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')

采纳的回答

Star Strider
Star Strider 2019-8-20
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 个评论
Hans123
Hans123 2019-9-13
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!

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