How to count points under the curve?Picking points in a plane under a curve given by a listHow to select points from a table that are within some regionSelecting points on either side of a curveApproximating an ornamental curveCount the number of regions made by Lissajous curvePlotting the image of a curve under a flowReducing the number of plotmarkers in listlineplotHow to calculate the control points of a Bézier curve?Plotting the sum of two points on an elliptic curveRegion bounded by the curveRandom points around the given curveFinding Area Under a Curveshowing the grid points on the curve
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How to count points under the curve?
Picking points in a plane under a curve given by a listHow to select points from a table that are within some regionSelecting points on either side of a curveApproximating an ornamental curveCount the number of regions made by Lissajous curvePlotting the image of a curve under a flowReducing the number of plotmarkers in listlineplotHow to calculate the control points of a Bézier curve?Plotting the sum of two points on an elliptic curveRegion bounded by the curveRandom points around the given curveFinding Area Under a Curveshowing the grid points on the curve
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
6
$begingroup$
I need a formula to calculate how many points are there under the curve $y=x$.
plotting graphics
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
$endgroup$
add a comment |
6
$begingroup$
I need a formula to calculate how many points are there under the curve $y=x$.
plotting graphics
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
$endgroup$
$begingroup$
Some related: (97299), (157149), (181620)...I think there's more, but couldn't find the right search terms.
$endgroup$
– Michael E2
Aug 1 at 20:46
add a comment |
6
6
6
$begingroup$
I need a formula to calculate how many points are there under the curve $y=x$.
plotting graphics
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
$endgroup$
I need a formula to calculate how many points are there under the curve $y=x$.
plotting graphics
plotting graphics
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
edited Aug 2 at 7:43
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
100k10 gold badges316 silver badges473 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
asked Aug 1 at 13:02
MathMath
313 bronze badges
313 bronze badges
$begingroup$
Some related: (97299), (157149), (181620)...I think there's more, but couldn't find the right search terms.
$endgroup$
– Michael E2
Aug 1 at 20:46
add a comment |
$begingroup$
Some related: (97299), (157149), (181620)...I think there's more, but couldn't find the right search terms.
$endgroup$
– Michael E2
Aug 1 at 20:46
$begingroup$
Some related: (97299), (157149), (181620)...I think there's more, but couldn't find the right search terms.
$endgroup$
– Michael E2
Aug 1 at 20:46
$begingroup$
Some related: (97299), (157149), (181620)...I think there's more, but couldn't find the right search terms.
$endgroup$
– Michael E2
Aug 1 at 20:46
add a comment |
3 Answers
3
active
oldest
votes
9
$begingroup$
Try
p = RandomReal[0, 1, 50, 2]
pu=Select[p, #[[2]] < #[[1]] &]
Length[pu] (* number of points yi<xi*)
Show[Plot[x, x, 0, 1], ListPlot[p],ListPlot[pu, PlotStyle -> Red]]
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
$endgroup$
2
$begingroup$
One could also usepu = Select[p, Apply[Greater]]in this case. A slower, but generalizable approach would bermf = RegionMember[ImplicitRegion[y <= x, x, y]]; pu = Select[p, rmf]
$endgroup$
– J. M. is away♦
Aug 1 at 13:33
$begingroup$
How about if i want to do that in sine curve?
$endgroup$
– Math
Aug 2 at 13:07
$begingroup$
@Math Something likeSelect[p, #[[2]] <Sin[ #[[1]] ] &]
$endgroup$
– Ulrich Neumann
Aug 2 at 13:13
add a comment |
9
$begingroup$
p = RandomReal[0, 1, 50, 2]
Count[p, x_, y_ /; y < x]
This is many times more efficient for long lists:
Total[UnitStep[Subtract @@ Transpose[p]]]
Edit:
A few timings under version 12.0 on macos:
p = RandomReal[0, 1, 1000000, 2];
Length[Select[p, #[[2]] < #[[1]] &]] // RepeatedTiming
Count[p, x_, y_ /; y < x] // RepeatedTiming
Total[UnitStep[Subtract @@ Transpose[p]]] // RepeatedTiming
1.31, 499894
0.757, 499894
0.0089, 499894
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
$endgroup$
$begingroup$
Very interesting alternatives. By the way the last one is the slowest (MMA v.11.0.1)
$endgroup$
– Ulrich Neumann
Aug 1 at 14:19
$begingroup$
@UlrichNeumann Really? That is indeed surprising... Unfornately, I cannot test it under v. 11.0.1; I have removed it recently.
$endgroup$
– Henrik Schumacher
Aug 1 at 16:26
$begingroup$
Similar timings under v10.4 and 11.3.
$endgroup$
– corey979
Aug 1 at 16:33
2
$begingroup$
@UlrichNeumann That is surprising. What if you try the compiled versionCompile[p,_Real,2,Total[UnitStep[p.1,-1]]]?
$endgroup$
– Silvia
Aug 2 at 3:20
$begingroup$
@Silvia Great idea to useDot! And I am also surprised thatCompilemakes a difference here.
$endgroup$
– Henrik Schumacher
Aug 2 at 3:21
|
show 1 more comment
8
$begingroup$
You can use RegionMember for this:
reg = RegionMember[ImplicitRegion[y<x, x,y]];
Tally @ reg[p]
False, 23, True, 27
This will not be as fast as Henrik's answer.
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
$begingroup$
Try
p = RandomReal[0, 1, 50, 2]
pu=Select[p, #[[2]] < #[[1]] &]
Length[pu] (* number of points yi<xi*)
Show[Plot[x, x, 0, 1], ListPlot[p],ListPlot[pu, PlotStyle -> Red]]
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
$endgroup$
2
$begingroup$
One could also usepu = Select[p, Apply[Greater]]in this case. A slower, but generalizable approach would bermf = RegionMember[ImplicitRegion[y <= x, x, y]]; pu = Select[p, rmf]
$endgroup$
– J. M. is away♦
Aug 1 at 13:33
$begingroup$
How about if i want to do that in sine curve?
$endgroup$
– Math
Aug 2 at 13:07
$begingroup$
@Math Something likeSelect[p, #[[2]] <Sin[ #[[1]] ] &]
$endgroup$
– Ulrich Neumann
Aug 2 at 13:13
add a comment |
9
$begingroup$
Try
p = RandomReal[0, 1, 50, 2]
pu=Select[p, #[[2]] < #[[1]] &]
Length[pu] (* number of points yi<xi*)
Show[Plot[x, x, 0, 1], ListPlot[p],ListPlot[pu, PlotStyle -> Red]]
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
$endgroup$
2
$begingroup$
One could also usepu = Select[p, Apply[Greater]]in this case. A slower, but generalizable approach would bermf = RegionMember[ImplicitRegion[y <= x, x, y]]; pu = Select[p, rmf]
$endgroup$
– J. M. is away♦
Aug 1 at 13:33
$begingroup$
How about if i want to do that in sine curve?
$endgroup$
– Math
Aug 2 at 13:07
$begingroup$
@Math Something likeSelect[p, #[[2]] <Sin[ #[[1]] ] &]
$endgroup$
– Ulrich Neumann
Aug 2 at 13:13
add a comment |
9
9
9
$begingroup$
Try
p = RandomReal[0, 1, 50, 2]
pu=Select[p, #[[2]] < #[[1]] &]
Length[pu] (* number of points yi<xi*)
Show[Plot[x, x, 0, 1], ListPlot[p],ListPlot[pu, PlotStyle -> Red]]
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
$endgroup$
Try
p = RandomReal[0, 1, 50, 2]
pu=Select[p, #[[2]] < #[[1]] &]
Length[pu] (* number of points yi<xi*)
Show[Plot[x, x, 0, 1], ListPlot[p],ListPlot[pu, PlotStyle -> Red]]
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
answered Aug 1 at 13:17
Ulrich NeumannUlrich Neumann
13.5k7 silver badges22 bronze badges
13.5k7 silver badges22 bronze badges
2
$begingroup$
One could also usepu = Select[p, Apply[Greater]]in this case. A slower, but generalizable approach would bermf = RegionMember[ImplicitRegion[y <= x, x, y]]; pu = Select[p, rmf]
$endgroup$
– J. M. is away♦
Aug 1 at 13:33
$begingroup$
How about if i want to do that in sine curve?
$endgroup$
– Math
Aug 2 at 13:07
$begingroup$
@Math Something likeSelect[p, #[[2]] <Sin[ #[[1]] ] &]
$endgroup$
– Ulrich Neumann
Aug 2 at 13:13
add a comment |
2
$begingroup$
One could also usepu = Select[p, Apply[Greater]]in this case. A slower, but generalizable approach would bermf = RegionMember[ImplicitRegion[y <= x, x, y]]; pu = Select[p, rmf]
$endgroup$
– J. M. is away♦
Aug 1 at 13:33
$begingroup$
How about if i want to do that in sine curve?
$endgroup$
– Math
Aug 2 at 13:07
$begingroup$
@Math Something likeSelect[p, #[[2]] <Sin[ #[[1]] ] &]
$endgroup$
– Ulrich Neumann
Aug 2 at 13:13
2
2
$begingroup$
One could also use
pu = Select[p, Apply[Greater]] in this case. A slower, but generalizable approach would be rmf = RegionMember[ImplicitRegion[y <= x, x, y]]; pu = Select[p, rmf]$endgroup$
– J. M. is away♦
Aug 1 at 13:33
$begingroup$
One could also use
pu = Select[p, Apply[Greater]] in this case. A slower, but generalizable approach would be rmf = RegionMember[ImplicitRegion[y <= x, x, y]]; pu = Select[p, rmf]$endgroup$
– J. M. is away♦
Aug 1 at 13:33
$begingroup$
How about if i want to do that in sine curve?
$endgroup$
– Math
Aug 2 at 13:07
$begingroup$
How about if i want to do that in sine curve?
$endgroup$
– Math
Aug 2 at 13:07
$begingroup$
@Math Something like
Select[p, #[[2]] <Sin[ #[[1]] ] &] $endgroup$
– Ulrich Neumann
Aug 2 at 13:13
$begingroup$
@Math Something like
Select[p, #[[2]] <Sin[ #[[1]] ] &] $endgroup$
– Ulrich Neumann
Aug 2 at 13:13
add a comment |
9
$begingroup$
p = RandomReal[0, 1, 50, 2]
Count[p, x_, y_ /; y < x]
This is many times more efficient for long lists:
Total[UnitStep[Subtract @@ Transpose[p]]]
Edit:
A few timings under version 12.0 on macos:
p = RandomReal[0, 1, 1000000, 2];
Length[Select[p, #[[2]] < #[[1]] &]] // RepeatedTiming
Count[p, x_, y_ /; y < x] // RepeatedTiming
Total[UnitStep[Subtract @@ Transpose[p]]] // RepeatedTiming
1.31, 499894
0.757, 499894
0.0089, 499894
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
$endgroup$
$begingroup$
Very interesting alternatives. By the way the last one is the slowest (MMA v.11.0.1)
$endgroup$
– Ulrich Neumann
Aug 1 at 14:19
$begingroup$
@UlrichNeumann Really? That is indeed surprising... Unfornately, I cannot test it under v. 11.0.1; I have removed it recently.
$endgroup$
– Henrik Schumacher
Aug 1 at 16:26
$begingroup$
Similar timings under v10.4 and 11.3.
$endgroup$
– corey979
Aug 1 at 16:33
2
$begingroup$
@UlrichNeumann That is surprising. What if you try the compiled versionCompile[p,_Real,2,Total[UnitStep[p.1,-1]]]?
$endgroup$
– Silvia
Aug 2 at 3:20
$begingroup$
@Silvia Great idea to useDot! And I am also surprised thatCompilemakes a difference here.
$endgroup$
– Henrik Schumacher
Aug 2 at 3:21
|
show 1 more comment
9
$begingroup$
p = RandomReal[0, 1, 50, 2]
Count[p, x_, y_ /; y < x]
This is many times more efficient for long lists:
Total[UnitStep[Subtract @@ Transpose[p]]]
Edit:
A few timings under version 12.0 on macos:
p = RandomReal[0, 1, 1000000, 2];
Length[Select[p, #[[2]] < #[[1]] &]] // RepeatedTiming
Count[p, x_, y_ /; y < x] // RepeatedTiming
Total[UnitStep[Subtract @@ Transpose[p]]] // RepeatedTiming
1.31, 499894
0.757, 499894
0.0089, 499894
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
$endgroup$
$begingroup$
Very interesting alternatives. By the way the last one is the slowest (MMA v.11.0.1)
$endgroup$
– Ulrich Neumann
Aug 1 at 14:19
$begingroup$
@UlrichNeumann Really? That is indeed surprising... Unfornately, I cannot test it under v. 11.0.1; I have removed it recently.
$endgroup$
– Henrik Schumacher
Aug 1 at 16:26
$begingroup$
Similar timings under v10.4 and 11.3.
$endgroup$
– corey979
Aug 1 at 16:33
2
$begingroup$
@UlrichNeumann That is surprising. What if you try the compiled versionCompile[p,_Real,2,Total[UnitStep[p.1,-1]]]?
$endgroup$
– Silvia
Aug 2 at 3:20
$begingroup$
@Silvia Great idea to useDot! And I am also surprised thatCompilemakes a difference here.
$endgroup$
– Henrik Schumacher
Aug 2 at 3:21
|
show 1 more comment
9
9
9
$begingroup$
p = RandomReal[0, 1, 50, 2]
Count[p, x_, y_ /; y < x]
This is many times more efficient for long lists:
Total[UnitStep[Subtract @@ Transpose[p]]]
Edit:
A few timings under version 12.0 on macos:
p = RandomReal[0, 1, 1000000, 2];
Length[Select[p, #[[2]] < #[[1]] &]] // RepeatedTiming
Count[p, x_, y_ /; y < x] // RepeatedTiming
Total[UnitStep[Subtract @@ Transpose[p]]] // RepeatedTiming
1.31, 499894
0.757, 499894
0.0089, 499894
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
$endgroup$
p = RandomReal[0, 1, 50, 2]
Count[p, x_, y_ /; y < x]
This is many times more efficient for long lists:
Total[UnitStep[Subtract @@ Transpose[p]]]
Edit:
A few timings under version 12.0 on macos:
p = RandomReal[0, 1, 1000000, 2];
Length[Select[p, #[[2]] < #[[1]] &]] // RepeatedTiming
Count[p, x_, y_ /; y < x] // RepeatedTiming
Total[UnitStep[Subtract @@ Transpose[p]]] // RepeatedTiming
1.31, 499894
0.757, 499894
0.0089, 499894
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
edited Aug 1 at 16:24
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
answered Aug 1 at 14:07
Henrik SchumacherHenrik Schumacher
67.1k5 gold badges96 silver badges185 bronze badges
67.1k5 gold badges96 silver badges185 bronze badges
$begingroup$
Very interesting alternatives. By the way the last one is the slowest (MMA v.11.0.1)
$endgroup$
– Ulrich Neumann
Aug 1 at 14:19
$begingroup$
@UlrichNeumann Really? That is indeed surprising... Unfornately, I cannot test it under v. 11.0.1; I have removed it recently.
$endgroup$
– Henrik Schumacher
Aug 1 at 16:26
$begingroup$
Similar timings under v10.4 and 11.3.
$endgroup$
– corey979
Aug 1 at 16:33
2
$begingroup$
@UlrichNeumann That is surprising. What if you try the compiled versionCompile[p,_Real,2,Total[UnitStep[p.1,-1]]]?
$endgroup$
– Silvia
Aug 2 at 3:20
$begingroup$
@Silvia Great idea to useDot! And I am also surprised thatCompilemakes a difference here.
$endgroup$
– Henrik Schumacher
Aug 2 at 3:21
|
show 1 more comment
$begingroup$
Very interesting alternatives. By the way the last one is the slowest (MMA v.11.0.1)
$endgroup$
– Ulrich Neumann
Aug 1 at 14:19
$begingroup$
@UlrichNeumann Really? That is indeed surprising... Unfornately, I cannot test it under v. 11.0.1; I have removed it recently.
$endgroup$
– Henrik Schumacher
Aug 1 at 16:26
$begingroup$
Similar timings under v10.4 and 11.3.
$endgroup$
– corey979
Aug 1 at 16:33
2
$begingroup$
@UlrichNeumann That is surprising. What if you try the compiled versionCompile[p,_Real,2,Total[UnitStep[p.1,-1]]]?
$endgroup$
– Silvia
Aug 2 at 3:20
$begingroup$
@Silvia Great idea to useDot! And I am also surprised thatCompilemakes a difference here.
$endgroup$
– Henrik Schumacher
Aug 2 at 3:21
$begingroup$
Very interesting alternatives. By the way the last one is the slowest (MMA v.11.0.1)
$endgroup$
– Ulrich Neumann
Aug 1 at 14:19
$begingroup$
Very interesting alternatives. By the way the last one is the slowest (MMA v.11.0.1)
$endgroup$
– Ulrich Neumann
Aug 1 at 14:19
$begingroup$
@UlrichNeumann Really? That is indeed surprising... Unfornately, I cannot test it under v. 11.0.1; I have removed it recently.
$endgroup$
– Henrik Schumacher
Aug 1 at 16:26
$begingroup$
@UlrichNeumann Really? That is indeed surprising... Unfornately, I cannot test it under v. 11.0.1; I have removed it recently.
$endgroup$
– Henrik Schumacher
Aug 1 at 16:26
$begingroup$
Similar timings under v10.4 and 11.3.
$endgroup$
– corey979
Aug 1 at 16:33
$begingroup$
Similar timings under v10.4 and 11.3.
$endgroup$
– corey979
Aug 1 at 16:33
2
2
$begingroup$
@UlrichNeumann That is surprising. What if you try the compiled version
Compile[p,_Real,2,Total[UnitStep[p.1,-1]]]?$endgroup$
– Silvia
Aug 2 at 3:20
$begingroup$
@UlrichNeumann That is surprising. What if you try the compiled version
Compile[p,_Real,2,Total[UnitStep[p.1,-1]]]?$endgroup$
– Silvia
Aug 2 at 3:20
$begingroup$
@Silvia Great idea to use
Dot! And I am also surprised that Compile makes a difference here.$endgroup$
– Henrik Schumacher
Aug 2 at 3:21
$begingroup$
@Silvia Great idea to use
Dot! And I am also surprised that Compile makes a difference here.$endgroup$
– Henrik Schumacher
Aug 2 at 3:21
|
show 1 more comment
8
$begingroup$
You can use RegionMember for this:
reg = RegionMember[ImplicitRegion[y<x, x,y]];
Tally @ reg[p]
False, 23, True, 27
This will not be as fast as Henrik's answer.
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
$endgroup$
add a comment |
8
$begingroup$
You can use RegionMember for this:
reg = RegionMember[ImplicitRegion[y<x, x,y]];
Tally @ reg[p]
False, 23, True, 27
This will not be as fast as Henrik's answer.
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
$endgroup$
add a comment |
8
8
8
$begingroup$
You can use RegionMember for this:
reg = RegionMember[ImplicitRegion[y<x, x,y]];
Tally @ reg[p]
False, 23, True, 27
This will not be as fast as Henrik's answer.
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
$endgroup$
You can use RegionMember for this:
reg = RegionMember[ImplicitRegion[y<x, x,y]];
Tally @ reg[p]
False, 23, True, 27
This will not be as fast as Henrik's answer.
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
answered Aug 1 at 15:51
Carl WollCarl Woll
89k3 gold badges117 silver badges228 bronze badges
89k3 gold badges117 silver badges228 bronze badges
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$begingroup$
Some related: (97299), (157149), (181620)...I think there's more, but couldn't find the right search terms.
$endgroup$
– Michael E2
Aug 1 at 20:46