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 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 9 hours ago
ATomekATomek
1098 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 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 9 hours ago
ATomekATomek
1098 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
8 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 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 9 hours ago
ATomekATomek
1098 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 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 9 hours ago
ATomekATomek
1098 bronze badges
edited 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
asked 9 hours ago
ATomekATomek
1098 bronze badges
edited 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
edited 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
edited 8 hours ago
MelaGo
2,0161 gold badge1 silver badge7 bronze badges
2,0161 gold badge1 silver badge7 bronze badges
asked 9 hours ago
ATomekATomek
1098 bronze badges
asked 9 hours ago
ATomekATomek
1098 bronze badges
asked 9 hours ago
ATomekATomek
1098 bronze badges
1098 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
8 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
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
$begingroup$
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
8 hours ago
$begingroup$
z values of the plot along the line projected onto the x-y plane
$endgroup$
– ATomek
8 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 8 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 1 hour 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 8 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 8 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 8 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 8 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
edited 8 hours ago
answered 8 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
answered 8 hours ago
kglrkglr
204k10 gold badges233 silver badges463 bronze badges
answered 8 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
answered 1 hour ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
answered 1 hour ago
Michael E2Michael E2
155k12 gold badges213 silver badges502 bronze badges
answered 1 hour 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
8 hours ago