# Tag Info

25

None of the solutions so far makes use of the fact that the data are percentages and hence add op to (nearly) 100. (* Add the rows of the data list *) Total[Rest /@ data, {2}] (* out *) {99.96, 99.98, 99.98, 99.99, 99.99, 99.99, 99.99, 99.97, 100., 99.99,100., 100.} Borrowing from Pinguin Dirk: BarChart[Rest /@ data ,ChartLayout -> "Stacked" ,...

22

One More way! The means of each data is the blue dot. bars are color coded according to the standard deviation within each sub list. ListLinePlot[Mean /@ data, Prolog -> MapThread[{Thickness[.04], ColorData["SandyTerrain"][#3], Line[{{#2, First@#1}, {#2, Last@#1}}], Opacity[0.7], White, Dashed, Thickness[0.003], Arrowheads[0.025], ...

20

One posibility is to use Tally with BarChart. For example: hist = Tally[{"A", "A", "B", "B", "B"}]; BarChart[hist[[All, 2]], ChartLabels -> hist[[All, 1]]]

20

I was sure Histogram can be modified to have the hatching style. Little late but what about this! g[{{xmin_, xmax_}, {ymin_, ymax_}}, ___] := Module[{yval, line}, yval = Range[ymin, ymax, 15]; line = Line /@ Transpose@{Most@({xmin, #} & /@ yval),Rest@({xmax, #} & /@ yval)}; {FaceForm[White],Polygon[{{xmin, ymin}, {xmax, ymin}, {xmax, ymax}, {...

19

I think you could achieve what you want by overlaying two Histogram objects, one with no Edges but colored bars, the second one with Edges, but transparent bars, as in this example: data = RandomVariate[NormalDistribution[10, 2], 500]; Show[{ Histogram[data, ChartStyle -> Directive[Opacity[0], EdgeForm[AbsoluteThickness[4]]]], Histogram[data, ...

18

This site has exactly what you want here, already in Mathematica code. One example here:

18

Here's my contribution for the first type of chart: bins = Tally @ Ceiling[data, 5]; labelfn = Text[HoldForm[# °], 8 {Sin[# °], Cos[# °]}] &; ptfn = Rotate[Point @ Thread @ {10 + Range@#2, 0}, Pi/2 - # °, {0, 0}] &; linefn = Rotate[Line[{{9.5, 0}, {10, 0}}], Pi/2 - # °, {0, 0}] &; Graphics[{ Circle[{0, 0}, 10], labelfn /@ {0, 90, 180, ...

18

Out of curiosity I tried this: DistributionChart[Rest /@ data, ChartLabels -> {data[[All, 1]]}, ChartElementFunction -> "HistogramDensity", ChartStyle -> {LightRed, LightGreen, LightBlue}, BarOrigin -> Left] As for 'interpretation', here's my attempt. This type of chart tries to show the distribution of the values in each 'row'. The ...

17

Without looking at performance, but only on understanding: First, you create a function f which takes a value and a count and which reproduces the value exactly count times. In the simplest case f[val_, count_] := ConstantArray[val, count] and you can call f[3,4] to get {3,3,3,3}. Now, you combine your input arrays so that you can call f directly for each ...

17

This may be what you are looking for. img = Import["http://i.imgur.com/Wd9lPRa.jpg"] Now use DominantColors. Graphics[{#, Disk[]}] & /@ DominantColors[img, 4]

17

The underlying problem is that Histogram creates a set of Rectangles which represent the bars. You can inspect this by looking at the underlying form of the created graphics h = Histogram[RandomVariate[NormalDistribution[10, 2], 500], Automatic, "PDF", PerformanceGoal -> "Speed"]; InputForm[h] This reveals that each bar in the histogram is ...

16

Under mma 8 you can use the undocumented {"Raw", n} bin specification to get exactly the number of bins you would like. Otherwise the bin widths and boundaries are chosen to be "nice" numbers. Here is an example: data = RandomVariate[NormalDistribution[0, 1], 200]; Histogram[data, {"Raw", 5}] (I saw this first in a comment by Brett Champion to the ...

16

Since nobody has used this function yet, I will place it here. Your data seems to be organised almost perfectly for ArrayPlot. First I removed the first column from the rest of the values and added to the axes ticks. The rest is just displayed via ArrayPlot, with a particular color scale. {xs, values} = {First[#], Transpose@Rest[#]} &@Thread@data; ...

15

You may be able to use the new WeightedData in version 9 with HistogramDistribution to create a weighted histogram. I've reduced the number of points for speed but it should hopefully scale to your actual problem. raw = RandomReal[1, {10000000, 2}]; binned = Table[{.01*Ceiling[raw[[i, 1]]/.01], .01* Ceiling[raw[[i, 2]]/.01], RandomInteger[1000]}, {i, ...

15

The function f takes as arguments the raw, unbinned data, the number of sectors, and a boolean parameter to indicate whether polar gridlines are to be drawn. data = {8, 9, 13, 13, 14, 18, 22, 27, 30, 34, 38, 38, 40, 44, 45, 47, 48, 48, 48, 48, 50, 53, 56, 57, 58, 58, 61, 63, 64, 64, 64, 65, 65, 68, 70, 73, 78, 78, 78, 83, 83, 88, 88, 88, 90, 92, 92, 93, 95,...

14

For those without v9, here's another attempt based on FindClusters, but using a different colour space. The idea is to reduce the effect of overall brightness on the "distance" between colours, so that the clustering gives more weight to differences of hue and is less likely to pick out different shades of gray. newspace[{r_, g_, b_}] := {r - g, b - g, (r + ...

14

I would do something like: t = RandomVariate[NormalDistribution[0, 1], 10000]; a = Histogram[t, 30, ChartStyle -> White]; b = Histogram[t, 30, ChartElements -> Graphics[{Black, Line[{{0, 0}, {1, 1}}]}]]; Show[a, b]

14

Here is a bit clumsy (had very little time) approach виа combination of new functionality Entity and regions. (* get the states *) divisions = EntityValue[Entity["AdministrativeDivision", {_, "UnitedStates"}], "Entities"]; (* get polygons of borders *) dat = EntityValue[ divisions, {"Population", "Polygon"}] /. {GeoPosition -> Identity, ...

14

data1 = RandomVariate[NormalDistribution[0, 1], 1000]; data2 = RandomVariate[NormalDistribution[0.3, 1], 1000]; One way is PairedHistogram PairedHistogram[data1, data2] A different layout can be achieved by hacking the height specification of Histogram. Show[ Histogram[data1, Automatic], Histogram[data2, Automatic, -#2 &] ] Update In ...

13

Is this what you mean? I put different colors in the background, to make clear what I added: Show[Histogram3D[dataHistogramSet, FaceGrids -> {Bottom, Front, Left}, ChartStyle -> "GrayTones", ViewPoint -> {2.78, 1.3, 1.43}, PlotLabel -> "Histogram of Dataset 1"], Graphics3D[ Translate[{EdgeForm[], {Red, Polygon[{{1995, 0, 0}, {...

13

Add the argument "Probability" to the Histogram command. To be precise, if list is your list of data, then Histogram[list,Automatic,"Probability"] should do the trick. The Automatic argument specifies the bin size.

13

Another potentially useful command of this kind is the CommonestFilter which looks locally about each pixel and chooses the most common value to display. Setting the neighborhood large causes large regions of constant color. For example img = Import["http://i.imgur.com/Wd9lPRa.jpg"] CommonestFilter[img, n] where img is the image from the OPs question ...

13

Here's my take at replicating your circular plot. To get the ticks right isn't as easy as one might think, there is no option to put the ticks on the inside of the circle. The ticks will have to be produced manually... CircularDotHistogram[data_, n_, clockwise_: True] := Module[{hist, pts, deg2rad, angdata}, hist = HistogramList[data, n][[2]]; deg2rad[...

13

As discussed in the comment, it seems you want: BarChart[Rest /@ data, ChartLabels -> {data[[All, 1]], None}, BarSpacing -> {0, 2}] see other options in BarChart to format as you desire (as I do not know what it is for, it's hard to suggest other things), bonne chance! or a version with labels for the bars, placed above the bars (see ...

12

If I understand correctly, this question is about complete automation for uniform look. Hence the specifics of solution below. Import data: data = Import["http://pastebin.com/download.php?i=fZKMqxK9"]; Define filter that automatically separates data by gender: filter[gender_] := Rest[Transpose[Select[data, #[[1]] == gender &]]] For complete ...

12

The FrameTicks specification changed in v6. Previously, the form {right, bottom, left, top} was used, but in v6, it was changed to {{left, right}, {bottom, top}} To maintain compatibility between the versions, the older form is still allowed for the older plots. But, quite a few plots were added since that point (e.g. Histogram was added in v7), and ...

12

You can simply get the SmoothKernelDistribution and build the plot as you'd like: data = Table[Sin[x]^3 + 1, {x, 0, 6 Pi, 0.1}]; dist = SmoothKernelDistribution[data]; LogLogPlot[PDF[dist, x], {x, 0.01, 2}]

11

So this generates the heatmap: << Calendar` year = 1990; yearLen = DaysBetween[{year, 1, 1}, {year, 12, 31}] + 1; data = RandomReal[1, yearLen]; days = Map[DayOfWeek[{year, 1, #}] &, Range[3, 9]]; day1 = Position[days, DayOfWeek[{year, 1, 1}]][[1, 1]]; dayn = Position[days, DayOfWeek[{year, 12, 1}]][[1, 1]]; Paddata = Join[ConstantArray[100, day1 -...

11

You can use a custom ChartElementDataFunction as follows: cedF[{{xmin_, xmax_}, {ymin_, ymax_}}, ___] :={If[xmax <= 8, RGBColor[1, 0, 0], Sequence[]], Dynamic@EdgeForm[Directive[Thickness[.015], Lighter@CurrentValue["Color"]]], Rectangle[{xmin, ymin}, {xmax, ymax}, RoundingRadius -> 5]}; Histogram[RandomVariate[NormalDistribution[10, 2], ...

11

Try use DominantColors on particular selections instead of the whole. After import select regions you want to analyze and copy that (optionally) multiple selections as a list of images. (New in 9?) img = Import["http://oaadonline.oxfordlearnersdictionaries.com" <> "/media/oaad8/fullsize/f/fru/fruit/fruit_fruit.jpg"] Paste it and apply dominants ...

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