Meaning of InterpolationOrder -> All for multidimensional interpolation Planned maintenance...
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Meaning of InterpolationOrder -> All for multidimensional interpolation
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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 21 hours ago
SzabolcsSzabolcs
164k14448950
$endgroup$
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 21 hours ago
SzabolcsSzabolcs
164k14448950
$endgroup$
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo
$endgroup$
– Henrik Schumacher
21 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
21 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
21 hours ago
$begingroup$
@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
21 hours ago
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
21 hours ago
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 21 hours ago
SzabolcsSzabolcs
164k14448950
$endgroup$
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 21 hours ago
SzabolcsSzabolcs
164k14448950
asked 21 hours ago
SzabolcsSzabolcs
164k14448950
asked 21 hours ago
SzabolcsSzabolcs
164k14448950
asked 21 hours ago
SzabolcsSzabolcs
164k14448950
asked 21 hours ago
SzabolcsSzabolcs
164k14448950
164k14448950
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo
$endgroup$
– Henrik Schumacher
21 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
21 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
21 hours ago
$begingroup$
@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
21 hours ago
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
21 hours ago
add a comment |
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo
$endgroup$
– Henrik Schumacher
21 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
21 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
21 hours ago
$begingroup$
@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
21 hours ago
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
21 hours ago
$begingroup$
Dunno, but the return value of
if["InterpolationOrder"] that I get is {9223372036854775806, 9223372036854775806}. Oo$endgroup$
– Henrik Schumacher
21 hours ago
$begingroup$
Dunno, but the return value of
if["InterpolationOrder"] that I get is {9223372036854775806, 9223372036854775806}. Oo$endgroup$
– Henrik Schumacher
21 hours ago
1
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
21 hours ago
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
21 hours ago
1
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
21 hours ago
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
21 hours ago
$begingroup$
@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
21 hours ago
$begingroup$
@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
21 hours ago
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
21 hours ago
$begingroup$
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
21 hours ago
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 19 hours ago
user21user21
20.2k45487
$endgroup$
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 19 hours ago
user21user21
20.2k45487
$endgroup$
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 19 hours ago
user21user21
20.2k45487
$endgroup$
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 19 hours ago
user21user21
20.2k45487
$endgroup$
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 19 hours ago
user21user21
20.2k45487
answered 19 hours ago
user21user21
20.2k45487
answered 19 hours ago
user21user21
20.2k45487
answered 19 hours ago
user21user21
20.2k45487
20.2k45487
add a comment |
add a comment |
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$begingroup$
Dunno, but the return value of
if["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo$endgroup$
– Henrik Schumacher
21 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
21 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
21 hours ago
$begingroup$
@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
21 hours ago
1
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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
21 hours ago