Easy Question, update scatter point in loop
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Hi:
I just want to get the 4 vectors scatter update every time in my loop. Delete the last time points then plot then hold on to new loop.
How to do that??
Thank you.
x1{1} = [.2; .3]; x2{1} = [.3; .4]; x3{1} = [.5; .6]; x4{1} = [.6; .7]; k = 1;
for k = 1:1:50
scatter(x1{k}(1,1), x1{k}(2,1));
hold on
scatter(x2{k}(1,1), x2{k}(2,1));
hold on
scatter(x3{k}(1,1), x3{k}(2,1));
hold on
scatter(x4{k}(1,1), x4{k}(2,1));
x1{k+1} = (x1{k} + x2{k} + x4{k})/3;
x2{k+1} = (x1{k} + x2{k} + x3{k})/3;
x3{k+1} = (x2{k} + x3{k} + x4{k})/3;
x4{k+1} = (x1{k} + x3{k} + x4{k})/3;
end
采纳的回答
Ben11
2014-6-24
What if you try this:
x1{1} = [.2; .3]; x2{1} = [.3; .4]; x3{1} = [.5; .6]; x4{1} = [.6; .7]; k = 1;
figure
for k = 1:1:50
hold all
scatter(x1{k}(1,1), x1{k}(2,1))
scatter(x2{k}(1,1), x2{k}(2,1));
scatter(x3{k}(1,1), x3{k}(2,1));
scatter(x4{k}(1,1), x4{k}(2,1));
x1{k+1} = (x1{k} + x2{k} + x4{k})/3;
x2{k+1} = (x1{k} + x2{k} + x3{k})/3;
x3{k+1} = (x2{k} + x3{k} + x4{k})/3;
x4{k+1} = (x1{k} + x3{k} + x4{k})/3;
hold off
drawnow
end
First you create a figure, then you force the plot to update with drawnow. You can keep the original "hold on" calls and see what it does as well.
12 个评论
Yuang
2014-6-24
Yuang
2014-6-24
Ben11
2014-6-24
ok. So is it something like this?:
x1{1} = [.2; .3]; x2{1} = [.3; .4]; x3{1} = [.5; .6]; x4{1} = [.6; .7]; k = 1;
figure
for k = 1:1:50
scatter(x1{k}(1,1), x1{k}(2,1))
hold on
scatter(x2{k}(1,1), x2{k}(2,1));
hold on
scatter(x3{k}(1,1), x3{k}(2,1));
hold on
scatter(x4{k}(1,1), x4{k}(2,1));
x1{k+1} = (x1{k} + x2{k} + x4{k})/3;
x2{k+1} = (x1{k} + x2{k} + x3{k})/3;
x3{k+1} = (x2{k} + x3{k} + x4{k})/3;
x4{k+1} = (x1{k} + x3{k} + x4{k})/3;
hold off
drawnow
end
Yuang
2014-6-24
Ben11
2014-6-24
You could add this line:
pause(n)
right after drawnow, where n is the time in seconds during which execution is paused.
Yuang
2014-6-24
Ben11
2014-6-24
Great! You're welcome.
Yuang
2014-6-24
Ben11
2014-6-24
Hum I'm don't know what a trial for points is sorry. Do you have an example?
Yuang
2014-6-24
Ben11
2014-6-24
ok. Do you mean something like this:
for k = 1:50
scatter(x1{k}(1,1), x1{k}(2,1))
if k > 1
line([x1{k-1}(1,1) x1{k}(1,1)],[x1{k-1}(2,1) x1{k}(2,1)],'Color','r');
end
drawnow
hold on
scatter(x2{k}(1,1), x2{k}(2,1));
if k > 1
line([x2{k-1}(1,1) x2{k}(1,1)],[x2{k-1}(2,1) x2{k}(2,1)],'Color','g');
end
drawnow
hold on
scatter(x3{k}(1,1), x3{k}(2,1));
if k > 1
line([x3{k-1}(1,1) x3{k}(1,1)],[x3{k-1}(2,1) x3{k}(2,1)],'Color','b');
end
drawnow
hold on
scatter(x4{k}(1,1), x4{k}(2,1));
if k > 1
line([x4{k-1}(1,1) x4{k}(1,1)],[x4{k-1}(2,1) x4{k}(2,1)],'Color','k');
end
drawnow
hold on
x1{k+1} = (x1{k} + x2{k} + x4{k})/3;
x2{k+1} = (x1{k} + x2{k} + x3{k})/3;
x3{k+1} = (x2{k} + x3{k} + x4{k})/3;
x4{k+1} = (x1{k} + x3{k} + x4{k})/3;
hold off
pause(.1)
end
I seems odd is it what you would expect?
更多回答（1 个）
Geoff Hayes
2014-6-24
Try grabbing the handle returned from each scatter call and then deleting that on subsequent iterations. Outside the for loop declare an array to manage the scatter plot handles
% declare an array for the handles to the scatter plots
scatterHandles = [];
Now within the for loop, delete the handles if any exist
for k = 1:1:50
% delete scatter plot data from previous iteration if it exists
if ~isempty(scatterHandles)
for m=1:length(scatterHandles)
delete(scatterHandles(m));
end
end
% get the scatter plot handles for each call to scatter
hold on
scatterHandles (1) = scatter(x1{k}(1,1), x1{k}(2,1));
scatterHandles (2) = scatter(x2{k}(1,1), x2{k}(2,1));
scatterHandles (3) = scatter(x3{k}(1,1), x3{k}(2,1));
scatterHandles (4) = scatter(x4{k}(1,1), x4{k}(2,1));
% etc.
end
Try the above and see what happens!
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