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added 572 characters in body

edited Apr 22, 2015 at 21:15

george2079

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Edit, based on the added info in the question:

 data1 = RandomVariate[BinormalDistribution[.75], 10]; distribution1 = SmoothKernelDistribution[data1]; data2 = RandomVariate[BinormalDistribution[.75], 10]; distribution2 = SmoothKernelDistribution[data2]; ContourPlot[ {PDF[distribution1, {x, y}] == 10^(-2), PDF[distribution2, {x, y}] == 10^(-2)}, {x, -3, 3}, {y, -3, 3}, ContourStyle -> {Red,Blue}, ContourShading -> None]

enter image description here

something like this?

 r = RandomVariate[BinormalDistribution[.5], 100]; hg = Histogram3D[r]

enter image description here

 hl = HistogramList[r]; Show[hg, Graphics3D[{Thick, Red, Map[Append[#, 5] & , First@Cases[Normal@First@ Cases[ListContourPlot[ Flatten[Table[{ Mean@hl[[1, 1, j ;; j + 1]], Mean@hl[[1, 2, i ;; i + 1]], hl[[2, j, i]]}, {i, Length@hl[[1, 2]] - 1}, {j, Length@hl[[1, 1]] - 1}], 1], Contours -> {5}], _GraphicsComplex, Infinity], _Line, Infinity], {2}]}]]

enter image description here

By the way, It would be cleaner to work with SmoothDensityHistogram, but I can't figure how to coax it to give a single contour line at a specified level..

something like this?

 r = RandomVariate[BinormalDistribution[.5], 100]; hg = Histogram3D[r]

enter image description here

 hl = HistogramList[r]; Show[hg, Graphics3D[{Thick, Red, Map[Append[#, 5] & , First@Cases[Normal@First@ Cases[ListContourPlot[ Flatten[Table[{ Mean@hl[[1, 1, j ;; j + 1]], Mean@hl[[1, 2, i ;; i + 1]], hl[[2, j, i]]}, {i, Length@hl[[1, 2]] - 1}, {j, Length@hl[[1, 1]] - 1}], 1], Contours -> {5}], _GraphicsComplex, Infinity], _Line, Infinity], {2}]}]]

enter image description here

By the way, It would be cleaner to work with SmoothDensityHistogram, but I can't figure how to coax it to give a single contour line at a specified level..

Edit, based on the added info in the question:

 data1 = RandomVariate[BinormalDistribution[.75], 10]; distribution1 = SmoothKernelDistribution[data1]; data2 = RandomVariate[BinormalDistribution[.75], 10]; distribution2 = SmoothKernelDistribution[data2]; ContourPlot[ {PDF[distribution1, {x, y}] == 10^(-2), PDF[distribution2, {x, y}] == 10^(-2)}, {x, -3, 3}, {y, -3, 3}, ContourStyle -> {Red,Blue}, ContourShading -> None]

enter image description here

something like this?

 r = RandomVariate[BinormalDistribution[.5], 100]; hg = Histogram3D[r]

enter image description here

 hl = HistogramList[r]; Show[hg, Graphics3D[{Thick, Red, Map[Append[#, 5] & , First@Cases[Normal@First@ Cases[ListContourPlot[ Flatten[Table[{ Mean@hl[[1, 1, j ;; j + 1]], Mean@hl[[1, 2, i ;; i + 1]], hl[[2, j, i]]}, {i, Length@hl[[1, 2]] - 1}, {j, Length@hl[[1, 1]] - 1}], 1], Contours -> {5}], _GraphicsComplex, Infinity], _Line, Infinity], {2}]}]]

enter image description here

By the way, It would be cleaner to work with SmoothDensityHistogram, but I can't figure how to coax it to give a single contour line at a specified level..

added 33 characters in body

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edited Apr 22, 2015 at 20:47

george2079

39.3k
1
45
115

something like this?

 r = RandomVariate[BinormalDistribution[.5], 100]; hg = Histogram3D[r]

enter image description here

 hl = HistogramList[r]; Show[hg, Graphics3D[{Thick, Red, Map[Append[#, 5] & , First@Cases[Normal@First@ Cases[ListContourPlot[ Flatten[Table[{ Mean@hl[[1, 1, j ;; j + 1]], Mean@hl[[1, 2, i ;; i + 1]],  hl[[2, j, i]]}, {i,  {i, Length@hl[[1, 2]] - 1}, {j, Length@hl[[1, 1]] - 1}], 1],   Contours -> {5}], _GraphicsComplex, Infinity], _Line, Infinity], {2}]}]]

enter image description here

By the way, It would be cleaner to work with SmoothDensityHistogram, but I can't figure how to coax it to give a single contour line at a specified level..

something like this?

 r = RandomVariate[BinormalDistribution[.5], 100]; hg = Histogram3D[r]

enter image description here

 hl = HistogramList[r]; Show[hg, Graphics3D[{Thick, Red, Map[Append[#, 5] & , First@Cases[Normal@First@ Cases[ListContourPlot[ Flatten[Table[{Mean@hl[[1, 1, j ;; j + 1]], Mean@hl[[1, 2, i ;; i + 1]], hl[[2, j, i]]}, {i,  Length@hl[[1, 2]] - 1}, {j, Length@hl[[1, 1]] - 1}], 1], Contours -> {5}], _GraphicsComplex, Infinity], _Line, Infinity], {2}]}]]

enter image description here

By the way, It would be cleaner to work with SmoothDensityHistogram, but I can't figure how to coax it to give a single contour line at a specified level..

something like this?

 r = RandomVariate[BinormalDistribution[.5], 100]; hg = Histogram3D[r]

enter image description here

 hl = HistogramList[r]; Show[hg, Graphics3D[{Thick, Red, Map[Append[#, 5] & , First@Cases[Normal@First@ Cases[ListContourPlot[ Flatten[Table[{ Mean@hl[[1, 1, j ;; j + 1]], Mean@hl[[1, 2, i ;; i + 1]],  hl[[2, j, i]]}, {i, Length@hl[[1, 2]] - 1}, {j, Length@hl[[1, 1]] - 1}], 1],   Contours -> {5}], _GraphicsComplex, Infinity], _Line, Infinity], {2}]}]]

enter image description here

By the way, It would be cleaner to work with SmoothDensityHistogram, but I can't figure how to coax it to give a single contour line at a specified level..

Source Link

answered Apr 22, 2015 at 20:42

george2079

39.3k
1
45
115

something like this?

 r = RandomVariate[BinormalDistribution[.5], 100]; hg = Histogram3D[r]

enter image description here

 hl = HistogramList[r]; Show[hg, Graphics3D[{Thick, Red, Map[Append[#, 5] & , First@Cases[Normal@First@ Cases[ListContourPlot[ Flatten[Table[{Mean@hl[[1, 1, j ;; j + 1]], Mean@hl[[1, 2, i ;; i + 1]], hl[[2, j, i]]}, {i, Length@hl[[1, 2]] - 1}, {j, Length@hl[[1, 1]] - 1}], 1], Contours -> {5}], _GraphicsComplex, Infinity], _Line, Infinity], {2}]}]]

enter image description here

By the way, It would be cleaner to work with SmoothDensityHistogram, but I can't figure how to coax it to give a single contour line at a specified level..