Skip to main content

list manipulation - Identifying and isolating sections of overlap in a set of integer intervals



I have a very large array like the following:


{{1,10},{8,9},{9,20},{34,40},{41,42},{43,50},{47,53},...};

Now, imagine that every element in the array corresponds to an integer interval. I'd like to delete elements in the array that fall within a larger interval, and whenever there is an interval-interval overlap, I'd like to shrink the intersecting intervals to eliminate the overlap and then add the former section where the overlap occurred as a new element to a separate array (here I'm calling this overlapArray).


Perhaps this is best explained by example. Looking at the smaller case:


{{1,10},{8,9},{9,20},{34,40}...}

We notice that there is an overlap between the first three elements of the array, and that the element {8,9} falls entirely in this overlap section. So we this array becomes the following after our procedure:


{{1,7},{11,20},{34,40}...}


overlapArray = {{8,10}};

Now for the full example, this:


{{1,10},{8,9},{9,20},{34,40},{41,42},{43,50},{47,53},...};

Becomes:


{{1,7},{11,20},{34,40},{41,42},{43,46},{51,53}

overlapArray = {{8,10},{47,50}};


Is there an elegant way to do this with list operations in Mathematica v9?



Answer



I like this one :P


list = {{1, 10}, {8, 9}, {9, 20}, {34, 40}, {41, 42}, {43, 50}, {47, 53}}

Composition[
Split[#, #2 - #1 == 1 &][[ All , {1, -1}]] &,
Sort,
DeleteCases[#, {_, 1}][[ All , 1]] &,
Tally,

Flatten,
Range @@@ # &
][list]


 {{8, 10}, {47, 50}}

Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....