Hàm bốn trong python 2024

Xem Hàm bốn trong python 2024

Tôi muốn sử dụng tích hợp. quad với một mảng nhưng nó trả về cho tôi. “Lỗi loại. chỉ các mảng có kích thước 1 mới có thể được chuyển đổi thành vô hướng Python”

Tôi hiểu rằng đối số đầu tiên được yêu cầu phải là một số vô hướng. Nhưng đối số tôi muốn sử dụng đến từ một hàm phụ thuộc vào một tham số và trả về một mảng và tôi không thể khắc phục sự cố

Đây là kịch bản của tôi trong Python

from scipy import integrate as intg

DfH_ref_jp = np.array([0,-393.52,-110.53,-74.87,-241.83,0]) *1000
n_jp = np.array([1.2,0.2,0.15,0.001,0.49,0.30])
Tref = 1298
Ta = 1310

def c_jp(T):
    t = T/1000
    XC = np.array([1,t,t**2,t**3,t**(-2)])
    Mah = M_ah(T)                            # a matrix defined before
    c = np.dot(Mah[:,:5],XC)

    return c

H1_jp = n_jp * (DfH_ref_jp + intg.quad(c_jp,Tref,Ta)[0])   # this where it doesn't work / [0] because intg.quad returns an array and I want the first value

Vì vậy, tôi đã thử với một hàm trả về vô hướng

def c_jp0(T):
    t = T/1000
    XC = np.array([1,t,t**2,t**3,t**(-2)])
    Mah = M_ah(T)
    c = np.dot(Mah[0,:5],XC)

    return c

H1_jp = np.zeros(6, dtype=float)
H1_jp[0] = n_jp[0] * (DfH_ref_jp[0] + intg.quad(c_jp0,Tref,Ta)[0]) 

Nó hoạt động nhưng tôi không muốn chỉ định 6 chức năng. c_jp0. c_jp6. Có ai có một ý tưởng làm thế nào để làm điều đó?

help(scipy.integrate.odeint)

sẽ hiển thị

Help on function odeint in module scipy.integrate.odepack:

odeint(func, y0, t, args=(), Dfun=None, col_deriv=0, full_output=0, ml=None, mu=None, rtol=None, atol=None, tcrit=None, h0=0.0, hmax=0.0, hmin=0.0, ixpr=0, mxstep=0, mxhnil=0, mxordn=12, mxords=5, printmessg=0)
    Integrate a system of ordinary differential equations.

    Solve a system of ordinary differential equations using lsoda from the
    FORTRAN library odepack.

    Solves the initial value problem for stiff or non-stiff systems
    of first order ode-s::

        dy/dt = func(y, t0, ...)

    where y can be a vector.

    *Note*: The first two arguments of ``func(y, t0, ...)`` are in the
    opposite order of the arguments in the system definition function used
    by the `scipy.integrate.ode` class.

    Parameters
    ----------
    func : callable(y, t0, ...)
        Computes the derivative of y at t0.
    y0 : array
        Initial condition on y (can be a vector).
    t : array
        A sequence of time points for which to solve for y.  The initial
        value point should be the first element of this sequence.
    args : tuple, optional
        Extra arguments to pass to function.
    Dfun : callable(y, t0, ...)
        Gradient (Jacobian) of `func`.
    col_deriv : bool, optional
        True if `Dfun` defines derivatives down columns (faster),
        otherwise `Dfun` should define derivatives across rows.
    full_output : bool, optional
        True if to return a dictionary of optional outputs as the second output
    printmessg : bool, optional
        Whether to print the convergence message

    Returns
    -------
    y : array, shape (len(t), len(y0))
        Array containing the value of y for each desired time in t,
        with the initial value `y0` in the first row.
    infodict : dict, only returned if full_output == True
        Dictionary containing additional output information

        =======  ============================================================
        key      meaning
        =======  ============================================================
        'hu'     vector of step sizes successfully used for each time step.
        'tcur'   vector with the value of t reached for each time step.
                 (will always be at least as large as the input times).
        'tolsf'  vector of tolerance scale factors, greater than 1.0,
                 computed when a request for too much accuracy was detected.
        'tsw'    value of t at the time of the last method switch
                 (given for each time step)
        'nst'    cumulative number of time steps
        'nfe'    cumulative number of function evaluations for each time step
        'nje'    cumulative number of jacobian evaluations for each time step
        'nqu'    a vector of method orders for each successful step.
        'imxer'  index of the component of largest magnitude in the
                 weighted local error vector (e / ewt) on an error return, -1
                 otherwise.
        'lenrw'  the length of the double work array required.
        'leniw'  the length of integer work array required.
        'mused'  a vector of method indicators for each successful time step:
                 1: adams (nonstiff), 2: bdf (stiff)
        =======  ============================================================

    Other Parameters
    ----------------
    ml, mu : int, optional
        If either of these are not None or non-negative, then the
        Jacobian is assumed to be banded.  These give the number of
        lower and upper non-zero diagonals in this banded matrix.
        For the banded case, `Dfun` should return a matrix whose
        rows contain the non-zero bands (starting with the lowest diagonal).
        Thus, the return matrix `jac` from `Dfun` should have shape
        ``(ml + mu + 1, len(y0))`` when ``ml >=0`` or ``mu >=0``.
        The data in `jac` must be stored such that ``jac[i - j + mu, j]``
        holds the derivative of the `i`th equation with respect to the `j`th
        state variable.  If `col_deriv` is True, the transpose of this
        `jac` must be returned.
    rtol, atol : float, optional
        The input parameters `rtol` and `atol` determine the error
        control performed by the solver.  The solver will control the
        vector, e, of estimated local errors in y, according to an
        inequality of the form ``max-norm of (e / ewt) <= 1``,
        where ewt is a vector of positive error weights computed as
        ``ewt = rtol * abs(y) + atol``.
        rtol and atol can be either vectors the same length as y or scalars.
        Defaults to 1.49012e-8.
    tcrit : ndarray, optional
        Vector of critical points (e.g. singularities) where integration
        care should be taken.
    h0 : float, (0: solver-determined), optional
        The step size to be attempted on the first step.
    hmax : float, (0: solver-determined), optional
        The maximum absolute step size allowed.
    hmin : float, (0: solver-determined), optional
        The minimum absolute step size allowed.
    ixpr : bool, optional
        Whether to generate extra printing at method switches.
    mxstep : int, (0: solver-determined), optional
        Maximum number of (internally defined) steps allowed for each
        integration point in t.
    mxhnil : int, (0: solver-determined), optional
        Maximum number of messages printed.
    mxordn : int, (0: solver-determined), optional
        Maximum order to be allowed for the non-stiff (Adams) method.
    mxords : int, (0: solver-determined), optional
        Maximum order to be allowed for the stiff (BDF) method.

    See Also
    --------
    ode : a more object-oriented integrator based on VODE.
    quad : for finding the area under a curve.

    Examples
    --------
    The second order differential equation for the angle `theta` of a
    pendulum acted on by gravity with friction can be written::

        theta''(t) + b*theta'(t) + c*sin(theta(t)) = 0

    where `b` and `c` are positive constants, and a prime (') denotes a
    derivative.  To solve this equation with `odeint`, we must first convert
    it to a system of first order equations.  By defining the angular
    velocity ``omega(t) = theta'(t)``, we obtain the system::

        theta'(t) = omega(t)
        omega'(t) = -b*omega(t) - c*sin(theta(t))

    Let `y` be the vector [`theta`, `omega`].  We implement this system
    in python as:

    >>> def pend(y, t, b, c):
    ..     theta, omega = y
    ..     dydt = [omega, -b*omega - c*np.sin(theta)]
    ..     return dydt
    ...

    We assume the constants are `b` = 0.25 and `c` = 5.0:

    >>> b = 0.25
    >>> c = 5.0

    For initial conditions, we assume the pendulum is nearly vertical
    with `theta(0)` = `pi` - 0.1, and it initially at rest, so
    `omega(0)` = 0.  Then the vector of initial conditions is

    >>> y0 = [np.pi - 0.1, 0.0]

    We generate a solution 101 evenly spaced samples in the interval
    0 <= `t` <= 10.  So our array of times is:

    >>> t = np.linspace(0, 10, 101)

    Call `odeint` to generate the solution.  To pass the parameters
    `b` and `c` to `pend`, we give them to `odeint` using the `args`
    argument.

    >>> from scipy.integrate import odeint
    >>> sol = odeint(pend, y0, t, args=(b, c))

    The solution is an array with shape (101, 2).  The first column
    is `theta(t)`, and the second is `omega(t)`.  The following code
    plots both components.

    >>> import matplotlib.pyplot as plt
    >>> plt.plot(t, sol[:, 0], 'b', label='theta(t)')
    >>> plt.plot(t, sol[:, 1], 'g', label='omega(t)')
    >>> plt.legend(loc='best')
    >>> plt.xlabel('t')
    >>> plt.grid()
    >>> plt.show()

Scipy quad sử dụng phương pháp nào?

Gói phụ và chức năng nào được sử dụng để tích hợp đơn lẻ trong Scipy?

Gói phụ và chức năng tương ứng nào được sử dụng để tích hợp kép trong Scipy?

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