I have the following 2n*π-periodic function F(x) = sin(x/n) and I need to graph the dx/dt = γ - F(x) on the segment from 0 to 2pi. So it should look like this. I tried to do it matlab this way:
gamma = 1.01;
n=3;
[t,phi] = ode45(#(t,x)gamma-sin(x/n), [0,400], pi);
[t1,phi1] = ode45(#(t,x)gamma-sin(x/n), [112,400], 0);
[t2,phi2] = ode45(#(t,x)gamma-sin(x/n), [231,250], 0);
figure();
plot(t, phi, 'k', t1, phi1, 'k', t2, phi2, 'k');
ylim([0 2*pi]);
yticks([0 pi 2*pi]);
yticklabels(["0" "\pi" "2\pi"]);
grid on; grid minor;
title('\itsin(x/n)')
but I only got something like this. So there the lines are not transferred, but "begin anew". does anyone here know how to do that?
I get a plot similar to your first sketch, and based on your code in the comments (in future, put such additions into the question itself, use formatting to mark it as addition, and cite it then in the comment) with the changes
use pi as initial point as seen in the drawing,
use the options of the ODE solver to restrict the step size, directly and by imposing error tolerances
your original time span covers about 3 periods, reduce this to [0, 200] to get the same features as the drawing.
gamma = 1.01; n=3;
opts = odeset('AbsTol',1e-6,'RelTol',1e-9,'MaxStep',0.1);
[t, phi] = ode45(#(t,x)gamma-sin(x/n), [0,200], pi, opts);
phi = mod(phi, 2*pi);
plot(t, phi, 'k');
ylim([0 2*pi]); yticks([0 pi 2*pi]); yticklabels(["0" "\pi" "2\pi"]);
grid on; grid minor;
title('\itsin(x/n)')
To get more elaborate, use events to get points on the numerical solution where it exactly crosses the 2*pi periods, then use that to segment the solution plot (styling left out)
function [ res, term, dir ] = event(t,y)
y = mod(y+pi,2*pi)-pi;
res = [ y ];
dir = [1]; % only crossing upwards
term = [0]; % do not terminate
end%function
opts = odeset(opts,'Events',#(t,y)event(t,y));
sol = ode45(#(t,x)gamma-sin(x/n), [0,200], pi, opts);
tfs = [ sol.xe; sol.x(end) ]
N = length(tfs)
clf;
t0 = 0;
for i=1:N
tf = tfs(i);
t = linspace(t0+1e-2,tf-1e-2,150);
y = deval(sol,t); % octave: deval=#(res,t) interp1(res.x, res.y,t)
y = mod(y,2*pi);
plot(t, y);
hold on;
t0=tf;
end;
hold off;
Related
I'm trying to create a figure where the user can select cells to turn on or off. Then, the user can click a button 'Enter' to save the board as an array. I successfully found a way to create interactivity in my plot thanks to a very useful explanation I found here. I just made some changes to suit my needs.
However, I can't find a way to save the board. The button is working (or at least isn't not working), but the data isn't saved. And I don't know how to fix that. Any help would be appreciated.
Here is my code:
function CreatePattern
hFigure = figure;
hAxes = axes;
axis equal;
axis off;
hold on;
% create a button to calculate the difference between 2 points
h = uicontrol('Position',[215 5 150 30],'String','Enter','Callback', #SaveArray);
function SaveArray(ButtonH, eventdata)
global initial
initial = Board;
close(hFigure)
end
N = 1; % for line width
M = 20; % board size
squareEdgeSize = 1;
% create the board of patch objects
hPatchObjects = zeros(M,M);
for j = M:-1:1
for k = 1:M
hPatchObjects(M - j+ 1, k) = rectangle('Position', [k*squareEdgeSize,j*squareEdgeSize,squareEdgeSize,squareEdgeSize], 'FaceColor', [0 0 0],...
'EdgeColor', 'w', 'LineWidth', N, 'HitTest', 'on', 'ButtonDownFcn', {#OnPatchPressedCallback, M - j+ 1, k});
end
end
%global Board
Board = zeros(M,M);
playerColours = [1 1 1; 0 0 0];
xlim([squareEdgeSize M*squareEdgeSize]);
ylim([squareEdgeSize M*squareEdgeSize]);
function OnPatchPressedCallback(hObject, eventdata, rowIndex, colIndex)
% change FaceColor to player colour
value = Board(rowIndex,colIndex);
if value == 1
set(hObject, 'FaceColor', playerColours(2, :));
Board(rowIndex,colIndex) = 0; % update board
else
set(hObject, 'FaceColor', playerColours(1, :));
Board(rowIndex,colIndex) = 1; % update board
end
end
end
%imwrite(~pattern,'custom_pattern.jpeg')
I am trying to solve a 3 Degree Of Freedom vessel-crane system in MATLAB using ode45. In my ode function, the vessel motion is determined by simple spring and damper constants, but the device below the main hoist system, a cranemaster, has a displacement-dependent stiffness and thus depends on one of the Degrees Of Freedom. To determine this value, I use the interp1 function to determine the stiffness from a dataset:
vq = interp1(x,v,xq,method,extrapolation)
This all works fine, but only when I add the extrapolation option in the interp1 function, even though the xq values all fall inside the dataset x-values. If I do not specify extrapolation, the function returns NAN.
I have checked if the stiffness values in the dataset make sense by taking the median value as a constant value and then the ode45 solver works just fine and the displacement never falls outside of the dataset. I have also noticed that if I switch to ode23 as a solver, it also works. This all makes me believe it has something to do with the solving method and the stepsize the ode solver takes as its initial value.
I have provided (a simplified version of) my code and a picture of the system to get a feel for the problem. I know that switching to ode23 solves the problem but I would like to understand why and, if possible, stay with ode45 as it is a more general solver.
Any help would be appreciated!
Regards,
Diederik
Overview of the system I am solving
close all
tspan = [0 100];
x0 = [0.0; 0; 0.0; 0; 0.0; 0]; % initial disp. and vel. is 0. for all 3 DOF.
% x = [z; zdot; theta; thetadot; u; udot]
[t,x] = ode45(#Three_DOF,tspan,x0);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Plots
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure()
subplot(2,1,1)
plot(t,x(:,5),'color','b','LineWidth',1)
grid on
xlabel('Time (s)')
ylabel('Displacement (m)')
title('Displacement Vs Time')
subplot(2,1,2)
plot(t,x(:,6),'color','#D95319','LineWidth',1)
grid on
xlabel('Time (s)')
ylabel('Velocity (m/s)')
title('Velocity Vs Time')
function [dxdt] = Three_DOF(t,x)
% vessel parameters
m = 570.5e6;
m55 = 4.591e12;
mh = 200e3;
c33 = 3.83952e8;
c55 = 3.66116e12;
wnh = 0.360;
wnp = 0.596;
a33 = (c33-wnh^2*m)/wnh^2;
a55 = (c55-wnp^2*m55)/wnp^2;
zeta33 = 0.386;
zeta55 = 0.121;
b33 = zeta33*2*sqrt(c33*(m+a33));
b55 = zeta55*2*sqrt(c55*(m55+a55));
Fa = 153e6;
Ma = 3.013e9;
w1 = 0.21/0.36;
w2 = 0.18/0.28;
%%%% Cable parameters
k_c = 496e6;
b_c = 5.01e6;
%%% CRANEMASTER parameters
x_data = [-2.0; -1.87; -1.73; -1.60; -1.47; -1.33; -1.20; -1.07; -0.93; -0.8; -0.67; -0.53; -0.4; -0.27; -0.13; 0.0; 0.13; 0.27; 0.40; 0.53; 0.67; 0.80; 0.93; 1.07; 1.20; 1.33; 1.47; 1.60; 1.73; 1.87; 2.0];
k_data = [760; 792.1; 784.8; 813.9; 828.4; 850.2; 879.3; 901.1; 922.9; 959.2; 981.0; 1010.1; 1039.1; 1068.2; 1104.5; 1133.6; 1177.2; 1206.3; 1242.6; 1286.2; 1329.8; 1373.4; 1409.7; 1460.6; 1511.5; 1562.3; 1620.5; 1671.3; 1729.5; 1794.9; 1853];
k_cm = 30000*interp1(x_data,k_data,x(5));%,'linear','extrap');
%k_cm = 30000*1133.61; % middle value of dataset k_data
b_cm = 6e4;
l = 217.5;
% Matrices & solve
Mass = [m+a33 0 0; 0 m55+a55 0; 0 0 mh];
k = [c33+k_c -k_c*l -k_c; -k_c*l c55+k_c*l^2 k_c*l; -k_c k_c*l k_c+k_cm];
b = [b33+b_c -b_c*l -b_c; -b_c*l b55+b_c*l^2 b_c*l; -b_c b_c*l b_c+b_cm];
F = [Fa*sin(w1*t); Ma*sin(w2*t); 0];
EOM = Mass\(F-b*([x(2); x(4); x(6)])-k*([x(1); x(3); x(5)]));
dxdt = [x(2); EOM(1); x(4); EOM(2); x(6); EOM(3)];
% x = [z; zdot; theta; thetadot; u; udot]
end
I'm new to reinforcement learning, and I would like to process audio signal using this technique. I built a basic step function that I wish to flatten to get my hands on Gym OpenAI and reinforcement learning in general.
To do so, I am using the GoalEnv provided by OpenAI since I know what the target is, the flat signal.
That is the image with input and desired signal :
The step function calls _set_action which performs achieved_signal = convolution(input_signal,low_pass_filter) - offset, low_pass_filter takes a cutoff frequency as input as well.
Cutoff frequency and offset are the parameters that act on the observation to get the output signal.
The designed reward function returns the frame to frame L2-norm between the input signal and the desired signal, to the negative, to penalize a large norm.
Following is the environment I created:
def butter_lowpass(cutoff, nyq_freq, order=4):
normal_cutoff = float(cutoff) / nyq_freq
b, a = signal.butter(order, normal_cutoff, btype='lowpass')
return b, a
def butter_lowpass_filter(data, cutoff_freq, nyq_freq, order=4):
b, a = butter_lowpass(cutoff_freq, nyq_freq, order=order)
y = signal.filtfilt(b, a, data)
return y
class `StepSignal(gym.GoalEnv)`:
def __init__(self, input_signal, sample_rate, desired_signal):
super(StepSignal, self).__init__()
self.initial_signal = input_signal
self.signal = self.initial_signal.copy()
self.sample_rate = sample_rate
self.desired_signal = desired_signal
self.distance_threshold = 10e-1
max_offset = abs(max( max(self.desired_signal) , max(self.signal))
- min( min(self.desired_signal) , min(self.signal)) )
self.action_space = spaces.Box(low=np.array([10e-4,-max_offset]),\
high=np.array([self.sample_rate/2-0.1,max_offset]), dtype=np.float16)
obs = self._get_obs()
self.observation_space = spaces.Dict(dict(
desired_goal=spaces.Box(-np.inf, np.inf, shape=obs['achieved_goal'].shape, dtype='float32'),
achieved_goal=spaces.Box(-np.inf, np.inf, shape=obs['achieved_goal'].shape, dtype='float32'),
observation=spaces.Box(-np.inf, np.inf, shape=obs['observation'].shape, dtype='float32'),
))
def step(self, action):
range = self.action_space.high - self.action_space.low
action = range / 2 * (action + 1)
self._set_action(action)
obs = self._get_obs()
done = False
info = {
'is_success': self._is_success(obs['achieved_goal'], self.desired_signal),
}
reward = -self.compute_reward(obs['achieved_goal'],self.desired_signal)
return obs, reward, done, info
def reset(self):
self.signal = self.initial_signal.copy()
return self._get_obs()
def _set_action(self, actions):
actions = np.clip(actions,a_max=self.action_space.high,a_min=self.action_space.low)
cutoff = actions[0]
offset = actions[1]
print(cutoff, offset)
self.signal = butter_lowpass_filter(self.signal, cutoff, self.sample_rate/2) - offset
def _get_obs(self):
obs = self.signal
achieved_goal = self.signal
return {
'observation': obs.copy(),
'achieved_goal': achieved_goal.copy(),
'desired_goal': self.desired_signal.copy(),
}
def compute_reward(self, goal_achieved, goal_desired):
d = np.linalg.norm(goal_desired-goal_achieved)
return d
def _is_success(self, achieved_goal, desired_goal):
d = self.compute_reward(achieved_goal, desired_goal)
return (d < self.distance_threshold).astype(np.float32)
The environment can then be instantiated into a variable, and flattened through the FlattenDictWrapper as advised here https://openai.com/blog/ingredients-for-robotics-research/ (end of the page).
length = 20
sample_rate = 30 # 30 Hz
in_signal_length = 20*sample_rate # 20sec signal
x = np.linspace(0, length, in_signal_length)
# Desired output
y = 3*np.ones(in_signal_length)
# Step signal
in_signal = 0.5*(np.sign(x-5)+9)
env = gym.make('stepsignal-v0', input_signal=in_signal, sample_rate=sample_rate, desired_signal=y)
env = gym.wrappers.FlattenDictWrapper(env, dict_keys=['observation','desired_goal'])
env.reset()
The agent is a DDPG Agent from keras-rl, since the actions can take any values in the continuous action_space described in the environment.
I wonder why the actor and critic nets need an input with an additional dimension, in input_shape=(1,) + env.observation_space.shape
nb_actions = env.action_space.shape[0]
# Building Actor agent (Policy-net)
actor = Sequential()
actor.add(Flatten(input_shape=(1,) + env.observation_space.shape, name='flatten'))
actor.add(Dense(128))
actor.add(Activation('relu'))
actor.add(Dense(64))
actor.add(Activation('relu'))
actor.add(Dense(nb_actions))
actor.add(Activation('linear'))
actor.summary()
# Building Critic net (Q-net)
action_input = Input(shape=(nb_actions,), name='action_input')
observation_input = Input(shape=(1,) + env.observation_space.shape, name='observation_input')
flattened_observation = Flatten()(observation_input)
x = Concatenate()([action_input, flattened_observation])
x = Dense(128)(x)
x = Activation('relu')(x)
x = Dense(64)(x)
x = Activation('relu')(x)
x = Dense(1)(x)
x = Activation('linear')(x)
critic = Model(inputs=[action_input, observation_input], outputs=x)
critic.summary()
# Building Keras agent
memory = SequentialMemory(limit=2000, window_length=1)
policy = BoltzmannQPolicy()
random_process = OrnsteinUhlenbeckProcess(size=nb_actions, theta=0.6, mu=0, sigma=0.3)
agent = DDPGAgent(nb_actions=nb_actions, actor=actor, critic=critic, critic_action_input=action_input,
memory=memory, nb_steps_warmup_critic=2000, nb_steps_warmup_actor=10000,
random_process=random_process, gamma=.99, target_model_update=1e-3)
agent.compile(Adam(lr=1e-3, clipnorm=1.), metrics=['mae'])
Finally, the agent is trained:
filename = 'mem20k_heaviside_flattening'
hist = agent.fit(env, nb_steps=10, visualize=False, verbose=2, nb_max_episode_steps=5)
with open('./history_dqn_test_'+ filename + '.pickle', 'wb') as handle:
pickle.dump(hist.history, handle, protocol=pickle.HIGHEST_PROTOCOL)
agent.save_weights('h5f_files/dqn_{}_weights.h5f'.format(filename), overwrite=True)
Now here is the catch: the agent seems to always be stuck to the same neighborhood of output values across all episodes for a same instance of my env:
The cumulated reward is negative since I just allowed the agent to get negative rewards. I used it from https://github.com/openai/gym/blob/master/gym/envs/robotics/fetch_env.py which is part of OpenAI code as example.
Across one episode, I should get varying sets of actions converging towards a (cutoff_final, offset_final) that would get my input step signal close to my output flat signal, which is clearly not the case. In addition, I thought, for successive episodes, I should get different actions.
I wonder why the actor and critic nets need an input with an additional dimension, in input_shape=(1,) + env.observation_space.shape
I think the GoalEnv is designed with HER (Hindsight Experience Replay) in mind, since it will use the "sub-spaces" inside the observation_space to learn from sparse reward signals (there is a paper in OpenAI website that explains how HER works). Haven't look at the implementation, but my guess is that there needs to be an additional input since HER also process the "goal" parameter.
Since it seems you are not using HER (works with any off-policy algorithm, including DQN, DDPG, etc), you should handcraft an informative reward function (rewards are not binary, eg, 1 if objective achieved, 0 otherwise) and use the base Env class. The reward should be calculated inside the step method, since rewards in MDP's are functions like r(s, a, s`) you probably will have all the information you need. Hope it helps.
I'm trying to plot a function that contains a definite integral. My code uses all anonymous functions. When I run the file, it gives me an error. My code is below:
%%% List of Parameters %%%
gamma_sp = 1;
cap_gamma = 15;
gamma_ph = 0;
omega_0 = -750;
d_omega_0 = 400;
omega_inh = 100;
d_omega_inh = 1000;
%%% Formulae %%%
gamma_t = gamma_sp/2 + cap_gamma/2 + gamma_ph;
G = #(x) exp(-(x-omega_inh).^2./(2*d_omega_inh.^2))./(sqrt(2*pi)*d_omega_inh);
F = #(x) exp(-(x-omega_0).^2./(2*d_omega_0.^2))./(sqrt(2*pi)*d_omega_0);
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
Q_integral = #(x,y) F(x)./(y - x + 1i*gamma_t);
A = #(y) integral(#(x)A_integral(x,y),-1000,1000);
Q = #(y) integral(#(x)Q_integral(x,y),-3000,0);
P1 = #(y) -1./(1i.*(gamma_sp + cap_gamma)).*(1./(y + 2.*1i.*gamma_t)*(A(y)-conj(A(0)))-1./y.*(A(y)-A(0))+cap_gamma./gamma_sp.*Q(y).*(A(0)-conj(A(0))));
P2 = #(y) conj(P1(y));
P = #(y) P1(y) - P2(y);
sig = #(y) abs(P(y)).^2;
rng = -2000:0.05:1000;
plot(rng,sig(rng))
It seems to me that when the program runs the plot command, it should put each value of rng into sig(y), and that value will be used as the y value in A_integral and Q_integral. However, matlab throws an error when I try to run the program.
Error using -
Matrix dimensions must agree.
Error in #(x,y)G(x)./(y-x+1i*gamma_t)
Error in #(x)A_integral(x,y)
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 89)
Q = integralCalc(fun,a,b,opstruct);
Error in #(y)integral(#(x)A_integral(x,y),-1000,1000)
Error in
#(y)-1./(1i.*(gamma_sp+cap_gamma)).*(1./(y+2.*1i.*gamma_t)*(A(y)-conj(A(0)))-1. /y.*(A(y)-A(0))+cap_gamma./gamma_sp.*Q(y).*(A(0)-conj(A(0))))
Error in #(y)P1(y)-P2(y)
Error in #(y)abs(P(y)).^2
Error in fwm_spec_diff_paper_eqn (line 26)
plot(rng,sig(rng))
Any ideas about what I'm doing wrong?
You have
>> rng = -2000:0.05:1000;
>> numel(rng)
ans =
60001
all 60001 elements get passed down to
A = #(y) integral(#(x)A_integral(x,y),-1000,1000);
which calls
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
(similar for Q). The thing is, integral is an adaptive quadrature method, meaning (roughly) that the amount of x's it will insert into A_integral varies with how A_integral behaves at certain x.
Therefore, the amount of elements in y will generally be different from the elements in x in the call to A_integral. This is why y-x +1i*gamma_t fails.
Considering the complexity of what you're trying to do, I think it is best to re-define all anonymous functions as proper functions, and integrate a few of them into single functions. Look into the documentation of bsxfun to see if that can help (e.g., bsxfun(#minus, y.', x) instead of y-x could perhaps fix a few of these issues), otherwise, vectorize only in x and loop over y.
Thanks Rody, that made sense to me. I keep trying to use matlab like mathematica and I forget how matlab does things. I modified the code a bit, and it produces the right result. The integrals are evaluated very roughly, but it should be easy to fix that. I've posted my modified code below.
%%% List of Parameters %%%
gamma_sp = 1;
cap_gamma = 15;
gamma_ph = 0;
omega_0 = -750;
d_omega_0 = 400;
omega_inh = 100;
d_omega_inh = 1000;
%%% Formulae %%%
gamma_t = gamma_sp/2 + cap_gamma/2 + gamma_ph;
G = #(x) exp(-(x-omega_inh).^2./(2*d_omega_inh.^2))./(sqrt(2*pi)*d_omega_inh);
F = #(x) exp(-(x-omega_0).^2./(2*d_omega_0.^2))./(sqrt(2*pi)*d_omega_0);
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
Q_integral = #(x,y) F(x)./(y - x + 1i*gamma_t);
w = -2000:0.05:1000;
sigplot = zeros(size(w));
P1plot = zeros(size(w));
P2plot = zeros(size(w));
Pplot = zeros(size(w));
aInt_range = -1000:0.1:1200;
qInt_range = -2000:0.1:100;
A_0 = sum(A_integral(aInt_range,0).*0.1);
for k=1:size(w,2)
P1plot(k) = -1./(1i*(gamma_sp + cap_gamma)).*(1./(w(k)+2.*1i.*gamma_t).*(sum(A_integral(aInt_range,w(k)).*0.1)-conj(A_0))-1./w(k).*(sum(A_integral(aInt_range,w(k)).*0.1)-A_0)+cap_gamma./gamma_sp.*sum(Q_integral(qInt_range,w(k)).*0.1).*(A_0-conj(A_0)));
P2plot(k) = conj(P1plot(k));
Pplot(k) = P1plot(k) - P2plot(k);
sigplot(k) = abs(Pplot(k)).^2;
end
plot(w,sigplot)
I am using Octave.
My problem is this: I want to fill the bubbles of my scatter plot, as well as place a legend. But I get errors when I try to use 'filled', and no legend comes up when I use legend(...).
Part of my code looks like this:
%ALL SAMPLES, PHI(Signal) # THETA(Sample)=0
figure(5)
plot( Angles(:,1)([18:27]), ALL([18:27]), 10, [1 0 1]); %Magenta
hold on
scatter(Angles(:,1)([68:76]), ALL([68:76]), 10, [0 0 0]); %Black
scatter(Angles(:,1)([86:95]), ALL([86:95]), 10, [1 0 0]); %Red
scatter(Angles(:,1)([119:127]), ALL([119:127]), 10, [0 1 0]); %Green
scatter(Angles(:,1)([133:141]), ALL([133:141]), 10, [0 0 1]); %Blue
hold off
xlabel('Signal PMT angle (Sample angle at 0)');
ylabel('Normalized (signal/monitor) intensity');
legend('Control', 'Control', '1+2','Virgin','Cycle #1', 'Location','NorthEast');
title('Plot of All Samples, "-int Intensity"')
I know it should beplot( Angles(:,1)([18:27]), ALL([18:27]), 10, [1 0 1], 'filled');, but I receive errors when I do that. Also, a legend never seems to show up.
Apparently there is a problem with using legend with scatter in Octave. Based on this post:
http://octave.1599824.n4.nabble.com/Legend-in-scatter-plot-td3568032.html
the trick is to use the plot function to make scatter plot. I wrote the following function for plotting a bunch of scatter plots on the same axis.
This function takes in a bunch of cell arrays of the same length. Each element of the cell array corresponds to a separate series. The function returns a cell array of the same length containing the handle associated with each plot. The arguments of the function are explained below:
x_vals: a cell array of arrays of doubles corresponding to x values.
y_vals: a cell array of arrays of doubles corresponding to y values.
sizes: a cell array of doubles representing the size of the markers.
colors: a cell array of double arrays of length 3, representing [R, G, B] color values of the markers.
styles: a cell array of strings representing the shape of the markers.
function [handles] = scatter_series_set(x_vals, y_vals, sizes, colors, styles)
N = length(x_vals);
if ( (~ ( N == length(y_vals))) || (~ ( N == length(sizes))) || ...
(~ ( N == length(colors))) || (~ ( N == length(styles))) )
error('scatter_series_set: all arguments must be cell arrays of the same length');
end
%plot the first series
handles = cell([N, 1]);
handles{1} = plot(x_vals{1}, y_vals{1});
set(handles{1}, 'linestyle', 'none');
set(handles{1}, 'marker', styles{1});
set(handles{1}, 'markersize', sizes{1});
set(handles{1}, 'color', colors{1});
%plot additional series if present
if N > 1
hold on;
for ind = 2:N
handles{ind} = plot(x_vals{ind}, y_vals{ind});
set(handles{ind}, 'linestyle', 'none');
set(handles{ind}, 'marker', styles{ind});
set(handles{ind}, 'markersize', sizes{ind});
set(handles{ind}, 'color', colors{ind});
end
hold off;
end
end
The following example demonstrates how to use this function.
x1 = 0:(2*pi/100):(2*pi);
x2 = 2*x1;
y1 = sin(x1);
y2 = cos(x1);
y3 = sin(x2);
y4 = cos(x2);
names = {'a', 'b', 'c', 'd'};
x_vals = {x1, x1, x1, x1};
y_vals = {y1, y2, y3, y4};
sizes = {10, 10, 10, 10};
colors = {[1, 0, 0], [0, 0, 1], [0, 0, 0], [0.7071, 0, 0.7071]};
styles = {'^', 's', 'x', '+'}
scatter_series_set(x_vals, y_vals, sizes, colors, styles);
legend(names, 'location', 'southeast');
The example code produces the following plot:
The following works for me:
n = 100;
x = randn(n, 1);
y = randn(n, 1);
S = rand(n, 1)*20;
hold on
scatter(x(1:50), y(1:50), S(1:50), "red", "filled")
scatter(x(51:100), y(51:100), S(51:100), "green", "filled")
hold off
print('-depsc', 'bubbleplot.eps');
However, I'm not able to add a legend, and I didn't find any bug report or indication of a missing functionality for this. So, as an alternative, I would suggest adding marker and text to your plot.