storing script outputs as an array

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Erik Anderson
Erik Anderson 2017-3-15
评论： Erik Anderson 2017-3-15
I wanted to try identifying convex polygons by their projected widths. How can I get the outputs, D, stored in a compact array that I can save and work with?
%loop that will have 1440 iterations (each iteration
%representing a rotation of the plane by 0.5 degrees)
for t= 1:2000
format('long')
%imagine points in 2-dimensional euclidean space. These three vectors
%represent the three vertices of an equilateral triangle cetered at the
%origin
A=[0;1];
B=[-0.86603; -0.50000];
C=[0.86603; 0.50000];
%T is designed to be a 0.5 degree counter-clockwise rotation matrix, sending each
%of the previous vectors to their 0.5 degree rotated counterparts
T= [ 0.99996 -0.00873;0.00873 0.99996];
%P extracts the x values of each vector above, discarding the rest
P= [1 0];
%represent the rotated images of A,B,C, for some iteration 't'
tA=(T^t)*A;
tB=(T^t)*B;
tC=(T^t)*C;
%represent the projection of tA,tB,tC onto the x-axis.
imA=P*tA;
imB=P*tB;
imC=P*tC;
%distances between the respective projections
distAB=abs(imA-imB);
distBC=abs(imB-imC);
distCA=abs(imC-imA);
%making a vector V with elements 'distAB' 'distBC' 'distCA'
X=[1 0 0];
Y=[0 1 0];
Z=[0 0 1];
dX=distAB*X;
dY=distBC*Y;
dZ=distCA*Z;
V= dX+dY+dZ;
% D = max element in V is the width of the 'shadow' of the triangle on the
%x-axis
D=max(V)
end
  2 个评论
Erik Anderson
Erik Anderson 2017-3-15
my question is
"How can I get the outputs, D, stored in a compact array that I can save and work with?"
right now I am using
Output=zeros(1,10000); Output(t)=D; xlswrite('triangleout.xls',Output);
seems to work well

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回答（1 个）

per isakson
per isakson 2017-3-15
编辑：per isakson 2017-3-15
I guess: this is what you try to do.
D = nan( 1, 2000 );
format('long')
%imagine points in 2-dimensional euclidean space. These three vectors
%represent the three vertices of an equilateral triangle cetered at the
%origin
A=[0;1];
B=[-0.86603; -0.50000];
C=[0.86603; 0.50000];
%T is designed to be a 0.5 degree counter-clockwise rotation matrix, sending each
%of the previous vectors to their 0.5 degree rotated counterparts
T= [ 0.99996 -0.00873;0.00873 0.99996];
%P extracts the x values of each vector above, discarding the rest
P= [1 0];
for t= 1:2000
%represent the rotated images of A,B,C, for some iteration 't'
tA=(T^t)*A;
tB=(T^t)*B;
tC=(T^t)*C;
%
%represent the projection of tA,tB,tC onto the x-axis.
imA=P*tA;
imB=P*tB;
imC=P*tC;
%
%distances between the respective projections
distAB=abs(imA-imB);
distBC=abs(imB-imC);
distCA=abs(imC-imA);
%
%making a vector V with elements 'distAB' 'distBC' 'distCA'
X=[1 0 0];
Y=[0 1 0];
Z=[0 0 1];
%
dX=distAB*X;
dY=distBC*Y;
dZ=distCA*Z;
V= dX+dY+dZ;
%
% D = max element in V is the width of the 'shadow' of the triangle on the
%x-axis
D(t)=max(V);
end
especially note
D = nan( 1, 2000 );
...
D(t) = max(V);
in the original code D was overwritten two thousand times

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