Shepard interpolation algorithm
$begingroup$
Classic way to get a value based on other points in Shepard (Inverse distance weighting) method:
$$F(x,y) = Sigma_{k=1}^{N}w_{k}(x,y)f_{k} / Sigma_{k=1}^{N}w_{k}(x,y) $$
where $w$ can be compute as some expression with metric (euclidian for example).
The paper Scattered Data Interpolation: Tests of Some Methods by Franke tells (p.5) that another way to get needed value is compute follows:
$$F(x,y) = Sigma_{k=1}^{N}w_{k} [f_{k} + frac{partial f}{partial x}_{k}(x-x_{k}) + frac{partial f}{partial y}_{k}(y-y_{k})]/ Sigma_{k=1}^{N}w_{k}(x,y)$$
But I do not understand, how can I coumpute this. For example I have sample:
$$mathbf{x} = [0, 2, 4] \
mathbf{y} = [3, 6, 5]
$$
and I need to get interpolated value for point $a = 1$.
How can I do this?
interpolation
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
$endgroup$
|
show 2 more comments
$begingroup$
Classic way to get a value based on other points in Shepard (Inverse distance weighting) method:
$$F(x,y) = Sigma_{k=1}^{N}w_{k}(x,y)f_{k} / Sigma_{k=1}^{N}w_{k}(x,y) $$
where $w$ can be compute as some expression with metric (euclidian for example).
The paper Scattered Data Interpolation: Tests of Some Methods by Franke tells (p.5) that another way to get needed value is compute follows:
$$F(x,y) = Sigma_{k=1}^{N}w_{k} [f_{k} + frac{partial f}{partial x}_{k}(x-x_{k}) + frac{partial f}{partial y}_{k}(y-y_{k})]/ Sigma_{k=1}^{N}w_{k}(x,y)$$
But I do not understand, how can I coumpute this. For example I have sample:
$$mathbf{x} = [0, 2, 4] \
mathbf{y} = [3, 6, 5]
$$
and I need to get interpolated value for point $a = 1$.
How can I do this?
interpolation
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
$endgroup$
$begingroup$
The second formula assumes that you know the gradient at every data point. By the way, your example does not describe samples of a bivariate function.
$endgroup$
– Yves Daoust
Jan 10 at 10:22
$begingroup$
Ok, I understand that. Is it mean that I cant reduce the second formula for my example?
$endgroup$
– Vasya Pravdin
Jan 10 at 10:31
$begingroup$
What do you mean ?
$endgroup$
– Yves Daoust
Jan 10 at 11:16
$begingroup$
Remove from second formula df/dy(y-y_k) and use only with (f_k + df/dx(x-x_k))?
$endgroup$
– Vasya Pravdin
Jan 10 at 13:33
$begingroup$
Are you trying to perform 1D interpolation ?
$endgroup$
– Yves Daoust
Jan 10 at 13:54
|
show 2 more comments
$begingroup$
Classic way to get a value based on other points in Shepard (Inverse distance weighting) method:
$$F(x,y) = Sigma_{k=1}^{N}w_{k}(x,y)f_{k} / Sigma_{k=1}^{N}w_{k}(x,y) $$
where $w$ can be compute as some expression with metric (euclidian for example).
The paper Scattered Data Interpolation: Tests of Some Methods by Franke tells (p.5) that another way to get needed value is compute follows:
$$F(x,y) = Sigma_{k=1}^{N}w_{k} [f_{k} + frac{partial f}{partial x}_{k}(x-x_{k}) + frac{partial f}{partial y}_{k}(y-y_{k})]/ Sigma_{k=1}^{N}w_{k}(x,y)$$
But I do not understand, how can I coumpute this. For example I have sample:
$$mathbf{x} = [0, 2, 4] \
mathbf{y} = [3, 6, 5]
$$
and I need to get interpolated value for point $a = 1$.
How can I do this?
interpolation
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
$endgroup$
Classic way to get a value based on other points in Shepard (Inverse distance weighting) method:
$$F(x,y) = Sigma_{k=1}^{N}w_{k}(x,y)f_{k} / Sigma_{k=1}^{N}w_{k}(x,y) $$
where $w$ can be compute as some expression with metric (euclidian for example).
The paper Scattered Data Interpolation: Tests of Some Methods by Franke tells (p.5) that another way to get needed value is compute follows:
$$F(x,y) = Sigma_{k=1}^{N}w_{k} [f_{k} + frac{partial f}{partial x}_{k}(x-x_{k}) + frac{partial f}{partial y}_{k}(y-y_{k})]/ Sigma_{k=1}^{N}w_{k}(x,y)$$
But I do not understand, how can I coumpute this. For example I have sample:
$$mathbf{x} = [0, 2, 4] \
mathbf{y} = [3, 6, 5]
$$
and I need to get interpolated value for point $a = 1$.
How can I do this?
interpolation
interpolation
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
asked Jan 10 at 10:17
Vasya PravdinVasya Pravdin
83
83
$begingroup$
The second formula assumes that you know the gradient at every data point. By the way, your example does not describe samples of a bivariate function.
$endgroup$
– Yves Daoust
Jan 10 at 10:22
$begingroup$
Ok, I understand that. Is it mean that I cant reduce the second formula for my example?
$endgroup$
– Vasya Pravdin
Jan 10 at 10:31
$begingroup$
What do you mean ?
$endgroup$
– Yves Daoust
Jan 10 at 11:16
$begingroup$
Remove from second formula df/dy(y-y_k) and use only with (f_k + df/dx(x-x_k))?
$endgroup$
– Vasya Pravdin
Jan 10 at 13:33
$begingroup$
Are you trying to perform 1D interpolation ?
$endgroup$
– Yves Daoust
Jan 10 at 13:54
|
show 2 more comments
$begingroup$
The second formula assumes that you know the gradient at every data point. By the way, your example does not describe samples of a bivariate function.
$endgroup$
– Yves Daoust
Jan 10 at 10:22
$begingroup$
Ok, I understand that. Is it mean that I cant reduce the second formula for my example?
$endgroup$
– Vasya Pravdin
Jan 10 at 10:31
$begingroup$
What do you mean ?
$endgroup$
– Yves Daoust
Jan 10 at 11:16
$begingroup$
Remove from second formula df/dy(y-y_k) and use only with (f_k + df/dx(x-x_k))?
$endgroup$
– Vasya Pravdin
Jan 10 at 13:33
$begingroup$
Are you trying to perform 1D interpolation ?
$endgroup$
– Yves Daoust
Jan 10 at 13:54
$begingroup$
The second formula assumes that you know the gradient at every data point. By the way, your example does not describe samples of a bivariate function.
$endgroup$
– Yves Daoust
Jan 10 at 10:22
$begingroup$
The second formula assumes that you know the gradient at every data point. By the way, your example does not describe samples of a bivariate function.
$endgroup$
– Yves Daoust
Jan 10 at 10:22
$begingroup$
Ok, I understand that. Is it mean that I cant reduce the second formula for my example?
$endgroup$
– Vasya Pravdin
Jan 10 at 10:31
$begingroup$
Ok, I understand that. Is it mean that I cant reduce the second formula for my example?
$endgroup$
– Vasya Pravdin
Jan 10 at 10:31
$begingroup$
What do you mean ?
$endgroup$
– Yves Daoust
Jan 10 at 11:16
$begingroup$
What do you mean ?
$endgroup$
– Yves Daoust
Jan 10 at 11:16
$begingroup$
Remove from second formula df/dy(y-y_k) and use only with (f_k + df/dx(x-x_k))?
$endgroup$
– Vasya Pravdin
Jan 10 at 13:33
$begingroup$
Remove from second formula df/dy(y-y_k) and use only with (f_k + df/dx(x-x_k))?
$endgroup$
– Vasya Pravdin
Jan 10 at 13:33
$begingroup$
Are you trying to perform 1D interpolation ?
$endgroup$
– Yves Daoust
Jan 10 at 13:54
$begingroup$
Are you trying to perform 1D interpolation ?
$endgroup$
– Yves Daoust
Jan 10 at 13:54
|
show 2 more comments
0
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$begingroup$
The second formula assumes that you know the gradient at every data point. By the way, your example does not describe samples of a bivariate function.
$endgroup$
– Yves Daoust
Jan 10 at 10:22
$begingroup$
Ok, I understand that. Is it mean that I cant reduce the second formula for my example?
$endgroup$
– Vasya Pravdin
Jan 10 at 10:31
$begingroup$
What do you mean ?
$endgroup$
– Yves Daoust
Jan 10 at 11:16
$begingroup$
Remove from second formula df/dy(y-y_k) and use only with (f_k + df/dx(x-x_k))?
$endgroup$
– Vasya Pravdin
Jan 10 at 13:33
$begingroup$
Are you trying to perform 1D interpolation ?
$endgroup$
– Yves Daoust
Jan 10 at 13:54