Saving Data In A For Loop Into An Array

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This seems like a basic question, but I can't seem to figure it out.
I have data for multiple spirals and I want all the x,y and z co-ordinates for each spiral in seperate arrays. I cannot preallocate because I am not aware of the size before running the code.
Would really appreciate any help on how to save my co-ordinates for each spiral in seperate arrays. The code is pasted below
clc; clear; close all
spirals = (1:10:61);
dy = max(spirals);
indent = linspace(-255+dy, 255-dy,4);
num = 1:1:7;
cla
hold on
layerheight = [1 2 4 5];
for columns=1:4
for rows = 1:length(spirals)
center = -250 + (spirals(rows))*num(rows);
gap=layerheight(columns);
d=layerheight(columns);
r = spirals(rows);
z=(linspace(0,d,10000));
t=(40*pi/gap)*z;
x=round(r*cos(t)+center);
y=round(r*sin(t)+ indent(columns));
plot3(x,y,z,'o','MarkerSize',2)
% for counter = 1:length(z)
% rowdata(counter) = [x(rows) y(rows) z(counter)]; %this is how I thought of saving the data, but it doesn't work
% end
end
end
hold off
xlim([-255 255])
ylim([-255 255])
zlim([0 5])
grid on
xlabel('\it Increasing Radius \rightarrow')
ylabel('\it Increasing Layer Height \leftarrow')

Accepted Answer

Star Strider
Star Strider on 20 Aug 2019
The easiest way would probably be to save them all in separate cell arrays, here ‘xc’, ‘yc’, and ‘zc’:
for columns=1:4
for rows = 1:length(spirals)
center = -250 + (spirals(rows))*num(rows);
gap=layerheight(columns);
d=layerheight(columns);
r = spirals(rows);
z=(linspace(0,d,10000));
t=(40*pi/gap)*z;
x=round(r*cos(t)+center);
y=round(r*sin(t)+ indent(columns));
plot3(x,y,z,'o','MarkerSize',2)
% for counter = 1:length(z)
% rowdata(counter) = [x(rows) y(rows) z(counter)]; %this is how I thought of saving the data, but it doesn't work
% end
xc{rows,columns} = x;
yc{rows,columns} = y;
zc{rows,columns} = z;
end
end
Experiment to get the result you want.
  6 Comments
Hans123
Hans123 on 13 Sep 2019
StarStrider! My apologies for getting back to you this late. I figure out the answer and I forgot that I reached out for you for help!
Again I am really sorry.
Thanks for your help, I always appreciate it!
Star Strider
Star Strider on 13 Sep 2019
No worries!
As always, my pleasure!

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