Skip to main content

probability or statistics - Histograms with pre-counted data


I've got some large data sets which have been counted but not binned already - essentially, a list of pairs of values (not bins) and counts.* (Or, equivalently, it's been binned into too-small bins.) I want to plot histograms for them. I remember the deprecated version of Histogram from a separate package had a FrequencyData option, but that seems to have disappeared. Is there any built-in way to accomplish this now? (I'd like to still have all the fancy features of Histogram, i.e. I don't want to just rebin the data myself and plot it directly. Notably I'd like to still be able to use Histogram's automatic bin specification, or something like it.)


*That is, my data is represented as {{1, 6}, {2, 4}, {3, 2}, ...} instead of {1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, ...}. (And before anyone suggests just expanding the data to the latter form to pass to Histogram: there are over 100K values, and the total count is over 100M.)


Edit: okay, let me be really explicit. The perfect thing would be to be able to take the first representation of the data ({{1,6}, ...}), and get exactly what Histogram would have produced had I given it the second version ({1,1,1,1,...}), without having to actually expand it to that form. (This includes being able to specify various options and extra arguments to Histogram.) I do not want a bar chart with 100K bars. I do not want to have to decide how many bins to make every time I do this, because I may do it many times with many varieties of data.



Answer



Histogram doesn't have any built-in support for weighted data, although it's an interesting idea, and most of the binning algorithms should be amenable to working with it.




That being said, here's a WeightedHistogram function, with some feedback from Andy Ross. It accepts



  • weighted values (in the same format as RandomChoice and EmpiricalDistribution)


  • binning specifications

  • Histogram options.


It doesn't support the height functions, since they'd have to be manually implemented. (This isn't hard, just a bit tedious since there are several of them.)


The implementation creates a representative sample of the data to compute the bins from. This is combined with the list of actual values to make sure we cover the extremes, which might have low weights and otherwise not show up in the sample.


Options[WeightedHistogram] = 
Append[Options[Histogram], "SampleSize" -> 1000];

WeightedHistogram[weights_ -> values_, o : OptionsPattern[]] :=
WeightedHistogram[weights -> values, Automatic, o]


WeightedHistogram[weights_ -> values_, bins_, o : OptionsPattern[]] :=
Block[{sample, newbins, valuelists, partitions},
sample = Join[
RandomChoice[weights -> values, OptionValue["SampleSize"]],
values];
newbins = First[HistogramList[sample, bins]];
partitions = Partition[newbins, 2, 1];
valuelists =
Total[Pick[weights, Thread[# <= values < #2]]] & @@@ partitions;

Histogram[values, {newbins}, valuelists &,
FilterRules[Flatten[{o}], Options[Histogram]]]
]

Now let's try it out with some data that is easily weighted:


data = RandomVariate[PoissonDistribution[30], 10^5];
{values, weights} = Transpose[Tally[data]];

Here's the Histogram applied to the original data:


Histogram[data]


enter image description here


Here's the weighted data, in vanilla and rainbow flavors:


Row[{
WeightedHistogram[weights -> values],
WeightedHistogram[weights -> values, {1}, ChartStyle -> "Rainbow"]
}]

enter image description here


Comments

Popular posts from this blog

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 - 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 - 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.