This is a follow-up question to my previous post: Python/Scipy Interpolation (map_coordinates)

Let’s say I want to interpolate over a 2d rectangular area. My variable ‘z’ contains the data as shown below. Each column is at a constant value, however, each row of the array may be at a different value as shown in the comment below.

```
from scipy import interpolate
from numpy import array
import numpy as np
# # 0.0000, 0.1750, 0.8170, 1.0000
z = array([[-2.2818,-2.2818,-0.9309,-0.9309], # 0.0000, 0.0000, 0.0000, 0.0000
[-2.2818,-2.2818,-0.9309,-0.9309], # 0.2620, 0.2784, 0.3379, 0.3526
[-1.4891,-1.4891,-0.5531,-0.5531], # 0.6121, 0.6351, 0.7118, 0.7309
[-1.4891,-1.4891,-0.5531,-0.5531]]) # 1.0000, 1.0000, 1.0000, 1.0000
# Rows, Columns = z.shape
cols = array([0.0000, 0.1750, 0.8170, 1.0000])
rows = array([0.0000, 0.2620, 0.6121, 1.0000])
sp = interpolate.RectBivariateSpline(rows, cols, z, kx=1, ky=1, s=0)
xi = np.array([0.00000, 0.26200, 0.27840, 0.33790, 0.35260, 0.61210, 0.63510,
0.71180, 0.73090, 1.00000], dtype=np.float)
yi = np.array([0.000, 0.167, 0.815, 1.000], dtype=np.float)
print sp(xi, yi)
```

As another way of visualizing this, the array of values I KNOW would be:

```
rows = array([0.0000, 0.2620, 0.2784, 0.3379, 0.3526,
0.6121, 0.6351, 0.7118, 0.7309, 1.0000])
# # 0.0000, 0.1750, 0.8170, 1.0000
z = array([[-2.2818,-2.2818,-0.9309,-0.9309], # 0.0000
[-2.2818, ?, ?, ?], # 0.2620,
[ ?,-2.2818, ?, ?], # 0.2784
[ ?, ?,-0.9309, ?], # 0.3379
[ ? ,?, ?,-0.9309], # 0.3526
[-1.4891, ?, ?, ?], # 0.6121
[ ?,-1.4891, ?, ?], # 0.6351
[ ?, ?,-0.5531, ?], # 0.7118
[ ?, ?, ?,-0.5531], # 0.7309
[-1.4891,-1.4891,-0.5531,-0.5531]]) # 1.0000
```

I do not know the ‘?’ values, and they should be interpolated. I tried replacing them with None, but then get ‘nan’ for all of my results.

**EDIT:**

I think I need to use either ‘griddata’ or ‘interp2’. griddata seems to produce the result I expect, but ‘interp2’ does not.

```
from scipy import interpolate
from numpy import array
import numpy as np
z = array([[-2.2818,-2.2818,-0.9309,-0.9309],
[-2.2818,-2.2818,-0.9309,-0.9309],
[-1.4891,-1.4891,-0.5531,-0.5531],
[-1.4891,-1.4891,-0.5531,-0.5531]])
rows = array([0.0000, 0.0000, 0.0000, 0.0000,
0.2620, 0.2784, 0.3379, 0.3526,
0.6121, 0.6351, 0.7118, 0.7309,
1.0000, 1.0000, 1.0000, 1.0000])
cols = array([0.0000, 0.1750, 0.8180, 1.0000,
0.0000, 0.1750, 0.8180, 1.0000,
0.0000, 0.1750, 0.8180, 1.0000,
0.0000, 0.1750, 0.8180, 1.0000])
xi = array([0.0000, 0.2620, 0.2784, 0.3379, 0.3526, 0.6121, 0.6351, 0.7118,
0.7309, 1.0000], dtype=np.float)
yi = array([0.000, 0.175, 0.818, 1.000], dtype=np.float)
GD = interpolate.griddata((rows, cols), z.ravel(),
(xi[None,:], yi[:,None]), method='linear')
I2 = interpolate.interp2d(rows, cols, z, kind='linear')
print GD.reshape(4, 10).T
print '\n'
print I2(xi, yi).reshape(4, 10).T
import matplotlib.pyplot as plt
import numpy.ma as ma
plt.figure()
GD = interpolate.griddata((rows.ravel(), cols.ravel()), z.ravel(),
(xi[None,:], yi[:,None]), method='linear')
CS = plt.contour(xi,yi,GD,15,linewidths=0.5,colors='k')
CS = plt.contourf(xi,yi,GD,15,cmap=plt.cm.jet)
plt.colorbar()
plt.scatter(rows,cols,marker='o',c='b',s=5)
plt.figure()
I2 = I2(xi, yi)
CS = plt.contour(xi,yi,I2,15,linewidths=0.5,colors='k')
CS = plt.contourf(xi,yi,I2,15,cmap=plt.cm.jet)
plt.colorbar()
plt.scatter(rows,cols,marker='o',c='b',s=5)
plt.show()
```

## Here is Solutions:

We have many solutions to this problem, But we recommend you to use the first solution because it is tested & true solution that will 100% work for you.

### Solution 1

Looks like you got it.

In your upper code example and in your previous (linked) question you have *structured* data. Which can be interpolated using `RectBivariateSpline`

or `interp2d`

. This means you have data that can be described on a grid (all points on the grid have a known value). The grid doesn’t necessarily have to have all the same dx and dy. (if all dx’s and dy’s were equal, you’d have a Regular Grid)

Now, your current question asks what to do if not all the points are known. This is known as *unstructured* data. All you have are a selection of points in a field. You can’t necessarily construct rectangles where all vertices have known values. For this type of data, you can use (as you have) `griddata`

, or a flavor of `BivariateSpline`

.

Now which to choose?

The nearest analogy to the structured `RectBivariateSpline`

is one of the *unstructured* `BivariateSpline`

classes: `SmoothBivariateSpline`

or `LSQBivariateSpline`

. If you want to use splines to interpolate the data, go with these. this makes your function smooth and differentiable, but you can get a surface that swings outside Z.max() or Z.min().

Since you are setting `ky=1`

and `kx=1`

and are getting what I am pretty sure is just linear interpolation on the *structured* data, I’d personally just switch from the `RectBivariateSpline`

spline scheme to the `interp2d`

*structured* grid interpolation scheme. I know the documentation says it is for regular grids, but the example in the `__doc__`

itself is only *structured*, not regular.

I’d be curious if you found any significant differences between the methods if you do end up switching. Welcome to SciPy.

**Note: Use and implement solution 1 because this method fully tested our system.Thank you 🙂**

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0