Question

我使用Scipy的Griddata使用绘制的点填充此gridded data（显示为空）。有没有办法根据（x，y）坐标得到插值（？）？这是绘图的代码，x，y，z值都是系列。

    xi = np.linspace(min(lats),max(lats),360)
    yi = np.linspace(min(lons),max(lons),360)
    # grid the data.
    zi = griddata((lats, lons), nuits, (xi[None,:], yi[:,None]), method='cubic')
    # contour the gridded data.
    plt.contourf(xi,yi,zi,15,cmap=cMap)
    plt.colorbar()
    # plot data points.
    #plt.scatter(lats,lons,c=nuits,marker='o',s=26,cmap=cMap)
    plt.scatter(lats,lons,facecolors='none', edgecolors='k',s=26)
    plt.show()

Answer 1

这样可行：

xi_coords = {value: index for index, value in enumerate(xi)}
yi_coords = {value: index for index, value in enumerate(yi)}
xic = <your coordinate>
yic = <your coordinate>
zi[xi_coords[xic], yi_coords[yic]]

Answer 2

您可以通过以下方式从 (xi,yi) 坐标中获得内插的 zi 坐标：

# (xi,yi) coords to get the interpolated zi coords
# where len(xic) = len(yic)  
xic = <your coordinate>
yic = <your coordinate>

# sort these coordinates in an increasing order
s_xic = np.sort(xic)
s_yic = np.sort(yic)

# indices belonging to xic, yic, that would sort the array
ind_s_xic = np.argsort(xic)
ind_s_yic = np.argsort(yic)

 
dict_xic = dict(zip(ind_s_xic, np.array(range(len(xic))))
dict_yic = dict(zip(ind_s_yic, np.array(range(len(yic))))


xi,yi = np.meshgrid(s_xic, s_yic)

# zi_grid has dimensions ( len(yic), len(xic) )
zi_grid = griddata((lats, lons), nuits, (xi, yi), method='cubic')

# zic is the interpolated z-coordinate data with an arrangement order,
# corresponding to the x and y-coordinate data in xic and yic
zic =  [ zi_grid[dict_yic[i], dict_xic[i]] for i in range(len(xic)) ]

访问 How does one use numpy's Meshgrid function with a random interval rather than a equally spaced one? 以了解 meshgrid 的工作原理。

Meshgrid 可以从你的不均匀间隔 (xi,yi) 坐标创建，之后，griddata 可以使用你的 meshgrid 从基于点 = (lats, lons), values = nuits 创建的插值表面插入 z 坐标.

从scipy interpolate / griddata中检索数据点

2 个答案: