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I'm not a python expert but I think that's a good answer to your question :

f(0,0)=1, f(1,0)=3 f(0,1)=5, f(1,1)=4

import numpy as np
from numpy.linalg import inv
# Load an color image in grayscale
z = np.array([[1], [3], [5], [4]])
m=np.array([[0, 0 ,1],[1,0 ,1],[0 ,1 ,1],[1 ,1 ,1]]);
print ("z = ",z)
print ("m = ",m)
mm=np.dot(m.T,m)
pinvm = inv(mm)
a= np.dot(pinvm,np.dot(m.T,z))
print ("a = ",a )
zz = np.dot(np.array([[0.5, 0.5 ,1]]),a)
print ("z = ",zz )

results interolate point at 0.5,0.5 is 3.25 :

z =  [[1]
 [3]
 [5]
 [4]]
m =  [[0 0 1]
 [1 0 1]
 [0 1 1]
 [1 1 1]]
a =  [[0.5 ]
 [2.5 ]
 [1.75]]
z =  [[3.25]]

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this answer is off topics : no opencv function

I'm not a python expert but I think that's a good answer to your question :

f(0,0)=1, f(1,0)=3 f(0,1)=5, f(1,1)=4

import numpy as np
from numpy.linalg import inv
# Load an color image in grayscale
z = np.array([[1], [3], [5], [4]])
m=np.array([[0, 0 ,1],[1,0 ,1],[0 ,1 ,1],[1 ,1 ,1]]);
print ("z = ",z)
print ("m = ",m)
mm=np.dot(m.T,m)
pinvm = inv(mm)
a= np.dot(pinvm,np.dot(m.T,z))
print ("a = ",a )
zz = np.dot(np.array([[0.5, 0.5 ,1]]),a)
print ("z = ",zz )

results interolate point at 0.5,0.5 is 3.25 :

z =  [[1]
 [3]
 [5]
 [4]]
m =  [[0 0 1]
 [1 0 1]
 [0 1 1]
 [1 1 1]]
a =  [[0.5 ]
 [2.5 ]
 [1.75]]
z =  [[3.25]]

Appuyez sur une touche pour continuer...

this answer is off topics : no opencv function

I'm not a python expert but I think that's a good answer to your question :

f(0,0)=1, f(1,0)=3 f(0,1)=5, f(1,1)=4

import numpy as np
from numpy.linalg import inv
# Load an color image in grayscale
z = np.array([[1], [3], [5], [4]])
m=np.array([[0, 0 ,1],[1,0 ,1],[0 ,1 ,1],[1 ,1 ,1]]);
print ("z = ",z)
print ("m = ",m)
mm=np.dot(m.T,m)
pinvm = inv(mm)
a= np.dot(pinvm,np.dot(m.T,z))
print ("a = ",a )
zz = np.dot(np.array([[0.5, 0.5 ,1]]),a)
print ("z = ",zz )

results interolate interpolate point at 0.5,0.5 is 3.25 :

z =  [[1]
 [3]
 [5]
 [4]]
m =  [[0 0 1]
 [1 0 1]
 [0 1 1]
 [1 1 1]]
a =  [[0.5 ]
 [2.5 ]
 [1.75]]
z =  [[3.25]]

Appuyez sur une touche pour continuer...

this answer is off topics : no opencv function

In c++ it's here with opencv function

I'm not a python expert but I think that's a good answer to your question :

f(0,0)=1, f(1,0)=3 f(0,1)=5, f(1,1)=4

I want to solve f(i,j)=z = M a where p=[ a0 a1 a2] and a0 xi+ a1yi +a2 =zi

import numpy as np
from numpy.linalg import inv
z = np.array([[1], [3], [5], [4]])
m=np.array([[0, 0 ,1],[1,0 ,1],[0 ,1 ,1],[1 ,1 ,1]]);
print ("z = ",z)
print ("m = ",m)
mm=np.dot(m.T,m)
pinvm = inv(mm)
a= np.dot(pinvm,np.dot(m.T,z))
print ("a = ",a )
zz = np.dot(np.array([[0.5, 0.5 ,1]]),a)
print ("z = ",zz )

results interpolate point at 0.5,0.5 is 3.25 :

z =  [[1]
 [3]
 [5]
 [4]]
m =  [[0 0 1]
 [1 0 1]
 [0 1 1]
 [1 1 1]]
a =  [[0.5 ]
 [2.5 ]
 [1.75]]
z =  [[3.25]]

Appuyez sur une touche pour continuer...

this answer is off topics : no opencv function

In c++ it's here with opencv function