Find an envelope of the list of points

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

I have a list of points as you can see in the image below. From this list of points, I want to generate a filtered list of points, which is the envelope. Additionally, calculate the area under the envelope.

rawData = 1.`, 1.`, 0.6666666666666666`, 
 0.9316770186335404`, 0.5`, 0.906832298136646`, 0.5`, 
 0.8757763975155279`, 0.5`, 0.8633540372670807`, 0.5`, 
 0.8509316770186336`, 0.5`, 0.7888198757763976`, 0.5`, 
 0.7142857142857143`, 0.5`, 0.6645962732919255`, 0.5`, 
 0.577639751552795`, 0.3333333333333333`, 
 0.5341614906832298`, 0.3333333333333333`, 
 0.453416149068323`, 0.16666666666666666`, 
 0.36024844720496896`, 0.16666666666666666`, 
 0.2670807453416149`, 0.`, 0.21739130434782608`, 0.`, 
 0.18633540372670807`, 0.`, 0.13664596273291926`, 0.`, 
 0.09937888198757763`, 0.`, 0.062111801242236024`, 0.`, 
 0.055900621118012424`, 0.`, 0.`;

points=1.`,1.`,0.6666666666666666`,0.9316770186335404`,0.5`,0.906832298136646`,0.3333333333333333`,0.5341614906832298`,0.16666666666666666`,0.36024844720496896`,0.`,0.21739130434782608`;

Show[ListPlot[rawData, PlotStyle -> Red], ListLinePlot[points]]
f = Interpolation[points, InterpolationOrder -> 1];
NIntegrate[f[t], t, 0, 1, Method -> "GlobalAdaptive"]

list-manipulation numerical-integration

share|improve this question

asked 1 hour ago

Kiril Danilchenko

625315

add a comment | 

up vote
2
down vote

favorite

I have a list of points as you can see in the image below. From this list of points, I want to generate a filtered list of points, which is the envelope. Additionally, calculate the area under the envelope.

rawData = 1.`, 1.`, 0.6666666666666666`, 
 0.9316770186335404`, 0.5`, 0.906832298136646`, 0.5`, 
 0.8757763975155279`, 0.5`, 0.8633540372670807`, 0.5`, 
 0.8509316770186336`, 0.5`, 0.7888198757763976`, 0.5`, 
 0.7142857142857143`, 0.5`, 0.6645962732919255`, 0.5`, 
 0.577639751552795`, 0.3333333333333333`, 
 0.5341614906832298`, 0.3333333333333333`, 
 0.453416149068323`, 0.16666666666666666`, 
 0.36024844720496896`, 0.16666666666666666`, 
 0.2670807453416149`, 0.`, 0.21739130434782608`, 0.`, 
 0.18633540372670807`, 0.`, 0.13664596273291926`, 0.`, 
 0.09937888198757763`, 0.`, 0.062111801242236024`, 0.`, 
 0.055900621118012424`, 0.`, 0.`;

points=1.`,1.`,0.6666666666666666`,0.9316770186335404`,0.5`,0.906832298136646`,0.3333333333333333`,0.5341614906832298`,0.16666666666666666`,0.36024844720496896`,0.`,0.21739130434782608`;

Show[ListPlot[rawData, PlotStyle -> Red], ListLinePlot[points]]
f = Interpolation[points, InterpolationOrder -> 1];
NIntegrate[f[t], t, 0, 1, Method -> "GlobalAdaptive"]

list-manipulation numerical-integration

share|improve this question

asked 1 hour ago

Kiril Danilchenko

625315

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I have a list of points as you can see in the image below. From this list of points, I want to generate a filtered list of points, which is the envelope. Additionally, calculate the area under the envelope.

rawData = 1.`, 1.`, 0.6666666666666666`, 
 0.9316770186335404`, 0.5`, 0.906832298136646`, 0.5`, 
 0.8757763975155279`, 0.5`, 0.8633540372670807`, 0.5`, 
 0.8509316770186336`, 0.5`, 0.7888198757763976`, 0.5`, 
 0.7142857142857143`, 0.5`, 0.6645962732919255`, 0.5`, 
 0.577639751552795`, 0.3333333333333333`, 
 0.5341614906832298`, 0.3333333333333333`, 
 0.453416149068323`, 0.16666666666666666`, 
 0.36024844720496896`, 0.16666666666666666`, 
 0.2670807453416149`, 0.`, 0.21739130434782608`, 0.`, 
 0.18633540372670807`, 0.`, 0.13664596273291926`, 0.`, 
 0.09937888198757763`, 0.`, 0.062111801242236024`, 0.`, 
 0.055900621118012424`, 0.`, 0.`;

points=1.`,1.`,0.6666666666666666`,0.9316770186335404`,0.5`,0.906832298136646`,0.3333333333333333`,0.5341614906832298`,0.16666666666666666`,0.36024844720496896`,0.`,0.21739130434782608`;

Show[ListPlot[rawData, PlotStyle -> Red], ListLinePlot[points]]
f = Interpolation[points, InterpolationOrder -> 1];
NIntegrate[f[t], t, 0, 1, Method -> "GlobalAdaptive"]

list-manipulation numerical-integration

share|improve this question

asked 1 hour ago

Kiril Danilchenko

625315

I have a list of points as you can see in the image below. From this list of points, I want to generate a filtered list of points, which is the envelope. Additionally, calculate the area under the envelope.

rawData = 1.`, 1.`, 0.6666666666666666`, 
 0.9316770186335404`, 0.5`, 0.906832298136646`, 0.5`, 
 0.8757763975155279`, 0.5`, 0.8633540372670807`, 0.5`, 
 0.8509316770186336`, 0.5`, 0.7888198757763976`, 0.5`, 
 0.7142857142857143`, 0.5`, 0.6645962732919255`, 0.5`, 
 0.577639751552795`, 0.3333333333333333`, 
 0.5341614906832298`, 0.3333333333333333`, 
 0.453416149068323`, 0.16666666666666666`, 
 0.36024844720496896`, 0.16666666666666666`, 
 0.2670807453416149`, 0.`, 0.21739130434782608`, 0.`, 
 0.18633540372670807`, 0.`, 0.13664596273291926`, 0.`, 
 0.09937888198757763`, 0.`, 0.062111801242236024`, 0.`, 
 0.055900621118012424`, 0.`, 0.`;

points=1.`,1.`,0.6666666666666666`,0.9316770186335404`,0.5`,0.906832298136646`,0.3333333333333333`,0.5341614906832298`,0.16666666666666666`,0.36024844720496896`,0.`,0.21739130434782608`;

Show[ListPlot[rawData, PlotStyle -> Red], ListLinePlot[points]]
f = Interpolation[points, InterpolationOrder -> 1];
NIntegrate[f[t], t, 0, 1, Method -> "GlobalAdaptive"]

list-manipulation numerical-integration

list-manipulation numerical-integration

share|improve this question

asked 1 hour ago

Kiril Danilchenko

625315

share|improve this question

asked 1 hour ago

Kiril Danilchenko

625315

share|improve this question

share|improve this question

asked 1 hour ago

Kiril Danilchenko

625315

asked 1 hour ago

Kiril Danilchenko

625315

asked 1 hour ago

Kiril Danilchenko

625315

625315

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

l, u = Transpose[Through[First, Last@#] & /@ GatherBy[SortBy[rawData, Last], First]];
f1, f2 = Interpolation[#, InterpolationOrder -> 1]& /@ l, u
NIntegrate[f2[t] - f1[t], t, 0, 1, Method -> "GlobalAdaptive"]

0.101967

Alternatively,

RegionMeasure@BoundaryDiscretizeGraphics[Polygon[Join[Reverse@l, u]]]

0.101967

ListPlot[rawData, l, u, Joined -> False, True, True, 
 PlotStyle -> Blue, Red, Green, Filling -> 2 -> 3]

share|improve this answer

edited 51 secs ago

answered 1 hour ago

kglr

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should use InterpolationOrder -> 1; otherwise the integral will be wrong.
  – Henrik Schumacher
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Henrik.
  – kglr
  15 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

If you assume "envelope" means the "shrink wrap" of the points, the answer is:

ConvexHullMesh[rawData]

If you want to get the area under the curve, add a point at 1,0.

myRegion = ConvexHullMesh[rawData];

Get the area:

RegionMeasure[myRegion]

(* 0.757764 *)

Get the coordinates:

MeshCoordinates[myRegion]

(*

1., 1., 0.5, 0.906832, 0., 0.217391, 0., 0.186335, 
 0., 0.136646, 0., 0.0993789, 0., 0.0621118, 0., 0.0559006, 
 0., 0., 1., 0.

*)

Show[HighlightMesh[myRegion, Style[2, Opacity[0.5]]], 
 Graphics[PointSize[0.02], Red, Point[MeshCoordinates[myRegion]]]]

share|improve this answer

edited 57 mins ago

answered 1 hour ago

David G. Stork

21.8k11747

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you for the answer. How I get the boundary points from the ConvexHull
  – Kiril Danilchenko
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem with this is that by design, it will miss dips or troughs in the data.
  – J. M. is computer-less♦
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It all comes down to what the poser means by "envelope." I assumed the Convex Hull.
  – David G. Stork
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A bit hard to explain since I am on mobile, but I will try. Look at kglr's answer. In particular, notice the green line in his answer. A convex hull approach will not be able to reproduce that.
  – J. M. is computer-less♦
  59 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Here another way, using the three-argument version of GroupBy:

a = KeySort[GroupBy[rawData, First -> Last, MinMax]];
lower = Values[a][[All, 1]];
upper = Values[a][[All, 2]];
t = Keys[a];
Show[
 ListLinePlot[Transpose[t, upper], Transpose[t, lower]],
 ListPlot[rawData, PlotStyle -> Red]
 ]

And since these functions are piecewise-linear, we can apply Tai's method directly to obtain the integral exactly:

ω = 0.5 (Join[#, 0.] + Join[0., #]) &@Differences[t];
(upper - lower).ω

0.101967

share|improve this answer

answered 16 mins ago

Henrik Schumacher

41.9k260125

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

l, u = Transpose[Through[First, Last@#] & /@ GatherBy[SortBy[rawData, Last], First]];
f1, f2 = Interpolation[#, InterpolationOrder -> 1]& /@ l, u
NIntegrate[f2[t] - f1[t], t, 0, 1, Method -> "GlobalAdaptive"]

0.101967

Alternatively,

RegionMeasure@BoundaryDiscretizeGraphics[Polygon[Join[Reverse@l, u]]]

0.101967

ListPlot[rawData, l, u, Joined -> False, True, True, 
 PlotStyle -> Blue, Red, Green, Filling -> 2 -> 3]

share|improve this answer

edited 51 secs ago

answered 1 hour ago

kglr

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should use InterpolationOrder -> 1; otherwise the integral will be wrong.
  – Henrik Schumacher
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Henrik.
  – kglr
  15 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

l, u = Transpose[Through[First, Last@#] & /@ GatherBy[SortBy[rawData, Last], First]];
f1, f2 = Interpolation[#, InterpolationOrder -> 1]& /@ l, u
NIntegrate[f2[t] - f1[t], t, 0, 1, Method -> "GlobalAdaptive"]

0.101967

Alternatively,

RegionMeasure@BoundaryDiscretizeGraphics[Polygon[Join[Reverse@l, u]]]

0.101967

ListPlot[rawData, l, u, Joined -> False, True, True, 
 PlotStyle -> Blue, Red, Green, Filling -> 2 -> 3]

share|improve this answer

edited 51 secs ago

answered 1 hour ago

kglr

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should use InterpolationOrder -> 1; otherwise the integral will be wrong.
  – Henrik Schumacher
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Henrik.
  – kglr
  15 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

l, u = Transpose[Through[First, Last@#] & /@ GatherBy[SortBy[rawData, Last], First]];
f1, f2 = Interpolation[#, InterpolationOrder -> 1]& /@ l, u
NIntegrate[f2[t] - f1[t], t, 0, 1, Method -> "GlobalAdaptive"]

0.101967

Alternatively,

RegionMeasure@BoundaryDiscretizeGraphics[Polygon[Join[Reverse@l, u]]]

0.101967

ListPlot[rawData, l, u, Joined -> False, True, True, 
 PlotStyle -> Blue, Red, Green, Filling -> 2 -> 3]

share|improve this answer

edited 51 secs ago

answered 1 hour ago

kglr

165k8188388

l, u = Transpose[Through[First, Last@#] & /@ GatherBy[SortBy[rawData, Last], First]];
f1, f2 = Interpolation[#, InterpolationOrder -> 1]& /@ l, u
NIntegrate[f2[t] - f1[t], t, 0, 1, Method -> "GlobalAdaptive"]

0.101967

Alternatively,

RegionMeasure@BoundaryDiscretizeGraphics[Polygon[Join[Reverse@l, u]]]

0.101967

ListPlot[rawData, l, u, Joined -> False, True, True, 
 PlotStyle -> Blue, Red, Green, Filling -> 2 -> 3]

share|improve this answer

edited 51 secs ago

answered 1 hour ago

kglr

165k8188388

share|improve this answer

share|improve this answer

edited 51 secs ago

edited 51 secs ago

edited 51 secs ago

answered 1 hour ago

kglr

165k8188388

answered 1 hour ago

kglr

165k8188388

answered 1 hour ago

kglr

165k8188388

165k8188388

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should use InterpolationOrder -> 1; otherwise the integral will be wrong.
  – Henrik Schumacher
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Henrik.
  – kglr
  15 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should use InterpolationOrder -> 1; otherwise the integral will be wrong.
  – Henrik Schumacher
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Henrik.
  – kglr
  15 mins ago
  

  
  
  
  

You should use InterpolationOrder -> 1; otherwise the integral will be wrong.
– Henrik Schumacher
15 mins ago

You should use InterpolationOrder -> 1; otherwise the integral will be wrong.
– Henrik Schumacher
15 mins ago

Thank you @Henrik.
– kglr
15 mins ago

Thank you @Henrik.
– kglr
15 mins ago

add a comment | 

up vote
1
down vote

If you assume "envelope" means the "shrink wrap" of the points, the answer is:

ConvexHullMesh[rawData]

If you want to get the area under the curve, add a point at 1,0.

myRegion = ConvexHullMesh[rawData];

Get the area:

RegionMeasure[myRegion]

(* 0.757764 *)

Get the coordinates:

MeshCoordinates[myRegion]

(*

1., 1., 0.5, 0.906832, 0., 0.217391, 0., 0.186335, 
 0., 0.136646, 0., 0.0993789, 0., 0.0621118, 0., 0.0559006, 
 0., 0., 1., 0.

*)

Show[HighlightMesh[myRegion, Style[2, Opacity[0.5]]], 
 Graphics[PointSize[0.02], Red, Point[MeshCoordinates[myRegion]]]]

share|improve this answer

edited 57 mins ago

answered 1 hour ago

David G. Stork

21.8k11747

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you for the answer. How I get the boundary points from the ConvexHull
  – Kiril Danilchenko
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem with this is that by design, it will miss dips or troughs in the data.
  – J. M. is computer-less♦
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It all comes down to what the poser means by "envelope." I assumed the Convex Hull.
  – David G. Stork
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A bit hard to explain since I am on mobile, but I will try. Look at kglr's answer. In particular, notice the green line in his answer. A convex hull approach will not be able to reproduce that.
  – J. M. is computer-less♦
  59 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

If you assume "envelope" means the "shrink wrap" of the points, the answer is:

ConvexHullMesh[rawData]

If you want to get the area under the curve, add a point at 1,0.

myRegion = ConvexHullMesh[rawData];

Get the area:

RegionMeasure[myRegion]

(* 0.757764 *)

Get the coordinates:

MeshCoordinates[myRegion]

(*

1., 1., 0.5, 0.906832, 0., 0.217391, 0., 0.186335, 
 0., 0.136646, 0., 0.0993789, 0., 0.0621118, 0., 0.0559006, 
 0., 0., 1., 0.

*)

Show[HighlightMesh[myRegion, Style[2, Opacity[0.5]]], 
 Graphics[PointSize[0.02], Red, Point[MeshCoordinates[myRegion]]]]

share|improve this answer

edited 57 mins ago

answered 1 hour ago

David G. Stork

21.8k11747

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you for the answer. How I get the boundary points from the ConvexHull
  – Kiril Danilchenko
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem with this is that by design, it will miss dips or troughs in the data.
  – J. M. is computer-less♦
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It all comes down to what the poser means by "envelope." I assumed the Convex Hull.
  – David G. Stork
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A bit hard to explain since I am on mobile, but I will try. Look at kglr's answer. In particular, notice the green line in his answer. A convex hull approach will not be able to reproduce that.
  – J. M. is computer-less♦
  59 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

If you assume "envelope" means the "shrink wrap" of the points, the answer is:

ConvexHullMesh[rawData]

If you want to get the area under the curve, add a point at 1,0.

myRegion = ConvexHullMesh[rawData];

Get the area:

RegionMeasure[myRegion]

(* 0.757764 *)

Get the coordinates:

MeshCoordinates[myRegion]

(*

1., 1., 0.5, 0.906832, 0., 0.217391, 0., 0.186335, 
 0., 0.136646, 0., 0.0993789, 0., 0.0621118, 0., 0.0559006, 
 0., 0., 1., 0.

*)

Show[HighlightMesh[myRegion, Style[2, Opacity[0.5]]], 
 Graphics[PointSize[0.02], Red, Point[MeshCoordinates[myRegion]]]]

share|improve this answer

edited 57 mins ago

answered 1 hour ago

David G. Stork

21.8k11747

If you assume "envelope" means the "shrink wrap" of the points, the answer is:

ConvexHullMesh[rawData]

If you want to get the area under the curve, add a point at 1,0.

myRegion = ConvexHullMesh[rawData];

Get the area:

RegionMeasure[myRegion]

(* 0.757764 *)

Get the coordinates:

MeshCoordinates[myRegion]

(*

1., 1., 0.5, 0.906832, 0., 0.217391, 0., 0.186335, 
 0., 0.136646, 0., 0.0993789, 0., 0.0621118, 0., 0.0559006, 
 0., 0., 1., 0.

*)

Show[HighlightMesh[myRegion, Style[2, Opacity[0.5]]], 
 Graphics[PointSize[0.02], Red, Point[MeshCoordinates[myRegion]]]]

share|improve this answer

edited 57 mins ago

answered 1 hour ago

David G. Stork

21.8k11747

share|improve this answer

share|improve this answer

edited 57 mins ago

edited 57 mins ago

edited 57 mins ago

answered 1 hour ago

David G. Stork

21.8k11747

answered 1 hour ago

David G. Stork

21.8k11747

answered 1 hour ago

David G. Stork

21.8k11747

21.8k11747

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you for the answer. How I get the boundary points from the ConvexHull
  – Kiril Danilchenko
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem with this is that by design, it will miss dips or troughs in the data.
  – J. M. is computer-less♦
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It all comes down to what the poser means by "envelope." I assumed the Convex Hull.
  – David G. Stork
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A bit hard to explain since I am on mobile, but I will try. Look at kglr's answer. In particular, notice the green line in his answer. A convex hull approach will not be able to reproduce that.
  – J. M. is computer-less♦
  59 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you for the answer. How I get the boundary points from the ConvexHull
  – Kiril Danilchenko
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The problem with this is that by design, it will miss dips or troughs in the data.
  – J. M. is computer-less♦
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It all comes down to what the poser means by "envelope." I assumed the Convex Hull.
  – David G. Stork
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  A bit hard to explain since I am on mobile, but I will try. Look at kglr's answer. In particular, notice the green line in his answer. A convex hull approach will not be able to reproduce that.
  – J. M. is computer-less♦
  59 mins ago
  

  
  
  
  

thank you for the answer. How I get the boundary points from the ConvexHull
– Kiril Danilchenko
1 hour ago

thank you for the answer. How I get the boundary points from the ConvexHull
– Kiril Danilchenko
1 hour ago

The problem with this is that by design, it will miss dips or troughs in the data.
– J. M. is computer-less♦
1 hour ago

The problem with this is that by design, it will miss dips or troughs in the data.
– J. M. is computer-less♦
1 hour ago

It all comes down to what the poser means by "envelope." I assumed the Convex Hull.
– David G. Stork
1 hour ago

It all comes down to what the poser means by "envelope." I assumed the Convex Hull.
– David G. Stork
1 hour ago

1

1

A bit hard to explain since I am on mobile, but I will try. Look at kglr's answer. In particular, notice the green line in his answer. A convex hull approach will not be able to reproduce that.
– J. M. is computer-less♦
59 mins ago

A bit hard to explain since I am on mobile, but I will try. Look at kglr's answer. In particular, notice the green line in his answer. A convex hull approach will not be able to reproduce that.
– J. M. is computer-less♦
59 mins ago

add a comment | 

up vote
1
down vote

Here another way, using the three-argument version of GroupBy:

a = KeySort[GroupBy[rawData, First -> Last, MinMax]];
lower = Values[a][[All, 1]];
upper = Values[a][[All, 2]];
t = Keys[a];
Show[
 ListLinePlot[Transpose[t, upper], Transpose[t, lower]],
 ListPlot[rawData, PlotStyle -> Red]
 ]

And since these functions are piecewise-linear, we can apply Tai's method directly to obtain the integral exactly:

ω = 0.5 (Join[#, 0.] + Join[0., #]) &@Differences[t];
(upper - lower).ω

0.101967

share|improve this answer

answered 16 mins ago

Henrik Schumacher

41.9k260125

add a comment | 

up vote
1
down vote

Here another way, using the three-argument version of GroupBy:

a = KeySort[GroupBy[rawData, First -> Last, MinMax]];
lower = Values[a][[All, 1]];
upper = Values[a][[All, 2]];
t = Keys[a];
Show[
 ListLinePlot[Transpose[t, upper], Transpose[t, lower]],
 ListPlot[rawData, PlotStyle -> Red]
 ]

And since these functions are piecewise-linear, we can apply Tai's method directly to obtain the integral exactly:

ω = 0.5 (Join[#, 0.] + Join[0., #]) &@Differences[t];
(upper - lower).ω

0.101967

share|improve this answer

answered 16 mins ago

Henrik Schumacher

41.9k260125

add a comment | 

up vote
1
down vote

up vote
1
down vote

Here another way, using the three-argument version of GroupBy:

a = KeySort[GroupBy[rawData, First -> Last, MinMax]];
lower = Values[a][[All, 1]];
upper = Values[a][[All, 2]];
t = Keys[a];
Show[
 ListLinePlot[Transpose[t, upper], Transpose[t, lower]],
 ListPlot[rawData, PlotStyle -> Red]
 ]

And since these functions are piecewise-linear, we can apply Tai's method directly to obtain the integral exactly:

ω = 0.5 (Join[#, 0.] + Join[0., #]) &@Differences[t];
(upper - lower).ω

0.101967

share|improve this answer

answered 16 mins ago

Henrik Schumacher

41.9k260125

Here another way, using the three-argument version of GroupBy:

a = KeySort[GroupBy[rawData, First -> Last, MinMax]];
lower = Values[a][[All, 1]];
upper = Values[a][[All, 2]];
t = Keys[a];
Show[
 ListLinePlot[Transpose[t, upper], Transpose[t, lower]],
 ListPlot[rawData, PlotStyle -> Red]
 ]

And since these functions are piecewise-linear, we can apply Tai's method directly to obtain the integral exactly:

ω = 0.5 (Join[#, 0.] + Join[0., #]) &@Differences[t];
(upper - lower).ω

0.101967

share|improve this answer

answered 16 mins ago

Henrik Schumacher

41.9k260125

share|improve this answer

share|improve this answer

answered 16 mins ago

Henrik Schumacher

41.9k260125

answered 16 mins ago

Henrik Schumacher

41.9k260125

answered 16 mins ago

Henrik Schumacher

41.9k260125

41.9k260125

add a comment | 

add a comment | 

 

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