How to set a marker at a desired location on a plot

12 vues (au cours des 30 derniers jours)
absomnia48
absomnia48 le 25 Déc 2016
Modifié(e) : Star Strider le 26 Déc 2016
Hi,I'm trying to set a marker at integer values of x.I've used index method(x(3)) but couldn't do it.Because I used linspace function.How do I mark my graph when x is equal to an integer number?Thanks.
x=linspace(-4,4);
y=x.*exp(-x.^2);
plot(x,y)
end

Réponse acceptée

Star Strider
Star Strider le 25 Déc 2016
The easiest way is to define a second vector with integer values, since by inspection only the endpoints of ‘x’ are integers:
x = linspace(-4,4);
xi = min(x) : max(x); % Integer ‘x’ Vector
y = x.*exp(-x.^2);
yi = xi.*exp(-xi.^2);
plot(x,y)
hold on
plot(xi, yi, 'rp', 'MarkerFaceColor','g')
hold off
grid

Plus de réponses (2)

John BG
John BG le 25 Déc 2016
use a vector index, let it be k, instead of the x reference of the plot.
1.
plot and add a marker
x=linspace(-4,4);
y=x.*exp(-x.^2);
plot(x,y)
hold on
p = plot(x(1),y(1),'o','MarkerFaceColor','red');
2.
now you can simply place the marker on one of the plotted points of y by varying k between 1 and numel(y), for instance:
k=20;
p.XData = x(k);
p.YData = y(k);
or
hold on;
p = plot(x(k),y(k),'o','MarkerFaceColor','red');
hold off;
the figure window also has a marker called Data Cursor.
if you find these lines useful would you please mark my answer as Accepted Answer?
To any other reader, if you find this answer of any help, please click on the thumbs-up vote link,
thanks in advance for time and attention
John BG

absomnia48
absomnia48 le 25 Déc 2016
Hey,another question here.I'm supposed to find the minimum and maximum value on the graph and draw a line until the axis.I have to do it with 'fminbnd' function.I'm writing this;
[-0.6869,-0.4285]=fminbnd(@xx.*exp(-x.^2),-0.6,0.6)
%or this one
fminbnd(x.*exp(-x.^2),-4,4)
but didn't work.Matlab tells me this 'Error: An array for multiple LHS assignment cannot contain expressions.'Thank you for the help
  2 commentaires
Star Strider
Star Strider le 25 Déc 2016
Modifié(e) : Star Strider le 26 Déc 2016
You are almost there with the ‘%or this one’ version. You only need to add a function handle, here that would be ‘@(x)’:
fx = @(x) x.*exp(-x.^2);
Ymin = fx(fminbnd(fx, -4, 4))
Ymax = fx(fminbnd(@(x) -fx(x), -4, 4))
Ymin =
-428.8819e-003
Ymax =
428.8819e-003
Note that to get the maximum, take the minimum of the negative of the funciton.
John BG
John BG le 25 Déc 2016
Modifié(e) : John BG le 25 Déc 2016
Start Strider, Y maxmin are
Ymax=0.42
Ymin=-0.42
while
Xmin=-0.707
Xmax=0.707
because
x=[-4:.00001:4]; y=x.*exp(-x.^2);
min(y)
=
-0.428881942471466
max(y)
ans =
0.428881942471466
find(y==min(y))
=
329290
x(find(y==min(y)))
=
-0.707110000000000

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