How to customise Polar Plots

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Sam Hurrell
Sam Hurrell 2023-4-2
评论： Star Strider 2023-4-6
I have data typically graphed as 'value (B) against angle (A)' as a plot (for angles -90 to 90) that I wish to graph as a polar plot. What command(s) can I write that'll graph this data with polar axes for angle as: -90 (on the left) to 90 (on the right) via 0 (below in the middle)?
  2 个评论
Star Strider
Star Strider 2023-4-2
Ran in:
Some detail is missing.
How do you want the tick labels to appear on the left side?
What do you want the top value (currently 180°) to be?
th = linspace(0, 2*pi, 360);
r = sin(2*th).^2;
figure
polarplot(th,r)
Ax = gca;
Ax.ThetaZeroLocation = 'bottom';
TTL = Ax.ThetaTickLabel
TTL = 12×1 cell array
{'0°' } {'30°' } {'60°' } {'90°' } {'120°'} {'150°'} {'180°'} {'210°'} {'240°'} {'270°'} {'300°'} {'330°'}
.
Sam Hurrell
Sam Hurrell 2023-4-4
Where you have 270 to 0 to 90, I want -90 to 0 to 90. I also want the top half of the graph gone leaving a semicircular polar plot.

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采纳的回答

Star Strider
Star Strider 2023-4-4
Ran in:
Try something like this —
th = linspace(0, 2*pi, 360);
r = exp(-0.1*th).*sin(2*th).^2;
figure
polarplot(th,r)
title('Original')
figure
polarplot(th+pi/2,r) % It May Be Necessary To Shift The Plot Phase (+90° Or +π/2) For This To Work Correctly
Ax = gca;
Ax.ThetaZeroLocation = 'bottom';
Ax.ThetaLim = [270 270+180];
Ax.ThetaTickLabel = compose('%d°',-90:30:90);
title('Shifted')
I needed to make my original function radially asymmetric to test this to be certain it would work correctly. It may be necessary for you to experiment with shifting the angles as well.
.
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显示 1更早的评论
Star Strider
Star Strider 2023-4-5
Ran in:
‘One small thing though, how do I set custom gridlines (every 15^o rather than every 30^o)?’
Try this —
th = linspace(0, 2*pi, 360);
r = exp(-0.1*th).*sin(2*th).^2;
figure
polarplot(th,r)
title('Original')
figure
polarplot(th+pi/2,r) % It May Be Necessary To Shift The Plot Phase (+90° Or +π/2) For This To Work Correctly
Ax = gca;
Ax.ThetaZeroLocation = 'bottom';
Ax.ThetaLim = [270 270+180];
Ax.ThetaTick = Ax.ThetaLim(1) : 15 : Ax.ThetaLim(2);
Ax.ThetaTickLabel = compose('%d°',-90:15:90);
title('Shifted With Ticks Every 15°')
.
Star Strider
Star Strider 2023-4-6
If my Answer helped you solve your problem, please Accept it!
.

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更多回答（1 个）

the cyclist
the cyclist 2023-4-2
编辑：the cyclist 2023-4-2
Ran in:
% Create plot
figure
theta = linspace(0,2*pi,25);
rho = 2*theta;
polarplot(theta,rho);
% Define a variable with the tick labels
thetaTickLabels = {'0°';'30°';'60°';'90°';'120°';'150°';'180°';'-150°';'-120°';'-90°';'-60°';'-30°'};
% Set theta how you want
set(gca,"ThetaZeroLocation","bottom", ...
"ThetaDir","counterclockwise", ...
"ThetaTickLabels",thetaTickLabels)
See the documentation for polarplot for details.
Note that while the axes are labeled in degrees, the default input is in radians. (That same page shows how to convert, if you need to.)
  1 个评论
Sam Hurrell
Sam Hurrell 2023-4-4
编辑：Sam Hurrell 2023-4-4
That's really helpful thanks, but how do I edit it to remove the top half of the graph (-90, 180, 90) leaving a semi-circular plot?

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