从 Python 转换/翻译为 Octave 或 Matlab

Question

我有一个 Python 代码，想用 Octave 重写它，但在转换过程中遇到了很多问题。我为其中一些人找到了解决方案，其中一些人仍然需要你的帮助。现在我将从这部分代码开始：

INVOLUTE_FI = 0
INVOLUTE_FO = 1
INVOLUTE_OI = 2
INVOLUTE_OO = 3
 def coords_inv(phi, geo,  theta, inv):
        """
        Coordinates of the involutes

        Parameters
        ----------
        phi : float
            The involute angle
        geo : struct
            The structure with the geometry obtained from get_geo()
        theta : float
            The crank angle, between 0 and 2*pi
        inv : int
            The key for the involute to be considered
        """

        rb = geo.rb
        ro = rb*(pi - geo.phi_fi0 + geo.phi_oo0)
        Theta = geo.phi_fie - theta - pi/2.0

        if inv == INVOLUTE_FI:
            x = rb*cos(phi)+rb*(phi-geo.phi_fi0)*sin(phi)
            y = rb*sin(phi)-rb*(phi-geo.phi_fi0)*cos(phi)
        elif inv == INVOLUTE_FO:
            x = rb*cos(phi)+rb*(phi-geo.phi_fo0)*sin(phi)
            y = rb*sin(phi)-rb*(phi-geo.phi_fo0)*cos(phi)
        elif inv == INVOLUTE_OI:
            x = -rb*cos(phi)-rb*(phi-geo.phi_oi0)*sin(phi)+ro*cos(Theta)
            y = -rb*sin(phi)+rb*(phi-geo.phi_oi0)*cos(phi)+ro*sin(Theta)
        elif inv == INVOLUTE_OO:
            x = -rb*cos(phi)-rb*(phi-geo.phi_oo0)*sin(phi)+ro*cos(Theta)
            y = -rb*sin(phi)+rb*(phi-geo.phi_oo0)*cos(phi)+ro*sin(Theta)
        else:
            raise ValueError('flag not valid')

        return x,y
    def CVcoords(CVkey, geo, theta, N = 1000):
        """ 
        Return a tuple of numpy arrays for x,y coordinates for the lines which 
        determine the boundary of the control volume
    Parameters
        ----------
        CVkey : string
            The key for the control volume for which the polygon is desired
        geo : struct
            The structure with the geometry obtained from get_geo()
        theta : float
            The crank angle, between 0 and 2*pi
        N : int
            How many elements to include in each entry in the polygon

        Returns
        -------
        x : numpy array
            X-coordinates of the outline of the control volume
        y : numpy array 
            Y-coordinates of the outline of the control volume
        """

        Nc1 = Nc(theta, geo, 1)
        Nc2 = Nc(theta, geo, 2)

        if CVkey == 'sa':

            r = (2*pi*geo.rb-geo.t)/2.0

            xee,yee = coords_inv(geo.phi_fie,geo,0.0,'fi')
            xse,yse = coords_inv(geo.phi_foe-2*pi,geo,0.0,'fo')
            xoie,yoie = coords_inv(geo.phi_oie,geo,theta,'oi')
            xooe,yooe = coords_inv(geo.phi_ooe,geo,theta,'oo')
            x0,y0 = (xee+xse)/2,(yee+yse)/2 

            beta = atan2(yee-y0,xee-x0)
            t = np.linspace(beta,beta+pi,1000)
            x,y = x0+r*np.cos(t),y0+r*np.sin(t)
              return np.r_[x,xoie,xooe,x[0]],np.r_[y,yoie,yooe,y[0]]

https://docs.scipy.org/doc/numpy/reference/generated/numpy.r_.html我只是不明白最后一个输出，我仍然混淆这里的_r是什么意思，我怎么能用Octave写它？....我读了链接中写的内容，但对我来说仍然不清楚。

Answer 1

return np.r_[x,xoie,xooe,x[0]], np.r_[y,yoie,yooe,y[0]]

该函数返回 2 个值，两个数组均由 np.r_ 创建。

np.r_[....] 具有索引语法，并最终被转换为对np.r_ 对象的函数调用。结果只是参数的串联：

In [355]: np.r_[1, 3, 6:8, np.array([3,2,1])]
Out[355]: array([1, 3, 6, 7, 3, 2, 1])

使用[] 表示法，它可以接受slice 之类的对象（6:8），尽管我在这里看不到任何对象。我必须研究其余代码以确定其他参数是标量（单个值）还是数组。

我的 Octave 生锈了（尽管我可以尝试转换）。

t = np.lispace... # I think that exists in Octave, a 1000 values
x = x0+r*np.cos(t)  # a derived array of 1000 values

xoiecoords_inv返回的值之一；可以是标量或数组。 x[0]x 的第一个值。所以r_ 可能会产生一个由x 和后续值组成的一维数组。