Changing colour of regression line in a scatter plot
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Danny
el 27 de Dic. de 2023
Editada: Dyuman Joshi
el 27 de Dic. de 2023
I am trying to either change the colour of the line in my scatter graph linear regression graph or the colour of the plots to make it easier to read
This is my current code to try and change the line colour to blue except both the line and points stay in red each time?
plot(x, y, 'rx', 'MarkerSize', 2, "MarkerFaceColor","b");
ylabel('Crash Risk');
xlabel('Ln Market Value');
title('How Ln Market Value affects Crash Risk')
6 comentarios
Star Strider
el 27 de Dic. de 2023
Missing function definitions.
%% Initialization
clear all; close all; clc
%% ======================= Part 1: Plotting =======================
fprintf('Plotting Data ...\n')
data = load('LnMarketVal.txt');
X = data(:, 1); y = data(:, 2);
m = length(y); % number of training examples
% Plot Data
% Note: You have to complete the code in plotData.m
plotData(X, y);
% fprintf('Program paused. Press enter to continue.\n');
% pause;
%% =================== Part 2: Gradient descent ===================
fprintf('Running Gradient Descent ...\n')
X = [ones(m, 1), data(:,1)]; % Add a column of ones to x
theta = zeros(2, 1); % initialize fitting parameters
% Some gradient descent settings
iterations = 1500;
alpha = 0.01;
% compute and display initial cost
computeCost(X, y, theta)
% run gradient descent
theta = gradientDescent(X, y, theta, alpha, iterations);
% print theta to screen
fprintf('Theta found by gradient descent: ');
fprintf('%f %f \n', theta(1), theta(2));
% Plot the linear fit
hold on; % keep previous plot visible
plot(X(:,2), X*theta, '-')
legend('Training data', 'Linear regression')
hold off % don't overlay any more plots on this figure
% Predict values for population sizes of 35,000 and 70,000
predict1 = [1, 3.5] *theta;
fprintf('Program paused. Press enter to continue.\n');
pause;
%% ============= Part 3: Visualizing J(theta_0, theta_1) =============
fprintf('Visualizing J(theta_0, theta_1) ...\n')
% Grid over which we will calculate J
theta0_vals = linspace(-10, 10, 100);
theta1_vals = linspace(-1, 4, 100);
% initialize J_vals to a matrix of 0's
J_vals = zeros(length(theta0_vals), length(theta1_vals));
% Fill out J_vals
for i = 1:length(theta0_vals)
for j = 1:length(theta1_vals)
t = [theta0_vals(i); theta1_vals(j)];
J_vals(i,j) = computeCost(X, y, t);
end
end
J_vals = J_vals';
% Surface plot
figure;
surf(theta0_vals, theta1_vals, J_vals)
xlabel('\theta_0'); ylabel('\theta_1');
% Contour plot
figure;
% Plot J_vals as 15 contours spaced logarithmically between 0.01 and 100
contour(theta0_vals, theta1_vals, J_vals, logspace(-2, 3, 20))
xlabel('\theta_0'); ylabel('\theta_1');
hold on;
plot(theta(1), theta(2), 'rx', 'MarkerSize', 10, 'LineWidth', 2);
.
Dyuman Joshi
el 27 de Dic. de 2023
Editada: Dyuman Joshi
el 27 de Dic. de 2023
@Danny, Please supply the functions used/called in the script as well.
Respuesta aceptada
Star Strider
el 27 de Dic. de 2023
x = (0:99);
y = 3*x + randn(size(x))*25 + 15;
figure
plot(x, y, 'rx', 'MarkerSize', 2, "MarkerFaceColor","b");
ylabel('Crash Risk');
xlabel('Ln Market Value');
title('How Ln Market Value affects Crash Risk')
lr = lsline;
lr.Color = 'b'; % Change Colour To Blue
B = [lr.XData; 1 1].' \ lr.YData.'; % Parameters
m = B(1) % Slope
b = B(2) % Inmtercept
The only argumnent lsline takes is the axis it refers to, so all modifications are made to it after it is plotted.
.
1 comentario
Image Analyst
el 27 de Dic. de 2023
How does lsline know what the data is? What if there are lots of things plotted? Or does it require only 1 set of x data and 1 set of y data?
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