26
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"
,...
25
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],
...
22
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}, {...
22
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, ...
22
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, ...
20
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 ...
19
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 ...
19
Since nobody who supposedly thought this was a good question seems actually to have wanted to write an answer, here is one from me.
Let us define a bimodal distribution:
dist = MixtureDistribution[
{0.6, 0.4},
{NormalDistribution[-0.8, 1.3], NormalDistribution[2.7, 0.4]}
];
pdfplot = Plot[PDF[dist, x], {x, -5, 5}]
To simulate your data we draw some ...
19
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,...
18
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
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.
17
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;
...
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 ...
17
Here's a similar figure with made-up data:
dist1 = NormalDistribution[0.002, 1/100];
dist2 = NormalDistribution[-0.002, 1/10];
randomWalk[] :=
NestWhileList[# + 0.7 RandomVariate[dist1] +
0.3 RandomVariate[dist2] &, 0.5, 0 < # < 1 &];
SeedRandom[0]
{foo, res} =
Reap[Do[Sow[Length@#, Clip[Last@#, {0, 1}]] &@randomWalk[], {...
16
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 + ...
16
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]
16
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[...
16
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,
...
16
For the case where the height function is "Count", we can use the formula from the linked page in a custom ChartElementFunction with the sample size (Length[data]) passed as metadata:
ceF[d_: .2, nsd_: 3, color_: Automatic][cedf_: "Rectangle"] :=
Module[{e = nsd /2 Sqrt[#[[2, 2]] (1 - #[[2, 2]]/ #3[[1]])]},
{ChartElementData[cedf][##], Thick,
...
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
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 practice, ...
15
Here is something using a custom ChartElementFunction
Module[{c = 0},
half[{{xmin_, xmax_}, {ymin_, ymax_}}, data_, metadata_] := (c++;
Map[Reverse[({0, Mean[{xmin, xmax}]} + # {1, (-1)^c})] &,
First@Cases[
First@Cases[InputForm[SmoothHistogram[data, Filling -> Axis]],
gc_GraphicsComplex :> Normal[gc], ∞],
...
14
As of version 9 there is an undocumented extension to the kernel functions which allows you to bound the domain. You do this by specifying the kernel function as {"Bounded", c, ker} where c is the left boundary (0 in your case) and ker is the usual kernel function. You can also allow for both the left and right to be bounded via {"Bounded", {c1, c2}, ker}. ...
14
Here is extended version of PlatoManiac's solution which allows changing of the direction of the hatching and also tuning the distance between hatches:
g[step_?NumberQ][{{xmin_, xmax_}, {ymin_, ymax_}}, ___] :=
Module[{yval, lines, xstart, xend},
yval = Range[ymin, ymax, Abs[step]];
If[step > 0, {xstart, xend} = {xmin, xmax}, {xstart, xend} = {...
14
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 ...
13
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 ...
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
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 ...
13
Why not just assemble the chart from rectangles?
data = {{-6.65, 55}, {-6.45, 15}, {-6.27, 10}, {-6, 5}, {-5.85, 3},
{-6.46, 6}, {-6.25, 3}, {-6.17, 2}};
Graphics[{EdgeForm[{Thick, Black}], RGBColor[0.3, 0.4, 0.4],
Rectangle[{#1 - 0.05, 0}, {#1 + 0.05, #2}] & @@@ data},
Frame -> True, AspectRatio -> 0.7, FrameLabel -> {"Binding Energy", "...
13
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}]
Only top voted, non community-wiki answers of a minimum length are eligible
