如何从scipy.interpolate.RectSphereBivariateSpline反算经纬度?

时间:2013-01-30 14:28:38

标签: python scipy interpolation

使用RectSphereBivariateSpline子包中scipy.interpolate函数的THIS示例,我想回溯计算纬度和经度(以度为单位)的数组,以及带有插值数据的数组网格上每个坐标的值。

插值数据对象RectSphereBivariateSpline创建似乎是数据值的 u v 组件,每个网格程度更改一个值(此示例中,纬度= 180,经度= 360)。

是吗?

我如何反向计算纬度,经度及其各自的数据值以进行绘图?

import numpy as np
from scipy.interpolate import RectSphereBivariateSpline

def geo_interp(lats,lons,data,grid_size_deg):
        '''We want to interpolate it to a global one-degree grid'''
        deg2rad = np.pi/180.
        new_lats = np.linspace(grid_size_deg, 180, 180) * deg2rad
        new_lons = np.linspace(grid_size_deg, 360, 360) * deg2rad
        new_lats, new_lons = np.meshgrid(new_lats, new_lons)

        '''We need to set up the interpolator object'''
        lut = RectSphereBivariateSpline(lats*deg2rad, lons*deg2rad, data)

        '''Finally we interpolate the data. The RectSphereBivariateSpline 
        object only takes 1-D arrays as input, therefore we need to do some reshaping.'''
        data_interp = lut.ev(new_lats.ravel(),
                        new_lons.ravel()).reshape((360, 180)).T
        return data_interp

if __name__ == '__main__':
        import matplotlib.pyplot as plt

        '''Suppose we have global data on a coarse grid'''
        lats = np.linspace(10, 170, 9) # in degrees
        lons = np.linspace(0, 350, 18) # in degrees
        data = np.dot(np.atleast_2d(90. - np.linspace(-80., 80., 18)).T,
                        np.atleast_2d(180. - np.abs(np.linspace(0., 350., 9)))).T

        '''Interpolate data to 1 degree grid'''
        data_interp = geo_interp(lats,lons,data,1)

        '''Looking at the original and the interpolated data, 
        one can see that the interpolant reproduces the original data very well'''
        fig = plt.figure()
        ax1 = fig.add_subplot(211)
        ax1.imshow(data, interpolation='nearest')
        ax2 = fig.add_subplot(212)
        ax2.imshow(data_interp, interpolation='nearest')
        plt.show()

我以为我可能会使用向量加法(即毕达哥拉斯定理),但这不起作用,因为我只有一个值用于每个度数变化,而不是点

pow((pow(data_interp[0,:],2.0)+pow(data_interp[:,0],2.0)),1/2.0)

1 个答案:

答案 0 :(得分:1)

似乎可以用于绘图的纬度和经度向量是在我们的“geo_interp”函数(“net_lats”和“new_lons”)中生成的。如果你在主程序中需要这些,你应该在函数之外声明这些向量,或者你应该让函数返回这些生成的向量以在主程序中使用。

希望这有帮助。