Extracting points from 3D plot that lie along an arbitrarily oriented lineHow to properly project a Graphics object consisting of line primitivesMapping Contour Plot onto ListPlot3D (or by using color variations)Region projection of multivariable interpolated functionParticle moving on curve which is the intersection of a surface and a planeRendering ListPlot3D SurfaceAnimate a circle “rolling” along a complicated 3D curveListPlot with a histogram of values on the vertical axisFinding optimal points in contours produced by ListContourPlotFinding average of attributed linesListPlot3D label is covered by surface in combined graphic
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Extracting points from 3D plot that lie along an arbitrarily oriented line
How to properly project a Graphics object consisting of line primitivesMapping Contour Plot onto ListPlot3D (or by using color variations)Region projection of multivariable interpolated functionParticle moving on curve which is the intersection of a surface and a planeRendering ListPlot3D SurfaceAnimate a circle “rolling” along a complicated 3D curveListPlot with a histogram of values on the vertical axisFinding optimal points in contours produced by ListContourPlotFinding average of attributed linesListPlot3D label is covered by surface in combined graphic
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Starting from i.e. the following 3d plot:
d = RandomReal[1, 100, 3];
ListPlot3D[d]
Is it possible to extract points that lay along an arbitrarily oriented line, i.e. like this:
Show[ListPlot3D[d], Graphics3D[Line[0, -.5, 1, 0.5, 1, 1]]]
?
EDIT: Z values of the plot along the line projected onto the x-y plane
plotting list-manipulation graphics3d mesh
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
$endgroup$
add a comment |
$begingroup$
Starting from i.e. the following 3d plot:
d = RandomReal[1, 100, 3];
ListPlot3D[d]
Is it possible to extract points that lay along an arbitrarily oriented line, i.e. like this:
Show[ListPlot3D[d], Graphics3D[Line[0, -.5, 1, 0.5, 1, 1]]]
?
EDIT: Z values of the plot along the line projected onto the x-y plane
plotting list-manipulation graphics3d mesh
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
$endgroup$
$begingroup$
Can you clarify a little, are you looking for the points that exactly intersect that line, or the z values of the plot along the line projected onto the x-y plane?
$endgroup$
– N.J.Evans
8 hours ago
$begingroup$
z values of the plot along the line projected onto the x-y plane
$endgroup$
– ATomek
7 hours ago
add a comment |
$begingroup$
Starting from i.e. the following 3d plot:
d = RandomReal[1, 100, 3];
ListPlot3D[d]
Is it possible to extract points that lay along an arbitrarily oriented line, i.e. like this:
Show[ListPlot3D[d], Graphics3D[Line[0, -.5, 1, 0.5, 1, 1]]]
?
EDIT: Z values of the plot along the line projected onto the x-y plane
plotting list-manipulation graphics3d mesh
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
$endgroup$
Starting from i.e. the following 3d plot:
d = RandomReal[1, 100, 3];
ListPlot3D[d]
Is it possible to extract points that lay along an arbitrarily oriented line, i.e. like this:
Show[ListPlot3D[d], Graphics3D[Line[0, -.5, 1, 0.5, 1, 1]]]
?
EDIT: Z values of the plot along the line projected onto the x-y plane
plotting list-manipulation graphics3d mesh
plotting list-manipulation graphics3d mesh
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
edited 7 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
2,0161 gold badge1 silver badge7 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
asked 8 hours ago
ATomekATomek
1048 bronze badges
1048 bronze badges
$begingroup$
Can you clarify a little, are you looking for the points that exactly intersect that line, or the z values of the plot along the line projected onto the x-y plane?
$endgroup$
– N.J.Evans
8 hours ago
$begingroup$
z values of the plot along the line projected onto the x-y plane
$endgroup$
– ATomek
7 hours ago
add a comment |
$begingroup$
Can you clarify a little, are you looking for the points that exactly intersect that line, or the z values of the plot along the line projected onto the x-y plane?
$endgroup$
– N.J.Evans
8 hours ago
$begingroup$
z values of the plot along the line projected onto the x-y plane
$endgroup$
– ATomek
7 hours ago
$begingroup$
Can you clarify a little, are you looking for the points that exactly intersect that line, or the z values of the plot along the line projected onto the x-y plane?
$endgroup$
– N.J.Evans
8 hours ago
$begingroup$
Can you clarify a little, are you looking for the points that exactly intersect that line, or the z values of the plot along the line projected onto the x-y plane?
$endgroup$
– N.J.Evans
8 hours ago
$begingroup$
z values of the plot along the line projected onto the x-y plane
$endgroup$
– ATomek
7 hours ago
$begingroup$
z values of the plot along the line projected onto the x-y plane
$endgroup$
– ATomek
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
SeedRandom[5]
d = RandomReal[1, 100, 3];
You can use -.5 + 3 # - #2 & (or Function[x, y, -.5 + 3 x - y]) as the setting for MeshFunctions in ListPlot3D:
Show[lp3d = ListPlot3D[d, MeshFunctions -> -.5 + 3 # - #2 &,
Mesh -> 0, MeshStyle -> Directive[Red, Thick], BoundaryStyle -> None],
Graphics3D[Thick , Blue, Line[0, -.5, 1, 0.5, 1, 1],
Opacity[.5, Yellow], EdgeForm @ None,
InfinitePlane[0, -.5, 0, 0, -.5, 1, 0.5, 1, 1]]]
To extract the points on the red line:
Cases[Normal @ lp3d, Line[x_] :> x, All][[1]]
0.492655, 0.977959, 0.559503, 0.491386, 0.973947,
0.484116, 0.477966, 0.933211, 0.303948, 0.476062, 0.92816,
0.451244, 0.459746, 0.878862, 0.640324, 0.457492,
0.872273,
0.586974, 0.454029, 0.861943, 0.568448, 0.441994,
0.825895,
0.39336, 0.396855, 0.690417, 0.20754, 0.395551, 0.686462,
0.255852, 0.392548, 0.677572, 0.17963, 0.350269, 0.550753,
0.612022, 0.341512, 0.524435, 0.760695, 0.313441,
0.440292,
0.52727, 0.304016, 0.411343, 0.164743, 0.300216, 0.400556,
0.123016, 0.293312, 0.379549, 0.431375, 0.279583, 0.33874,
0.571648, 0.267421, 0.302245, 0.850239, 0.266698,
0.299818,
0.85685, 0.226072, 0.177329, 0.737677, 0.217928, 0.152775,
0.650557, 0.192837, 0.0784445, 0.831587, 0.187349,
0.0619587,
0.854096, 0.174868, 0.024533, 0.602241
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
$endgroup$
add a comment |
$begingroup$
You can use the interpolation that ListPlot uses, which you can then evaluate at any point on the line (within the domain of the interpolation):
zFN = Interpolation[d, InterpolationOrder -> 1,
"ExtrapolationHandler" -> Indeterminate &, "WarningMessage" -> False];
ClearAll[xyline, zSect];
xyline[x_] = (1 - 2 x), 2 x.0, -.5, 0.5, 1;
zSect[x_] := zFN @@ xyline[x];
zSect[0.25] (* test a value *)
(* 0.654833 *)
Show[
ListPlot3D[d],
ParametricPlot3D[Append[xyline[x], zSect[x]], x, 0, 1]
]
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
SeedRandom[5]
d = RandomReal[1, 100, 3];
You can use -.5 + 3 # - #2 & (or Function[x, y, -.5 + 3 x - y]) as the setting for MeshFunctions in ListPlot3D:
Show[lp3d = ListPlot3D[d, MeshFunctions -> -.5 + 3 # - #2 &,
Mesh -> 0, MeshStyle -> Directive[Red, Thick], BoundaryStyle -> None],
Graphics3D[Thick , Blue, Line[0, -.5, 1, 0.5, 1, 1],
Opacity[.5, Yellow], EdgeForm @ None,
InfinitePlane[0, -.5, 0, 0, -.5, 1, 0.5, 1, 1]]]
To extract the points on the red line:
Cases[Normal @ lp3d, Line[x_] :> x, All][[1]]
0.492655, 0.977959, 0.559503, 0.491386, 0.973947,
0.484116, 0.477966, 0.933211, 0.303948, 0.476062, 0.92816,
0.451244, 0.459746, 0.878862, 0.640324, 0.457492,
0.872273,
0.586974, 0.454029, 0.861943, 0.568448, 0.441994,
0.825895,
0.39336, 0.396855, 0.690417, 0.20754, 0.395551, 0.686462,
0.255852, 0.392548, 0.677572, 0.17963, 0.350269, 0.550753,
0.612022, 0.341512, 0.524435, 0.760695, 0.313441,
0.440292,
0.52727, 0.304016, 0.411343, 0.164743, 0.300216, 0.400556,
0.123016, 0.293312, 0.379549, 0.431375, 0.279583, 0.33874,
0.571648, 0.267421, 0.302245, 0.850239, 0.266698,
0.299818,
0.85685, 0.226072, 0.177329, 0.737677, 0.217928, 0.152775,
0.650557, 0.192837, 0.0784445, 0.831587, 0.187349,
0.0619587,
0.854096, 0.174868, 0.024533, 0.602241
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
$endgroup$
add a comment |
$begingroup$
SeedRandom[5]
d = RandomReal[1, 100, 3];
You can use -.5 + 3 # - #2 & (or Function[x, y, -.5 + 3 x - y]) as the setting for MeshFunctions in ListPlot3D:
Show[lp3d = ListPlot3D[d, MeshFunctions -> -.5 + 3 # - #2 &,
Mesh -> 0, MeshStyle -> Directive[Red, Thick], BoundaryStyle -> None],
Graphics3D[Thick , Blue, Line[0, -.5, 1, 0.5, 1, 1],
Opacity[.5, Yellow], EdgeForm @ None,
InfinitePlane[0, -.5, 0, 0, -.5, 1, 0.5, 1, 1]]]
To extract the points on the red line:
Cases[Normal @ lp3d, Line[x_] :> x, All][[1]]
0.492655, 0.977959, 0.559503, 0.491386, 0.973947,
0.484116, 0.477966, 0.933211, 0.303948, 0.476062, 0.92816,
0.451244, 0.459746, 0.878862, 0.640324, 0.457492,
0.872273,
0.586974, 0.454029, 0.861943, 0.568448, 0.441994,
0.825895,
0.39336, 0.396855, 0.690417, 0.20754, 0.395551, 0.686462,
0.255852, 0.392548, 0.677572, 0.17963, 0.350269, 0.550753,
0.612022, 0.341512, 0.524435, 0.760695, 0.313441,
0.440292,
0.52727, 0.304016, 0.411343, 0.164743, 0.300216, 0.400556,
0.123016, 0.293312, 0.379549, 0.431375, 0.279583, 0.33874,
0.571648, 0.267421, 0.302245, 0.850239, 0.266698,
0.299818,
0.85685, 0.226072, 0.177329, 0.737677, 0.217928, 0.152775,
0.650557, 0.192837, 0.0784445, 0.831587, 0.187349,
0.0619587,
0.854096, 0.174868, 0.024533, 0.602241
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
$endgroup$
add a comment |
$begingroup$
SeedRandom[5]
d = RandomReal[1, 100, 3];
You can use -.5 + 3 # - #2 & (or Function[x, y, -.5 + 3 x - y]) as the setting for MeshFunctions in ListPlot3D:
Show[lp3d = ListPlot3D[d, MeshFunctions -> -.5 + 3 # - #2 &,
Mesh -> 0, MeshStyle -> Directive[Red, Thick], BoundaryStyle -> None],
Graphics3D[Thick , Blue, Line[0, -.5, 1, 0.5, 1, 1],
Opacity[.5, Yellow], EdgeForm @ None,
InfinitePlane[0, -.5, 0, 0, -.5, 1, 0.5, 1, 1]]]
To extract the points on the red line:
Cases[Normal @ lp3d, Line[x_] :> x, All][[1]]
0.492655, 0.977959, 0.559503, 0.491386, 0.973947,
0.484116, 0.477966, 0.933211, 0.303948, 0.476062, 0.92816,
0.451244, 0.459746, 0.878862, 0.640324, 0.457492,
0.872273,
0.586974, 0.454029, 0.861943, 0.568448, 0.441994,
0.825895,
0.39336, 0.396855, 0.690417, 0.20754, 0.395551, 0.686462,
0.255852, 0.392548, 0.677572, 0.17963, 0.350269, 0.550753,
0.612022, 0.341512, 0.524435, 0.760695, 0.313441,
0.440292,
0.52727, 0.304016, 0.411343, 0.164743, 0.300216, 0.400556,
0.123016, 0.293312, 0.379549, 0.431375, 0.279583, 0.33874,
0.571648, 0.267421, 0.302245, 0.850239, 0.266698,
0.299818,
0.85685, 0.226072, 0.177329, 0.737677, 0.217928, 0.152775,
0.650557, 0.192837, 0.0784445, 0.831587, 0.187349,
0.0619587,
0.854096, 0.174868, 0.024533, 0.602241
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
$endgroup$
SeedRandom[5]
d = RandomReal[1, 100, 3];
You can use -.5 + 3 # - #2 & (or Function[x, y, -.5 + 3 x - y]) as the setting for MeshFunctions in ListPlot3D:
Show[lp3d = ListPlot3D[d, MeshFunctions -> -.5 + 3 # - #2 &,
Mesh -> 0, MeshStyle -> Directive[Red, Thick], BoundaryStyle -> None],
Graphics3D[Thick , Blue, Line[0, -.5, 1, 0.5, 1, 1],
Opacity[.5, Yellow], EdgeForm @ None,
InfinitePlane[0, -.5, 0, 0, -.5, 1, 0.5, 1, 1]]]
To extract the points on the red line:
Cases[Normal @ lp3d, Line[x_] :> x, All][[1]]
0.492655, 0.977959, 0.559503, 0.491386, 0.973947,
0.484116, 0.477966, 0.933211, 0.303948, 0.476062, 0.92816,
0.451244, 0.459746, 0.878862, 0.640324, 0.457492,
0.872273,
0.586974, 0.454029, 0.861943, 0.568448, 0.441994,
0.825895,
0.39336, 0.396855, 0.690417, 0.20754, 0.395551, 0.686462,
0.255852, 0.392548, 0.677572, 0.17963, 0.350269, 0.550753,
0.612022, 0.341512, 0.524435, 0.760695, 0.313441,
0.440292,
0.52727, 0.304016, 0.411343, 0.164743, 0.300216, 0.400556,
0.123016, 0.293312, 0.379549, 0.431375, 0.279583, 0.33874,
0.571648, 0.267421, 0.302245, 0.850239, 0.266698,
0.299818,
0.85685, 0.226072, 0.177329, 0.737677, 0.217928, 0.152775,
0.650557, 0.192837, 0.0784445, 0.831587, 0.187349,
0.0619587,
0.854096, 0.174868, 0.024533, 0.602241
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
edited 7 hours ago
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
answered 7 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
204k10 gold badges233 silver badges463 bronze badges
add a comment |
add a comment |
$begingroup$
You can use the interpolation that ListPlot uses, which you can then evaluate at any point on the line (within the domain of the interpolation):
zFN = Interpolation[d, InterpolationOrder -> 1,
"ExtrapolationHandler" -> Indeterminate &, "WarningMessage" -> False];
ClearAll[xyline, zSect];
xyline[x_] = (1 - 2 x), 2 x.0, -.5, 0.5, 1;
zSect[x_] := zFN @@ xyline[x];
zSect[0.25] (* test a value *)
(* 0.654833 *)
Show[
ListPlot3D[d],
ParametricPlot3D[Append[xyline[x], zSect[x]], x, 0, 1]
]
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
$endgroup$
add a comment |
$begingroup$
You can use the interpolation that ListPlot uses, which you can then evaluate at any point on the line (within the domain of the interpolation):
zFN = Interpolation[d, InterpolationOrder -> 1,
"ExtrapolationHandler" -> Indeterminate &, "WarningMessage" -> False];
ClearAll[xyline, zSect];
xyline[x_] = (1 - 2 x), 2 x.0, -.5, 0.5, 1;
zSect[x_] := zFN @@ xyline[x];
zSect[0.25] (* test a value *)
(* 0.654833 *)
Show[
ListPlot3D[d],
ParametricPlot3D[Append[xyline[x], zSect[x]], x, 0, 1]
]
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
$endgroup$
add a comment |
$begingroup$
You can use the interpolation that ListPlot uses, which you can then evaluate at any point on the line (within the domain of the interpolation):
zFN = Interpolation[d, InterpolationOrder -> 1,
"ExtrapolationHandler" -> Indeterminate &, "WarningMessage" -> False];
ClearAll[xyline, zSect];
xyline[x_] = (1 - 2 x), 2 x.0, -.5, 0.5, 1;
zSect[x_] := zFN @@ xyline[x];
zSect[0.25] (* test a value *)
(* 0.654833 *)
Show[
ListPlot3D[d],
ParametricPlot3D[Append[xyline[x], zSect[x]], x, 0, 1]
]
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
$endgroup$
You can use the interpolation that ListPlot uses, which you can then evaluate at any point on the line (within the domain of the interpolation):
zFN = Interpolation[d, InterpolationOrder -> 1,
"ExtrapolationHandler" -> Indeterminate &, "WarningMessage" -> False];
ClearAll[xyline, zSect];
xyline[x_] = (1 - 2 x), 2 x.0, -.5, 0.5, 1;
zSect[x_] := zFN @@ xyline[x];
zSect[0.25] (* test a value *)
(* 0.654833 *)
Show[
ListPlot3D[d],
ParametricPlot3D[Append[xyline[x], zSect[x]], x, 0, 1]
]
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
answered 55 mins ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
155k12 gold badges213 silver badges502 bronze badges
add a comment |
add a comment |
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$begingroup$
Can you clarify a little, are you looking for the points that exactly intersect that line, or the z values of the plot along the line projected onto the x-y plane?
$endgroup$
– N.J.Evans
8 hours ago
$begingroup$
z values of the plot along the line projected onto the x-y plane
$endgroup$
– ATomek
7 hours ago