Meaning of InterpolationOrder -> All for multidimensional interpolation
6
$begingroup$
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 4 hours ago
SzabolcsSzabolcs
164k14448950
$endgroup$
add a comment |
6
$begingroup$
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 4 hours ago
SzabolcsSzabolcs
164k14448950
$endgroup$
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo
$endgroup$
– Henrik Schumacher
4 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
4 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
3 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
3 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
3 hours ago
add a comment |
6
6
6
$begingroup$
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 4 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 4 hours ago
SzabolcsSzabolcs
164k14448950
asked 4 hours ago
SzabolcsSzabolcs
164k14448950
asked 4 hours ago
SzabolcsSzabolcs
164k14448950
asked 4 hours ago
SzabolcsSzabolcs
164k14448950
asked 4 hours ago
SzabolcsSzabolcs
164k14448950
164k14448950
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo
$endgroup$
– Henrik Schumacher
4 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
4 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
3 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
3 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
3 hours ago
add a comment |
$begingroup$
Dunno, but the return value ofif["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo
$endgroup$
– Henrik Schumacher
4 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
4 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
3 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
3 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
3 hours ago
$begingroup$
Dunno, but the return value of
if["InterpolationOrder"] that I get is {9223372036854775806, 9223372036854775806}. Oo$endgroup$
– Henrik Schumacher
4 hours ago
$begingroup$
Dunno, but the return value of
if["InterpolationOrder"] that I get is {9223372036854775806, 9223372036854775806}. Oo$endgroup$
– Henrik Schumacher
4 hours ago
1
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
4 hours ago
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
4 hours ago
1
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
3 hours ago
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
3 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
3 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
3 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
3 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
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
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 1 hour ago
user21user21
20k45386
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195279%2fmeaning-of-interpolationorder-all-for-multidimensional-interpolation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
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 1 hour ago
user21user21
20k45386
$endgroup$
add a comment |
5
$begingroup$
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 1 hour ago
user21user21
20k45386
$endgroup$
add a comment |
5
5
5
$begingroup$
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 1 hour ago
user21user21
20k45386
$endgroup$
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 1 hour ago
user21user21
20k45386
answered 1 hour ago
user21user21
20k45386
answered 1 hour ago
user21user21
20k45386
answered 1 hour ago
user21user21
20k45386
20k45386
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195279%2fmeaning-of-interpolationorder-all-for-multidimensional-interpolation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Dunno, but the return value of
if["InterpolationOrder"]that I get is{9223372036854775806, 9223372036854775806}. Oo$endgroup$
– Henrik Schumacher
4 hours ago
1
$begingroup$
@HenrikSchumacher Oops ... It seems I tried this with M12.0 (it's available in the cloud).
$endgroup$
– Szabolcs
4 hours ago
1
$begingroup$
Anyways, very good questions. I am also curious what works there in the background.
$endgroup$
– Henrik Schumacher
3 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
3 hours ago
1
$begingroup$
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
3 hours ago