how to store the matrix from for loop?

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JAGADEESH JAGA
JAGADEESH JAGA am 1 Feb. 2017
Kommentiert: JAGADEESH JAGA am 3 Feb. 2017
m=3;
A=[ 1 1 0];
B=[0 1 0];
D=B-A;
d1=m*A;
for X=0:1:m
C=d1+D*X
end

Akzeptierte Antwort

KSSV
KSSV am 1 Feb. 2017
m=3;
A=[ 1 1 0];
B=[0 1 0];
D=B-A;
d1=m*A;
C = zeros(m+1,length(d1)) ;
for X=1:1:m+1
C(X,:)=d1+D*X ;
end
  1 Kommentar
JAGADEESH JAGA
JAGADEESH JAGA am 3 Feb. 2017
the answer is really helpful for me.thanks for the answer. could you answer my question m=3; t=[1 0 0; 0 (1/sqrt(2)) (1/sqrt(2))]; v1=0.24; v2=0.956; a=[ 1 1 0]; b=[0 1 0]; p=b-a; d1=m*a; for x=1:1:m+1 c=d1+p*(x-1) f=transpose(c); vndq=t*f; vnq=vndq(1,1); vnd=vndq(2,1); d(x)=abs(v2-vnq)+abs(v1-vnd) n=min(d) end In the given code c row matrix corresponding minimum value of d matrix must be output if the d changes corresponding c matrix should be abtained to that minimum distance using for loop

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Weitere Antworten (1)

Andrei Bobrov
Andrei Bobrov am 1 Feb. 2017
Bearbeitet: Andrei Bobrov am 2 Feb. 2017
m=3;
A=[ 1 1 0];
B=[0 1 0];
D=B-A;
d1=m*A;
C = d1 + (0:m)'*D; % R2016b and later
C = bsxfun(@plus,d1,(0:m)'*D); % R2016a and earlier
  2 Kommentare
Stephen23
Stephen23 am 2 Feb. 2017
+1 nice use of MATLAB.
Andrei Bobrov
Andrei Bobrov am 2 Feb. 2017
Thank you Stephen!

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