The function 'polarhistogram' can produce overlapping bins

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Alexandre Aksenov
Alexandre Aksenov 2022-1-11
编辑： Alexandre Aksenov 2022-9-18
Good day
I am experiencing an unexpected behavior of the function 'polarhistogram'. Here is my code:
thetaEx = [0.1 1.1 5.4 3.4 2.3 4.5 3.2 3.4 5.6 2.3 2.1 3.5 0.6 6.1];
%see https://www.mathworks.com/help/matlab/ref/polarhistogram.html
figure('Name','Plausible histogram');
polarhist1 = polarhistogram(thetaEx);
Edges1 = polarhist1.BinEdges; %-> (0:2*pi/3:2*pi) : 1x4 double
figure('Name','histogram with intersecting bins');
polarhist2 = polarhistogram(thetaEx+2*pi/3);
Edges2 = polarhist2.BinEdges; %-> (pi/2:3*pi/4:(2*pi+3*pi/4)) : 1x4 double
Overlap2 = Edges2(end)-Edges2(1)-2*pi; %-> pi/4
%These numbers confirm an overlap.
figure('Name','histogram made plausible');
polarhist3 = polarhistogram(mod(thetaEx+2*pi/3,2*pi));
Edges3 = polarhist3.BinEdges; %equals Edges1
Here are the figures.
The bins of the second histogram are clearly overlapping, which is confirmed by the property 'BinEdges'. This is not what one would expect of a histogram.
The third histogram shows that this unexpected behavior varies depending on shifts by multiples of 2*pi, although this transformation does not modify an angular distribution.
Could one indicate whether I am missing a necessary condition for using the function, or another important detail?
Thanks in advance,
Alexandre
The version of Matlab used: R2021b
Edit
The second histogram changes when the same code is run in Matlab R2022b. The new edges contain no overlap:
Edges2 = polarhist2.BinEdges; %R2022b: 2*pi/3:2*pi/3:8*pi/3
The new figure is a correct histogram, looking as follows:
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Adam Danz
Adam Danz 2022-1-17
I can't imagine a circumstance where wrapped radians to [0,2pi] would cause a problem since the range of radians in a polar plot are the same interval, [0,2pi].
You could contact tech support to report this issue. If they provide a better workaround or explain why this is not a bug, please share their feedback.

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Shivam Singh
Shivam Singh 2022-2-4
Hello,
It is my understanding that you are facing the issue of overlapping bins in some cases when you are using the “polarhistogram” function.
This issue is brought to the notice of our developers and they may be investigating further.
You may refer to the below example which is a workaround by using "mod" function:
thetaEx = [0.1 1.1 5.4 3.4 2.3 4.5 3.2 3.4 5.6 2.3 2.1 3.5 0.6 6.1];
%see https://www.mathworks.com/help/matlab/ref/polarhistogram.html
figure('Name','Plausible histogram');
polarhist1 = polarhistogram(thetaEx);
Edges1 = polarhist1.BinEdges; %-> (0:2*pi/3:2*pi) : 1x4 double
figure('Name','histogram with intersecting bins');
polarhist2 = polarhistogram(mod(thetaEx+2*pi/3,2*pi));% THIS LINE IS ONLY MODIFIED FROM YOUR CODE
Edges2 = polarhist2.BinEdges; %-> (pi/2:3*pi/4:(2*pi+3*pi/4)) : 1x4 double
Overlap2 = Edges2(end)-Edges2(1)-2*pi; %-> pi/4
%These numbers confirm an overlap.
figure('Name','histogram made plausible');
polarhist3 = polarhistogram(mod(thetaEx+2*pi/3,2*pi));
Edges3 = polarhist3.BinEdges; %equals Edges1
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Alexandre Aksenov
Alexandre Aksenov 2022-9-18
编辑：Alexandre Aksenov 2022-9-18
Thanks to Mathworks for your job! I am glad to be able to use the official instruction again, without workaround.

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