python - 具有 'linear' 和 'cubic' 的 Scipy griddata 产生 nan

Question

以下代码应生成网格数据。但如果我选择“三次”或“线性”插值类型，我会在 z 网格中得到 nan。我选择“最近”一切都运行良好。 下面是一个示例代码:

import numpy as np
from scipy.interpolate import griddata

x = np.array([0.03,0.05,0033])
y = np.array([0.004,0.01,0.02])
z = np.array([1,2,3])


xy = np.zeros((2,np.size(x)))
xy[0] = x
xy[1] = y
xy = xy.T

grid_x, grid_y = np.mgrid[0.0:0.09:250*1j, 0.0:0.03:250*1j] #generating the grid


i_type= 'cubic' #nearest, linear, cubic
grid_z = griddata(xy, z, (grid_x, grid_y), method=i_type)

#check if there is a nan in the z grid:
print np.isnan(grid_z).any()

我不知道为什么这不起作用..

Answer 1

您看到的区域比您的输入点大得多。这对于“最近”并不重要，因为这总是将最近的值放置到某个坐标。但“线性”和“三次”不会外推，而是默认使用 nan 填充不在输入区域内的值。

另请参阅griddata的文档:

fill_value : float, optional
Value used to fill in for requested points outside of the convex hull of the input points. If not provided, then the default is nan. This option has no effect for the ‘nearest’ method.

由 imshow 绘制时最容易理解:

绘图创建:

import numpy as np
from scipy.interpolate import griddata

x = np.array([0.03,0.05,0.033])
y = np.array([0.004,0.01,0.02])
z = np.array([1,2,3])


xy = np.zeros((2,np.size(x)))
xy[0] = x
xy[1] = y
xy = xy.T

grid_x, grid_y = np.mgrid[0.0:0.09:250*1j, 0.0:0.03:250*1j] #generating the grid

fig, axs = plt.subplots(3)
for i, i_type in enumerate(['cubic', 'nearest', 'linear']): #, cubic
    grid_z = griddata(xy, z, (grid_x, grid_y), method=i_type)

    #check if there is a nan in the z grid:
    axs[i].imshow(grid_z)
    axs[i].set_title(i_type)

plt.tight_layout()