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python - 具有“线性”和“立方”的 Scipy 网格数据产生 nan

问题描述

以下代码应生成网格数据。但如果我选择插值类型“立方”或“线性”，我会在 z 网格中得到 nan。温我选择'最近'一切运行良好。这是一个示例代码：

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..

标签： pythonnumpyscipyinterpolation

解决方案

您查看的区域比您的输入点大得多。这对于“最近”无关紧要，因为这总是将最近的值放在某个坐标上。但是 'linear' 和 'cubic' 不会外推，而是默认用 nan 填充不在输入区域内的值。

另请参阅以下文档griddata：

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.

绘制时最好理解imshow：

在此处输入图像描述

创建的情节：

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()

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