I want to change the color of the markers in scatter plot which are below the function line
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I have the following code
clc
clear
close all
d = 4;
N = 1000;
X = linspace(0,d,N);
func = @(X) 2*X.*cos(X.^2).*exp(sin(X.^2)) + 14;
limit = func(X);
x = 4*rand(1,N);
y = 25*rand(1,N);
figure(1),hold on
h = scatter(x,y,10,'markerfacecolor','b');
plot(X,limit,'k-','linew',1.3)
for i = 1:N
if any(y(i) < limit)
set(h,'YData','MarkerFaceColor','r','Markeredgecolor','r')
end
end
but it keeps giving the following error
Error using matlab.graphics.chart.primitive.Scatter/set
Invalid parameter/value pair arguments.
Any help would be appreciated
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Star Strider
2022년 4월 11일
d = 4;
N = 1000;
X = linspace(0,d,N);
func = @(X) 2*X.*cos(X.^2).*exp(sin(X.^2)) + 14;
limit = func(X);
x = 4*rand(1,N);
y = 25*rand(1,N);
figure(1)
hold on
scatter(x,y,10,'markerfacecolor','b');
Lv = y < func(x); % Logical Vector Based On 'func' Value At Each 'x'
scatter(x(Lv),y(Lv),10,'MarkerFaceColor','r','Markeredgecolor','r')
plot(X,limit,'k-','linew',1.3)
.
댓글 수: 2
Star Strider
2022년 4월 11일
Thank you!
Sure!
The ‘Lv’ variable is a logical vector that calculates the value of ‘func’ at each ‘x’ value and compares that result with the corresponding ‘y’ value. If that ‘y’ value is less than ‘func(x)’, that value of ‘Lv’ is set to true. (It definitely helps that ‘func’ is an anonymous function in your original code, making this straightforward.)
The scatter call just after that uses ‘Lv’ to refer to the individual ‘x’ and ‘y’ values, plotting only those that are true as defined by ‘Lv’.
추가 답변 (1개)
MJFcoNaN
2022년 4월 11일
You can try this:
clc
clear
close all
d = 4;
N = 1000;
X = linspace(0,d,N);
func = @(X) 2*X.*cos(X.^2).*exp(sin(X.^2)) + 14;
limit = func(X);
x = 4*rand(1,N);
y = 25*rand(1,N);
ind_r = y < limit;
c=zeros(N,3);
% red
c(ind_r, 1)=1;
% blue
c(~ind_r, 3)=1;
figure(1),hold on
h = scatter(x,y,10,c);
plot(X,limit,'k-','linew',1.3)
댓글 수: 2
MJFcoNaN
2022년 4월 11일
Why do you make both x and y random?
If your task allows one random value, for example x=X, it may give out a more "reasonable" result, which achieves the same algorithm as the other answer.
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