Skip to main content

plotting - How can I plot histogram with the same number of values in every bin?



For example I have 100 values sample. I'd like to build histogram in which every bin contains, for example, 10 values. How can i do that? Thanks.



Answer



You can use the values of the quantiles of your sample as bin delimiters for your histogram. You can think of $n$-quantiles as those threshold values that divide your data set into $n$ equal-sized subsets.


Let's generate some sample data and set your requirements, i.e. number of points per bin:


SeedRandom[10]
sample = RandomVariate[NormalDistribution[], 200];
datapointsperbin = 10;
numberofbins = IntegerPart[Length[sample]/datapointsperbin];

This is what a regular histogram with evenly spaced bins would look like for that sample:



Histogram[sample]

Even binned histogram


Now we use Quantile to calculate numberofbins quantiles for your distribution, then we use those values as bin delimiters for your histogram.


Histogram[
sample,
{Table[Quantile[sample, i/numberofbins], {i, 1, numberofbins - 1}]}
]

unevenly binned histogram



You can see from the vertical axis of the histogram that each bin contains 10 samples, as specified by the value of datapointperbin.




Having done this, however, I still wonder why you need such a histogram. Of course, if what you needed was to calculate the intervals that would accomplish such binning, given your sample, the magic is all in the Quantile function, so you can get those values directly as well:


Table[Quantile[sample, i/numberofbins], {i, 1, numberofbins - 1}]


{-1.8614, -1.42414, -1.21859, -0.971859, -0.905122, -0.707023, -0.470983, -0.274088, -0.163548, 0.0100698, 0.122639, 0.271601, 0.383704, 0.475579, 0.608299, 0.873699, 1.03975, 1.33463, 1.81741}



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.