I'd like to relabel the radial tick markers in the following polar log plot:
f = figure ;
t = 0:0.01: pi/2 ;
polar(t, 10 * log10(cos(t))/(50) + 1)
from 1, 0.8, 0.6, 0.4, 0.2 to 0, -10, -20, -30, -40 (i.e. radial dB ticks).
Trying some of the methods from Fixing the Radial Axis on MATLAB Polar Plots, I was able to relabel the markers provided my tick markers were positive and increasing.
I tried the following based on How to remove Rho labels from Matlab polar plot?
rho_labels = {'1' '0.8' '0.6' '0.4' '0.2'};
rho_labels2 = {'0' '-10' '-20' '-30' '-40'};
for r=1:length(rho_labels)
ff = findall(f, 'string', rho_labels{r}) ;
ff = rho_labels2{r} ;
end
but it also didn't work (seems to do nothing, so I suspect I'm operating on a copy of the find results not handled).
How can these tick markers be adjusted? Also, if I wanted a different number of concentric circles than 5, how can that be done (for example, 4 subdivisions with -40 dB at the "origin".)
Here is a way to rename the radial tick labels. Be warned that if there is a match between any radial and angular tick mark labels, both will be replaced and the angular labels will be wrong. But the angular tick labels are fixed as 0, 30, 60, ..., 330, so as long as the radial tick labels don't include these values, you should be fine.
What the code does is find all the text in the plot using findall, trim the blank spaces at the front of each string, then, for each entry in rho_labels, set the string entry corresponding to the tick label with that identifier to the corresponding entry in rho_labels2.
f = figure ;
t = 0:0.01: pi/2 ;
polar(t, 10 * log10(cos(t))/(50) + 1)
rho_labels = {'1' '0.8' '0.6' '0.4' '0.2'};
rho_labels2 = {'0' '-10' '-20' '-30' '-40'};
ff = findall(f,'type','text');
t=strtrim(get(ff,'String'));
for r=1:length(rho_labels)
set(ff(strcmp(t,rho_labels{r})),'String',rho_labels2{r})
end
To modify the number of rings, I think think of anything better than modifying polar.m and creating your own custom polarMOD.m function. If you do this you can also do the radial labels as well. In fact, there might be such a function on the MathWorks File Exchange.
Related
I'm representing a surface using "surf" function, with a colorbar. I would like to keep the default ticks of the colorbar, but add a custom tick on this colorbar, at a specific value (that I could make red to distinguish it from other ticks for example). Any idea on how to add a custom tick like that with keeping existing ticks on the colorbar ?
Thanks
As Luis mentioned in the comments, you can add an additional tick mark like so
h = colorbar;
newTick = 0.75;
h.Ticks = sort([h.Ticks newTick]);
If you want to add a line to the bar, the easiest thing (I think) is to use an annotation which is positioned relative to the figure (the same as the colorbar), so we can overlay it
pos = h.Position;
r = (newTick - min(h.Ticks))/(max(h.Ticks)-min(h.Ticks));
annotation( 'line', pos(1)+[0, pos(3)], [1, 1]*(pos(2)+pos(4)*r), ...
'color', [1,0,0], 'linewidth', 2 );
I'm setting the x position of the annotation to match the left and right sides of the colorbar, and the y position to match the bottom plus the relative % of the height according to the tick value.
Result:
Similarly, you could use annotatation exclusively to just get a red label, it's a little more convoluted to get everything lined up correctly, you have to make sure the text box is wide enough to be on a single line and vertically aligned to the middle to get the position right:
h = colorbar;
newTick = 0.75;
pos = h.Position;
r = (newTick - min(h.Ticks))/(max(h.Ticks)-min(h.Ticks));
h = 0.2;
annotation( 'textbox', [pos(1)+pos(3)/2, (pos(2)+pos(4)*r)-(h/2), pos(3)*2, h], ...
'color', [1,0,0], 'string', ['- ' num2str(newTick)], 'linestyle', 'none', ...
'VerticalAlignment', 'middle' );
How do I draw a 0.5 degree x 0.5 degree grid over the country map in a MATLAB figure?
The code below gives me a gridded figure but not with 0.5x0.5 degree spacing.
borders('Iran Islamic Republic of')
grid on
ax.GridLineStyle = '-';
Can anyone tell me how to add 0.5x0.5 grid along x and y-axis to this figure?
The borders function is taken from the MATLAB File Exchange
You can use xticks() and yticks() functions (matlab tutorial). Your code should be something like:
borders('Iran Islamic Republic of')
grid on
ax.GridLineStyle = '-';
% Modify the X and Y ticks positions
xticks([44:.5:65]);
yticks([25:.5:40]);
This creates ticks every 0.5 degrees (from degree 44 until 65 in x, and from 25 to 40 in y). If the tick labels are overlaping, you can delete some. For example for the x-axis:
%Delete some labels, otherwise overcrowded
xlabels = xticklabels();
for i=2:2:length(xticks())
xlabels(i)={''};
end
xticklabels(xlabels)
I would like to add labels to some points plotted using the command scatter. For the sake of simplicity, let's say I have only one point:
x = 10;
pointSize = 100;
fontSize = 20;
P = scatter(x, 0, pointSize, [0,0,0], 'filled');
text(x, 0, 'pointLabel',...
'HorizontalAlignment', 'center',...
'VerticalAlignment', 'bottom',...
'FontSize', fontSize);
The problem with the previous commands is that the text pointLabel overlaps with the point P depending on the values assigned to the properties pointsize and fontSize.
I have read the documentation of the text command, but the examples only show how to put a label horizontally aligned with a specific point in the diagram. If the alignment needs to be horizontal it is easy, but I could not find a general way to compute the y coordinate of the label pointLabel from the values of the other dimensions.
Clearly I can reach a good alignment by testing various combinations of values, but I am looking for a general solution.
Is there anyone who can help me?
This assumes you are using >=R2014b, though it can also be accomplished in older versions using set and get commands.
When a text object is created, its default units are data coordinates, but those can be changed. In your case, I'd go with points.
x = 10;
pointSize = 100;
fontSize = 20;
P = scatter(x, 0, pointSize, [0,0,0], 'filled');
t = text(x, 0, 'pointLabel',...
'HorizontalAlignment', 'center',...
'VerticalAlignment', 'bottom',...
'FontSize', fontSize);
% It's always a good idea to switch back to the default units, so remember them.
originalUnits = t.Units;
t.Units = 'points';
% Shift the text up by the sqrt(pi)/2 times the radius of the point
t.Position(2) = t.Position(2) + sqrt(pointSize)/2;
t.Units = originalUnits;
Check out Text Properties for more info. If you want to get really sophisticated, you can use the read-only property Extent and your known marker size and position to calculate when a label is overlapping one of your points. Since the default unit is in data space, no conversions are necessary.
If you're working with an older version of MATLAB, all of these options and properties are still available, you just have to work a little harder to use them. For instance, you can't direction set the position as above, but you would instead use get to assign it to a temporary variable, change it, and then use set to update. More lines of code, but ultimately the same effect.
I have data that records the x and y positions of an animal in a 2D assay over time stored in a matlab matrix. I can plot these co-ordinates over time, and extract the velocity information and plot this using cline.
The problem I am having at the moment is calculating the heading angle. It should be a trivial trigonometry question, but I am drawing a blank on the best way to start.
The data is stored in a matrix xy representing x and y co-ordinates:
796.995391705069 151.755760368664
794.490825688073 150.036697247706
788.098591549296 145.854460093897
786.617021276596 144.327659574468
781.125000000000 140.093750000000
779.297872340426 138.072340425532
775.294642857143 133.879464285714
What I would like to be able to do is know the angle of the line drawn from (796.995, 151.755) to (794.490, 150.036), and so on. My research suggests atan2 will be the appropriate function, but I am unsure how to call it correctly to give useful information.
difx = xy(1,1) - xy(2,1);
dify = xy(1,2) - xy(2,2);
angle = atan2(dify,difx);
angle = angle*180/pi % convert to degrees
The result is 34.4646. Is this correct?
If it is correct, how do I get the value to be in the range 0-360?
You can use the diff function to get all the differences at once:
dxy = diff(xy); % will contain [xy(2,1)-xy(1,1) xy(2,2)-xy(1,2); ...
Then you compute the angle using the atan2 function:
a = atan2(dxy(:,2), dxy(:,1));
You convert to degrees with
aDeg = 180 * a / pi;
And finally take the angle modulo 360 to get it between 0 and 360:
aDeg = mod(aDeg, 360);
So - you pretty much got it right, yes. Except that you have calculated the heading from point 2 to point 1, and I suspect you want to start at 1 and move towards 2. That would give you a negative number - or modulo 360, an angle of about 325 degrees.
Also, using the diff function gets you the entire array of headings all at once which is a slight improvement over your code.
[rc mi]=
EDIT the problem of "phase wrapping" - when the heading goes from 359 to 0 - is quite a common problem. If you are interested in knowing when a large change happens, you can try the following trick (using aDeg from above - angle in degrees).
dDeg1 = diff(aDeg); % the change in angle
dDeg2 = diff(mod(aDeg + 90, 360)); % we moved the phase wrap point by 180 degrees
dDeg12 = [dDeg1(:) dDeg2(:)]';
[rc mi]= min(abs(dDeg12));
indx = sub2ind(size(dDeg12), mi, 1:size(dDeg12, 2));
result = dDeg12(ii);
What I did there: one of the variables (dDeg or dDeg2) does not see the phase wrap, and the min function finds out which one (it will have a smaller absolute difference). The sub2ind looks up that number (it is either positive or negative - but it's the smaller one of the two), and that is the value that ends up in result.
You can verify the angle by plotting a little line that starts at the first point and end in the direction of the heading. If the angle is correct, it will point in the direction of the next point in xy. Everything depends on where yo define 0 degrees at (straight up, say) from and whether positive degrees is rotation counterclockwise (I do) or clockwise. In MATLAB you can get the numbers between 0 and 360 but using modulo---or you can just add 180 to your results but this will change the definition of where the 0 degree mark is.
I made the following script that is a bit complex but shows how to calculate the heading/angle for all points in vector format and then displays them.
xy =[ 796.995391705069 151.755760368664
794.490825688073 150.036697247706
788.098591549296 145.854460093897
786.617021276596 144.327659574468
781.125000000000 140.093750000000
779.297872340426 138.072340425532
775.294642857143 133.879464285714];
% t = linspace(0,3/2*pi, 14)';
% xy = [sin(t), cos(t)];
% calculate the angle:
myDiff = diff(xy);
myAngle = mod(atan2(myDiff(:,1), myDiff(:,2))*180/pi, 360);
% Plot the original Data:
figure(1);
clf;
subplot(1,3,1);
plot(xy(:,1), xy(:,2), '-bx', 'markersize', 12);
hold all
axis equal;grid on;
title('Original Data');
% Plot the calculated angle:
subplot(1,3,2);
plot(myAngle);
axis tight; grid on;
title('Heading');
% Now plot the result with little lines pointing int he heading:
subplot(1,3,3);
plot(xy(:,1), xy(:,2), '-bx', 'markersize', 12);
hold all
% Just for visualization:
vectorLength = max(.8, norm(xy(1,:)- xy(2,:)));
for ind = 1:length(xy)-1
startPoint = xy(ind,:)';
endPoint = startPoint + vectorLength*[sind(myAngle(ind)); cosd(myAngle(ind))];
myLine = [startPoint, endPoint];
plot(myLine(1,:), myLine(2, :), ':r ', 'linewidth', 2)
end
axis equal;grid on;
title('Original Data with Heading Drawn On');
For example, if you use my test data
t = linspace(0,3/2*pi, 14)';
xy = [sin(t), cos(t)];
You get the following:
and if you do yours you get
Note how the little red line starts at the original data point and moves in the direction of the next point---just like the original blue line connecting the points.
Also note that the use of diff in the code to difference all the points properly at once. This is faster and avoids any problems with the direction--looks like in your case it's swapped.
I have a figure of size 14 x 14 square drawn inside an axis of 20 x 20, in matlab.
I am trying to draw circles of radius 0.7 inside the square and need to arrange them evenly. I need to draw 233 circles. Please let me know how can I do it?
Currently I can draw them randomly but couldn't get 233 circle. Please see my below code.
Your reply is appreciated.
% Urban, sub urban, Rural areas
x_area =[3, 12, 6];
y_area = [6, 8, 16];
r_area = [1, 7, 2];
f = figure;
hAxs = axes('Parent',f);
hold on, box on, axis equal
xlabel('x')
ylabel('y','Rotation',0)
title('Compute the area of circles a vectorized way for several cicles')
axis([0 20 0 20])
rectangle('Position',[5,1,14,14])
rectangle('Position',[3,1,2,2])
rectangle('Position',[1,3,4,4])
hold on, box on, axis equal
a = 233;
x_base_urban = randi([6 18], 1, a);
b = rand([10 8], 1);
y_base_urban = randi([2 14],1, a);
r_base_urban = 0.9;
size_x = size(x_base_urban);
size_x = size_x(2);
size_y = size(y_base_urban);
size_y = size_y(2);
colour = rand(size_x,3);
for t = 1: size_x
plot(x_base_urban(t)+ r_base_urban.*cos(0:2*pi/100:2*pi),...
y_base_urban(t)+ r_base_urban.*sin(0:2*pi/100:2*pi),'parent',hAxs)
plot(x_base_urban(t),y_base_urban(t),'+','parent',hAxs)
end
Thanks
Randomly plotting everything won't work. Actually, if your circles can't overlap, nothing will work. To show that, just compare the results of following calculations:
lSquare = 14;
rCircle = 0.7;
nCircles = 233;
areaCircles = nCircles * pi * rCircle^2
areaSquare = lSquare^2
You will see that areaCircles > areaSquare, so it is impossible to fit them all in. On the other hand, if areaSquare >= areaCircles does not guarantee you that a solution exists!
Try your setup with a smaller example to come up with a solution. E.g. take a square box and a bunch of spherical objects (balls, marbles, oranges, apples, ... if need be) and try to fit as much of those in your box. If that works, you might even want to draw their positions on a sheet of paper before trying to implement it.
If you do this correctly, you will get an idea of how to stack round objects in a square container. That is also exactly what you need to do in your exercise. Then try to make a model/algorithm of what you did manually and implement that in MATLAB. It won't be hard, but you will need some small calculations: Pythagoras and intersection of circles.
I also suggest you use a function to draw a circle as #Andrey shows, so something of the form function drawCircle(center, radius). That allows you to keep complexity down.
If your circles can overlap, then the solution is quite easy: look at a circle as an object with a center point and distribute these center points evenly over the square. Don't use rand to do this, but calculate their positions yourself.
If you can't find a solution, I might expand my answer in a few days.
Without diving too deep into your code, I think that you need to add a hold function, after the first plot
for t = 1: size_x
plot(x_base_urban(t)+ r_base_urban.*cos(0:2*pi/100:2*pi),...
y_base_urban(t)+ r_base_urban.*sin(0:2*pi/100:2*pi),'parent',hAxs)
plot(x_base_urban(t),y_base_urban(t),'+','parent',hAxs)
hold(hAxs,'on');
end
By the way, the best way to draw circle is by using rectangle command.
rectangle('Curvature',[1 1],'Position',[1 3 4 5])
So you can create a PlotCircle function (like #EgonGeerardyn suggests) like this:
function plotCircle(x,y,r)
rectangle('Position',[x-r y-r 2*r 2*r],'Curvature',[1 1]);
end