Skip to main content

image processing - How do I transform a rasterized graphic's coordinates back to its original ListPlot data coordinates?


From this question I successfully made an elliptical fit for my data. However, when I try and collect the datapoints within the ellipse via this question the coordinates of my ellipse correspond to an ellipse of a rasterized image of my listplot data and not the original. How can I transform or scale the ellipse that fits the rasterized image to the original listplot.


Ok, with all that said heres an example.


data = RandomReal[NormalDistribution[], {100000, 2}]
p = ListPlot[data, ImageSize -> 4000];

f = FillingTransform@ColorNegate@Binarize@p // DeleteSmallComponents
{c, s, t} = 1 /. ComponentMeasurements[f, {"Centroid", "SemiAxes", "Orientation"}]
Show[Rasterize[p], Graphics[{Red, Rotate[Circle[c, s],t]}]]

where I get a nice image:


example data


however, the coordinates c, s, t of the ellipse are pixle coordiantes corresponding to the rasterized image rather than data coordinates.


So when I need the ellipses parameter to do any calculations I get bunk results.


The image processing approach would be the best as I am filtering out the most dense cluster of data.


Thanks so much.




Answer



(*Generate Data and fit*)
data1 = RandomReal[NormalDistribution[10, 1], {10^4}];(*test data*)
data2 = RandomReal[NormalDistribution[20, 5], {10^4}];(*test data*)
data = Transpose@{data1, data2};
r = RotationTransform[Pi/8];
data3 = r /@ data;

(*we need to specify PlotRange due to a kown bug in AbsoluteOptions[] *)
prange = {Min@#, Max@#} & /@ {First@#, Last@#} &@Transpose@data3;


p = ListPlot[data3, Axes -> None, PlotRange -> prange];
f = FillingTransform@ColorNegate@Binarize@p // DeleteSmallComponents;
{co, so, to} = 1 /. ComponentMeasurements[f, {"Centroid", "SemiAxes", "Orientation"}];

(*Transform Image to Graphic coordinates*)
c = Rescale[co[[#]], {1, ImageDimensions[f][[#]]}, prange[[#]]] & /@ {1, 2};
s = Rescale[so[[#]], {0, Norm@ImageDimensions@f}, {0, Norm@(Differences/@ prange)}] & /@ {1, 2};
t = -ArcTan@Rescale[Tan@to, {0, 1/Divide @@ ImageDimensions[f]},
{0, 1/First@(Divide @@ Differences /@ prange)}];

(* Replot graphic*)

{s1, s2} = s;
{cx, cy} = c;
f0 = Sqrt[s1 s1 - s2 s2];
f1 = {cx + f0 Cos[t], cy - f0 Sin[t]};
f2 = {cx - f0 Cos[t], cy + f0 Sin[t]};
r = 2 Sqrt[f0 f0 + s2 s2];

sd = Select[data3, EuclideanDistance[#, f1] + EuclideanDistance[#, f2] < r &];

Show[p, Graphics[{Red, PointSize[Large], Point@sd}], Axes -> True]

Mathematica graphics


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...