Making a histogram with bin spacing and exporting the data

Question

I want to create a histogram which has spacing between its bins (as shown below):

to create this I used the following method:

     data1 = RandomVariate[NormalDistribution[0, 1], 500];

     f2[d_][{{xmin_, xmax_}, {ymin_, ymax_}}, ___] := Rectangle[{xmin + d (xmax - xmin)/2, ymin}, {xmax - d (xmax - xmin)/2, ymax}]
     h1=Histogram[data1, Automatic, ChartElementFunction -> f2[0.1]]

However, I want to export this data as is to a list so I can apply some functions and try to see how much worse my fits get as I try to increase the space in between the bins. To export the data to a list I use:

     histData=HistogramList[h1]

which understandably gives me errors. Is there a way I can get the data as it is shown in the drawn histogram (image above) to a list so I can try to apply fits on it with a function etc. This is just a test example, my real function is for 10^7 points and has a custom probability distribution function

Answer 1

I think I now know what you want (but I certainly don't understand fitting objective).

The following will get you the coordinates of the "thinned" bins.

(* Generate some data *)
SeedRandom[12345];
data1 = RandomVariate[NormalDistribution[0, 1], 500];

(* Construct a histogram *)
b0 = HistogramList[data1]
(* {{-(7/2), -3, -(5/2), -2, -(3/2), -1, -(1/2), 0, 1/2, 1, 3/2, 2, 5/2, 3, 7/2},
    {1, 1, 14, 23, 54, 88, 97, 82, 75, 35, 21, 6, 2, 1}} *)



(* Get the binwidth, bin heights, and midpoints *)
binWidth = b0[[1, 2]] - b0[[1, 1]]
(* 1/2 *)
binHeights = b0[[2]]
(* {1, 1, 14, 23, 54, 88, 97, 82, 75, 35, 21, 6, 2, 1} *)
midPoints = (b0[[1, 2 ;; Length[b0[[1]]]]] + b0[[1, 1 ;; Length[b0[[1]]] - 1]])/2
(* {-(13/4), -(11/4), -(9/4), -(7/4), -(5/4), -(3/4), -(1/4), 1/4, 
    3/4, 5/4, 7/4, 9/4, 11/4, 13/4} *)

(* Function to get the bin boundaries of the thinned bins *)
thinnedBins[d_] := Table[{midPoints[[i]] - binWidth (1 - d)/2, 
   midPoints[[i]] + binWidth (1 - d)/2}, {i, Length[midPoints]}]

(* Create a dataset with the thinned bin boundaries, plot the correspoinding histogram *)
b = thinnedBins[0.3]
(* {{-3.425, -3.075}, {-2.925, -2.575}, {-2.425, -2.075}, {-1.925, -1.575},
    {-1.425, -1.075}, {-0.925, -0.575}, {-0.425, -0.075}, {0.075, 0.425},
    {0.575, 0.925}, {1.075, 1.425}, {1.575, 1.925}, {2.075, 2.425}, 
    {2.575, 2.925}, {3.075, 3.425}} *)
newHistogram = Table[{{b[[i, 1]], 0}, {b[[i, 1]], binHeights[[i]]}, {b[[i, 2]], 
    binHeights[[i]]}, {b[[i, 2]], 0}}, {i, Length[midPoints]}];
ListPlot[newHistogram, Joined -> True, PlotStyle -> Black]

Answer 2

You may use DiscretePlot with it's ExtentSize option.

With

SeedRandom[12345];
data1 = RandomVariate[NormalDistribution[0, 1], 500];
hList = HistogramList[data1];

Then

update to centre "bars"

Module[{i = 1},
 DiscretePlot[
  hList[[2, i++]],
  {bin, Differences@hList[[1]]/2 + Most@hList[[1]]},
  PlotStyle -> Directive[Thin, Blue, EdgeForm[{Thin, Blue}]],
  ExtentSize -> Scaled[.8]
  ]
 ]

Increasing the Scaled ExtentSize towards one will shrink the gap and decreasing it towards zero will widen the gap.

Note that with continuous data there should be no spaces between the bars as the data can appear within these spaces but your plot communicates that it cannot. If you have discrete data then you should see BarChart.

Hope this helps.