_ind=[q2], u_ind=[u2], q_dependent=[q1],
            u_dependent=[u1], configuration_constraints=f_c,
            velocity_constraints=f_v, acceleration_constraints=f_a, kd_eqs=kde)
    (fr, frstar) = KM.kanes_equations([(P, R)], [pP])

    # Set the operating point to be straight down, and non-moving
    q_op = {q1: L, q2: 0}
    u_op = {u1: 0, u2: 0}
    ud_op = {u1d: 0, u2d: 0}

    A, B, inp_vec = KM.linearize(op_point=[q_op, u_op, ud_op], A_and_B=True,
            new_method=True, simplify=True)

    assert A == Matrix([[0, 1], [-9.8/L, 0]])
    assert B == Matrix([])
开发者ID:Festy,项目名称:sympy,代码行数:65,代码来源:test_linearize.py

示例6: linearize

# 需要导入模块: from sympy import Matrix [as 别名]
# 或者: from sympy.Matrix import simplify [as 别名]
    def linearize(self, op_point=None, A_and_B=False, simplify=False):
        """Linearize the system about the operating point. Note that
        q_op, u_op, qd_op, ud_op must satisfy the equations of motion.
        These may be either symbolic or numeric.

        Parameters
        ----------
        op_point : dict or iterable of dicts, optional
            Dictionary or iterable of dictionaries containing the operating
            point conditions. These will be substituted in to the linearized
            system before the linearization is complete. Leave blank if you
            want a completely symbolic form. Note that any reduction in
            symbols (whether substituted for numbers or expressions with a
            common parameter) will result in faster runtime.

        A_and_B : bool, optional
            If A_and_B=False (default), (M, A, B) is returned for forming
            [M]*[q, u]^T = [A]*[q_ind, u_ind]^T + [B]r. If A_and_B=True,
            (A, B) is returned for forming dx = [A]x + [B]r, where
            x = [q_ind, u_ind]^T.

        simplify : bool, optional
            Determines if returned values are simplified before return.
            For large expressions this may be time consuming. Default is False.

        Note that the process of solving with A_and_B=True is computationally
        intensive if there are many symbolic parameters. For this reason,
        it may be more desirable to use the default A_and_B=False,
        returning M, A, and B. More values may then be substituted in to these
        matrices later on. The state space form can then be found as
        A = P.T*M.LUsolve(A), B = P.T*M.LUsolve(B), where
        P = Linearizer.perm_mat.
        """

        # Compose dict of operating conditions
        if isinstance(op_point, dict):
            op_point_dict = op_point
        elif isinstance(op_point, collections.Iterable):
            op_point_dict = {}
            for op in op_point:
                op_point_dict.update(op)
        else:
            op_point_dict = {}

        # Extract dimension variables
        l, m, n, o, s, k = self._dims

        # Rename terms to shorten expressions
        M_qq = self._M_qq
        M_uqc = self._M_uqc
        M_uqd = self._M_uqd
        M_uuc = self._M_uuc
        M_uud = self._M_uud
        M_uld = self._M_uld
        A_qq = self._A_qq
        A_uqc = self._A_uqc
        A_uqd = self._A_uqd
        A_qu = self._A_qu
        A_uuc = self._A_uuc
        A_uud = self._A_uud
        B_u = self._B_u
        C_0 = self._C_0
        C_1 = self._C_1
        C_2 = self._C_2

        # Build up Mass Matrix
        #     |M_qq    0_nxo   0_nxk|
        # M = |M_uqc   M_uuc   0_mxk|
        #     |M_uqd   M_uud   M_uld|
        if o != 0:
            col2 = Matrix([zeros(n, o), M_uuc, M_uud])
        if k != 0:
            col3 = Matrix([zeros(n + m, k), M_uld])
        if n != 0:
            col1 = Matrix([M_qq, M_uqc, M_uqd])
            if o != 0 and k != 0:
                M = col1.row_join(col2).row_join(col3)
            elif o != 0:
                M = col1.row_join(col2)
            else:
                M = col1
        elif k != 0:
            M = col2.row_join(col3)
        else:
            M = col2
        M_eq = _subs_keep_derivs(M, op_point_dict)

        # Build up state coefficient matrix A
        #     |(A_qq + A_qu*C_1)*C_0       A_qu*C_2|
        # A = |(A_uqc + A_uuc*C_1)*C_0    A_uuc*C_2|
        #     |(A_uqd + A_uud*C_1)*C_0    A_uud*C_2|
        # Col 1 is only defined if n != 0
        if n != 0:
            r1c1 = A_qq
            if o != 0:
                r1c1 += (A_qu * C_1)
            r1c1 = r1c1 * C_0
            if m != 0:
                r2c1 = A_uqc
                if o != 0:
#.........这里部分代码省略.........
开发者ID:Eskatrem,项目名称:sympy,代码行数:103,代码来源:linearize.py


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