Scipy griddata with linear and cubic yields nan

2024/11/17 17:40:48

the following code should produce griddata. But in case I choose as interpolation type 'cubic' or 'linear' I am getting nan's in the z grid. Wen i'm choosing 'nearest' everything is running fine. Here is an example code:

import numpy as np
from scipy.interpolate import griddatax = 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.Tgrid_x, grid_y = np.mgrid[0.0:0.09:250*1j, 0.0:0.03:250*1j] #generating the gridi_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()

I don't have any idea why this is not working..

Answer

Your area you look at is simply much larger than your input points. This doesn't matter for 'nearest' as this always puts the nearest value to a certain coordinate. But 'linear' and 'cubic' do not extrapolate but fill the values which are not within the input area with nan by default.

See also the docs of 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.

Best understandable when plotted by imshow:

enter image description here

plot created with:

import numpy as np
from scipy.interpolate import griddatax = 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.Tgrid_x, grid_y = np.mgrid[0.0:0.09:250*1j, 0.0:0.03:250*1j] #generating the gridfig, axs = plt.subplots(3)
for i, i_type in enumerate(['cubic', 'nearest', 'linear']): #, cubicgrid_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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