Meaning of InterpolationOrder -> All for multidimensional interpolation Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How to get zeroth-order (piecewise constant) interpolation of scattered data?Interpolation of multidimensional data organized logarithmicallyMultidimensional interpolation with duplicate abscissa valuesSeries expansion of InterpolatingFunction obtained from NDSolveMultidimensional Interpolation with 3 independent Variables with modfied data setHow does ListInterpolation work?Deleting mesh elements from a meshCustom interpolation on unstructured grid (2D, 3D)Interpolation order reduced to 1 due to unstructured grid error; yet proper syntax?Interpolation of a list defined on a region
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Meaning of InterpolationOrder -> All for multidimensional interpolation
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How to get zeroth-order (piecewise constant) interpolation of scattered data?Interpolation of multidimensional data organized logarithmicallyMultidimensional interpolation with duplicate abscissa valuesSeries expansion of InterpolatingFunction obtained from NDSolveMultidimensional Interpolation with 3 independent Variables with modfied data setHow does ListInterpolation work?Deleting mesh elements from a meshCustom interpolation on unstructured grid (2D, 3D)Interpolation order reduced to 1 due to unstructured grid error; yet proper syntax?Interpolation of a list defined on a region
$begingroup$
What specific method does Interpolation use for unstructured multi-dimensional data when we set InterpolationOrder -> All? Documentation links are welcome.
Example 2D data:
data = RandomReal[1, 20, 3];
When the data points are not on a grid, the only allowed settings for InterpolationOrder are 1 and All, according to the error message issued when trying something else.
With 1, it is clear how it works: a Delaunay triangulation is computed and linear interpolation is done over each triangle.
But how does All work, and what determines the actual order that is chosen?
if = Interpolation[data, InterpolationOrder -> All];
if["InterpolationOrder"]
(* 5 *)
Show[
Plot3D[if[x, y], x, 0, 1, y, 0, 1],
Graphics3D[PointSize[Large], Point[data]]
]
interpolation
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
$endgroup$
add a comment |
$begingroup$
What specific method does Interpolation use for unstructured multi-dimensional data when we set InterpolationOrder -> All? Documentation links are welcome.
Example 2D data:
data = RandomReal[1, 20, 3];
When the data points are not on a grid, the only allowed settings for InterpolationOrder are 1 and All, according to the error message issued when trying something else.
With 1, it is clear how it works: a Delaunay triangulation is computed and linear interpolation is done over each triangle.
But how does All work, and what determines the actual order that is chosen?
if = Interpolation[data, InterpolationOrder -> All];
if["InterpolationOrder"]
(* 5 *)
Show[
Plot3D[if[x, y], x, 0, 1, y, 0, 1],
Graphics3D[PointSize[Large], Point[data]]
]
interpolation
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
$endgroup$
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is9223372036854775806, 9223372036854775806. Oo
$endgroup$
– Henrik Schumacher
Apr 16 at 8:49
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
Apr 16 at 8:54
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
Apr 16 at 9:00
$begingroup$
@HenrikSchumacher If this gives a hint, starting from 4 data points, the first 3 data point counts get interpolation order 2, then the next 4 get 3, then the next 5 get 4, etc.
$endgroup$
– Szabolcs
Apr 16 at 9:01
1
$begingroup$
That sounds as if they were using straight-forward global interpolation by a polynomial of degree up ton. Then you haveBinomial[n, 2]basis functions. In that case, this should become nasty for higher point counts due to Runge's phenomenon and ill-conditioned linear systems (for solving for the coefficients). So I presume, that they will switch to another method when the point count becomes larger...
$endgroup$
– Henrik Schumacher
Apr 16 at 9:06
add a comment |
$begingroup$
What specific method does Interpolation use for unstructured multi-dimensional data when we set InterpolationOrder -> All? Documentation links are welcome.
Example 2D data:
data = RandomReal[1, 20, 3];
When the data points are not on a grid, the only allowed settings for InterpolationOrder are 1 and All, according to the error message issued when trying something else.
With 1, it is clear how it works: a Delaunay triangulation is computed and linear interpolation is done over each triangle.
But how does All work, and what determines the actual order that is chosen?
if = Interpolation[data, InterpolationOrder -> All];
if["InterpolationOrder"]
(* 5 *)
Show[
Plot3D[if[x, y], x, 0, 1, y, 0, 1],
Graphics3D[PointSize[Large], Point[data]]
]
interpolation
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
$endgroup$
What specific method does Interpolation use for unstructured multi-dimensional data when we set InterpolationOrder -> All? Documentation links are welcome.
Example 2D data:
data = RandomReal[1, 20, 3];
When the data points are not on a grid, the only allowed settings for InterpolationOrder are 1 and All, according to the error message issued when trying something else.
With 1, it is clear how it works: a Delaunay triangulation is computed and linear interpolation is done over each triangle.
But how does All work, and what determines the actual order that is chosen?
if = Interpolation[data, InterpolationOrder -> All];
if["InterpolationOrder"]
(* 5 *)
Show[
Plot3D[if[x, y], x, 0, 1, y, 0, 1],
Graphics3D[PointSize[Large], Point[data]]
]
interpolation
interpolation
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
asked Apr 16 at 8:43
SzabolcsSzabolcs
165k14450954
165k14450954
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is9223372036854775806, 9223372036854775806. Oo
$endgroup$
– Henrik Schumacher
Apr 16 at 8:49
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
Apr 16 at 8:54
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
Apr 16 at 9:00
$begingroup$
@HenrikSchumacher If this gives a hint, starting from 4 data points, the first 3 data point counts get interpolation order 2, then the next 4 get 3, then the next 5 get 4, etc.
$endgroup$
– Szabolcs
Apr 16 at 9:01
1
$begingroup$
That sounds as if they were using straight-forward global interpolation by a polynomial of degree up ton. Then you haveBinomial[n, 2]basis functions. In that case, this should become nasty for higher point counts due to Runge's phenomenon and ill-conditioned linear systems (for solving for the coefficients). So I presume, that they will switch to another method when the point count becomes larger...
$endgroup$
– Henrik Schumacher
Apr 16 at 9:06
add a comment |
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is9223372036854775806, 9223372036854775806. Oo
$endgroup$
– Henrik Schumacher
Apr 16 at 8:49
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
Apr 16 at 8:54
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
Apr 16 at 9:00
$begingroup$
@HenrikSchumacher If this gives a hint, starting from 4 data points, the first 3 data point counts get interpolation order 2, then the next 4 get 3, then the next 5 get 4, etc.
$endgroup$
– Szabolcs
Apr 16 at 9:01
1
$begingroup$
That sounds as if they were using straight-forward global interpolation by a polynomial of degree up ton. Then you haveBinomial[n, 2]basis functions. In that case, this should become nasty for higher point counts due to Runge's phenomenon and ill-conditioned linear systems (for solving for the coefficients). So I presume, that they will switch to another method when the point count becomes larger...
$endgroup$
– Henrik Schumacher
Apr 16 at 9:06
$begingroup$
Dunno, but the return value of
if["InterpolationOrder"] that I get is 9223372036854775806, 9223372036854775806. Oo$endgroup$
– Henrik Schumacher
Apr 16 at 8:49
$begingroup$
Dunno, but the return value of
if["InterpolationOrder"] that I get is 9223372036854775806, 9223372036854775806. Oo$endgroup$
– Henrik Schumacher
Apr 16 at 8:49
1
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
Apr 16 at 8:54
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
Apr 16 at 8:54
1
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
Apr 16 at 9:00
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
Apr 16 at 9:00
$begingroup$
@HenrikSchumacher If this gives a hint, starting from 4 data points, the first 3 data point counts get interpolation order 2, then the next 4 get 3, then the next 5 get 4, etc.
$endgroup$
– Szabolcs
Apr 16 at 9:01
$begingroup$
@HenrikSchumacher If this gives a hint, starting from 4 data points, the first 3 data point counts get interpolation order 2, then the next 4 get 3, then the next 5 get 4, etc.
$endgroup$
– Szabolcs
Apr 16 at 9:01
1
1
$begingroup$
That sounds as if they were using straight-forward global interpolation by a polynomial of degree up to
n. Then you have Binomial[n, 2] basis functions. In that case, this should become nasty for higher point counts due to Runge's phenomenon and ill-conditioned linear systems (for solving for the coefficients). So I presume, that they will switch to another method when the point count becomes larger...$endgroup$
– Henrik Schumacher
Apr 16 at 9:06
$begingroup$
That sounds as if they were using straight-forward global interpolation by a polynomial of degree up to
n. Then you have Binomial[n, 2] basis functions. In that case, this should become nasty for higher point counts due to Runge's phenomenon and ill-conditioned linear systems (for solving for the coefficients). So I presume, that they will switch to another method when the point count becomes larger...$endgroup$
– Henrik Schumacher
Apr 16 at 9:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is code that has been written many moons ago... first an example:
d = 0.4138352728412389, 0.02365673668161028, 0.5509946389658635,
0.7254061374370833, 0.14521595926324116,
0.6528630823305817, 0.48768962246740544,
0.22066264105073286, 0.8309710560928056,
0.3496966364384875, 0.4553589220242207,
0.9383446951847001, 0.2126873262146789,
0.017512080396716145, 0.967248982535015,
0.6211273372083488, 0.3548669163916416,
0.737108322193581, 0.6919974835480842, 0.9322403408098401;
f = 0.9953617542392983, 0.14070666511222818,
0.285662339441511, 0.7988192898854105, 0.3592646208757597,
0.565455746009103, 0.22110814761432618, 0.2735048548887764,
0.08792348530403005, 0.4202942851818514;
data = Join[d, f, 2];
if = Interpolation[data, InterpolationOrder -> All];
if[0.5, 0.5]
0.268157
And here is roughly what it does:
dt = Transpose[d];
temp = Join[ConstantArray[1., Length[d]], dt, dt[[1]]^2,
dt[[1]]*dt[[2]], dt[[2]]^2, dt[[1]]^3,
dt[[1]]^2*dt[[2]], dt[[1]]*dt[[2]]^2, dt[[2]]^3];
p = Transpose[temp];
ls = LinearSolve[p];
vals = ls[Flatten[f]];
System`Private`EvaluateListPolynomial[vals, 0.5, 0.5]
0.268157
answered Apr 16 at 10:59
user21user21
20.7k55997
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is code that has been written many moons ago... first an example:
d = 0.4138352728412389, 0.02365673668161028, 0.5509946389658635,
0.7254061374370833, 0.14521595926324116,
0.6528630823305817, 0.48768962246740544,
0.22066264105073286, 0.8309710560928056,
0.3496966364384875, 0.4553589220242207,
0.9383446951847001, 0.2126873262146789,
0.017512080396716145, 0.967248982535015,
0.6211273372083488, 0.3548669163916416,
0.737108322193581, 0.6919974835480842, 0.9322403408098401;
f = 0.9953617542392983, 0.14070666511222818,
0.285662339441511, 0.7988192898854105, 0.3592646208757597,
0.565455746009103, 0.22110814761432618, 0.2735048548887764,
0.08792348530403005, 0.4202942851818514;
data = Join[d, f, 2];
if = Interpolation[data, InterpolationOrder -> All];
if[0.5, 0.5]
0.268157
And here is roughly what it does:
dt = Transpose[d];
temp = Join[ConstantArray[1., Length[d]], dt, dt[[1]]^2,
dt[[1]]*dt[[2]], dt[[2]]^2, dt[[1]]^3,
dt[[1]]^2*dt[[2]], dt[[1]]*dt[[2]]^2, dt[[2]]^3];
p = Transpose[temp];
ls = LinearSolve[p];
vals = ls[Flatten[f]];
System`Private`EvaluateListPolynomial[vals, 0.5, 0.5]
0.268157
answered Apr 16 at 10:59
user21user21
20.7k55997
$endgroup$
add a comment |
$begingroup$
This is code that has been written many moons ago... first an example:
d = 0.4138352728412389, 0.02365673668161028, 0.5509946389658635,
0.7254061374370833, 0.14521595926324116,
0.6528630823305817, 0.48768962246740544,
0.22066264105073286, 0.8309710560928056,
0.3496966364384875, 0.4553589220242207,
0.9383446951847001, 0.2126873262146789,
0.017512080396716145, 0.967248982535015,
0.6211273372083488, 0.3548669163916416,
0.737108322193581, 0.6919974835480842, 0.9322403408098401;
f = 0.9953617542392983, 0.14070666511222818,
0.285662339441511, 0.7988192898854105, 0.3592646208757597,
0.565455746009103, 0.22110814761432618, 0.2735048548887764,
0.08792348530403005, 0.4202942851818514;
data = Join[d, f, 2];
if = Interpolation[data, InterpolationOrder -> All];
if[0.5, 0.5]
0.268157
And here is roughly what it does:
dt = Transpose[d];
temp = Join[ConstantArray[1., Length[d]], dt, dt[[1]]^2,
dt[[1]]*dt[[2]], dt[[2]]^2, dt[[1]]^3,
dt[[1]]^2*dt[[2]], dt[[1]]*dt[[2]]^2, dt[[2]]^3];
p = Transpose[temp];
ls = LinearSolve[p];
vals = ls[Flatten[f]];
System`Private`EvaluateListPolynomial[vals, 0.5, 0.5]
0.268157
answered Apr 16 at 10:59
user21user21
20.7k55997
$endgroup$
add a comment |
$begingroup$
This is code that has been written many moons ago... first an example:
d = 0.4138352728412389, 0.02365673668161028, 0.5509946389658635,
0.7254061374370833, 0.14521595926324116,
0.6528630823305817, 0.48768962246740544,
0.22066264105073286, 0.8309710560928056,
0.3496966364384875, 0.4553589220242207,
0.9383446951847001, 0.2126873262146789,
0.017512080396716145, 0.967248982535015,
0.6211273372083488, 0.3548669163916416,
0.737108322193581, 0.6919974835480842, 0.9322403408098401;
f = 0.9953617542392983, 0.14070666511222818,
0.285662339441511, 0.7988192898854105, 0.3592646208757597,
0.565455746009103, 0.22110814761432618, 0.2735048548887764,
0.08792348530403005, 0.4202942851818514;
data = Join[d, f, 2];
if = Interpolation[data, InterpolationOrder -> All];
if[0.5, 0.5]
0.268157
And here is roughly what it does:
dt = Transpose[d];
temp = Join[ConstantArray[1., Length[d]], dt, dt[[1]]^2,
dt[[1]]*dt[[2]], dt[[2]]^2, dt[[1]]^3,
dt[[1]]^2*dt[[2]], dt[[1]]*dt[[2]]^2, dt[[2]]^3];
p = Transpose[temp];
ls = LinearSolve[p];
vals = ls[Flatten[f]];
System`Private`EvaluateListPolynomial[vals, 0.5, 0.5]
0.268157
answered Apr 16 at 10:59
user21user21
20.7k55997
$endgroup$
This is code that has been written many moons ago... first an example:
d = 0.4138352728412389, 0.02365673668161028, 0.5509946389658635,
0.7254061374370833, 0.14521595926324116,
0.6528630823305817, 0.48768962246740544,
0.22066264105073286, 0.8309710560928056,
0.3496966364384875, 0.4553589220242207,
0.9383446951847001, 0.2126873262146789,
0.017512080396716145, 0.967248982535015,
0.6211273372083488, 0.3548669163916416,
0.737108322193581, 0.6919974835480842, 0.9322403408098401;
f = 0.9953617542392983, 0.14070666511222818,
0.285662339441511, 0.7988192898854105, 0.3592646208757597,
0.565455746009103, 0.22110814761432618, 0.2735048548887764,
0.08792348530403005, 0.4202942851818514;
data = Join[d, f, 2];
if = Interpolation[data, InterpolationOrder -> All];
if[0.5, 0.5]
0.268157
And here is roughly what it does:
dt = Transpose[d];
temp = Join[ConstantArray[1., Length[d]], dt, dt[[1]]^2,
dt[[1]]*dt[[2]], dt[[2]]^2, dt[[1]]^3,
dt[[1]]^2*dt[[2]], dt[[1]]*dt[[2]]^2, dt[[2]]^3];
p = Transpose[temp];
ls = LinearSolve[p];
vals = ls[Flatten[f]];
System`Private`EvaluateListPolynomial[vals, 0.5, 0.5]
0.268157
answered Apr 16 at 10:59
user21user21
20.7k55997
answered Apr 16 at 10:59
user21user21
20.7k55997
answered Apr 16 at 10:59
user21user21
20.7k55997
answered Apr 16 at 10:59
user21user21
20.7k55997
20.7k55997
add a comment |
add a comment |
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$begingroup$
Dunno, but the return value of
if["InterpolationOrder"]that I get is9223372036854775806, 9223372036854775806. Oo$endgroup$
– Henrik Schumacher
Apr 16 at 8:49
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
Apr 16 at 8:54
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
Apr 16 at 9:00
$begingroup$
@HenrikSchumacher If this gives a hint, starting from 4 data points, the first 3 data point counts get interpolation order 2, then the next 4 get 3, then the next 5 get 4, etc.
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– Szabolcs
Apr 16 at 9:01
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That sounds as if they were using straight-forward global interpolation by a polynomial of degree up to
n. Then you haveBinomial[n, 2]basis functions. In that case, this should become nasty for higher point counts due to Runge's phenomenon and ill-conditioned linear systems (for solving for the coefficients). So I presume, that they will switch to another method when the point count becomes larger...$endgroup$
– Henrik Schumacher
Apr 16 at 9:06