Obtaning a matrix with the right indexes

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Nuno
Nuno 2013-4-15
My question:
A want to obtain a matrix A(4xN)=[A11 A21 A31 A41; A12 A22 A32 A42; ...; A1N A2N A3N A4N].
So I think that I have to do something like this:
syms A1 A2 A3 A4;
N=10; %for example
A=sym(zeros(4,N));
B=sym(zeros(4,N));
for i=1:N
A(:,i)=[A1; A2; A3; A4]
end
for j=1:N
B(:,j)=[j; j; j; j]
end
C=char(A)
D=char(B)
%And now I have to do a concatenation:
E=strcat([C,D])
%But this won't work of course and I don't know if some part of my program is correct or there is a different approach for this.
Thank you for your atention.
  2 个评论
Nuno
Nuno 2013-4-17
编辑：Walter Roberson 2013-4-18
O.K. I have almost what I want, but now matlab is giving me an error and I can't seem to go around it.
My program:
syms A1 A2 A3 A4;
N=20; %for example
A=sym(zeros(4,N));
B=sym(zeros(4,N));
H=sym(zeros(4,N));
A='A1A2A3A4'
B=1:1:N
C=mat2str(B)
%Concatenation
M=0;
s1=0:N-10
s2=1:N-9
for m=1:N
if m<10
H(1,m)=[A(1:2),C(2*m)];
H(2,m)=[A(3:4),C(2*m)];
H(3,m)=[A(5:6),C(2*m)];
H(4,m)=[A(7:8),C(2*m)];
end
if m>=10
for k=1:N
H(1,m)=[A(1:2),C(2*m+s1(k):2*m+s2(k))];
H(2,m)=[A(3:4),C(2*m+s1(k):2*m+s2(k))];
H(3,m)=[A(5:6),C(2*m+s1(k):2*m+s2(k))];
H(4,m)=[A(7:8),C(2*m+s1(k):2*m+s2(k))];
end
end
end
disp(H)
And the error:
Error using sym>convertExpression (line 2256)
Conversion to 'sym' returned the MuPAD error: Error: Unexpected 'integer'. [line 1, col 4]
Error in sym>convertChar (line 2167)
s = convertExpression(x);
Error in sym>convertCharWithOption (line 2150)
s = convertChar(x);
Error in sym>tomupad (line 1881)
S = convertCharWithOption(x,a);
Error in sym (line 108)
S.s = tomupad(x,'');
Error in sym/subsasgn (line 1614)
if ~isa(B,'sym'), B = sym(B); end
Error in la1 (line 27)
H(1,m)=[A(1:2),C(2*m+s1(k):2*m+s2(k))];
Nuno
Nuno 2013-4-17
Can anyone help me with this?
Thank you.

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采纳的回答

Walter Roberson
Walter Roberson 2013-4-18
Which MATLAB version are you using? If you are using R2011b or later, try just
A = sym('A', [4 N]);
  1 个评论
Nuno
Nuno 2013-4-18
Thank you Walter Roberson! It makes a lot easier to read the program and it works fine too.

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更多回答（1 个）

Ahmed A. Selman
Ahmed A. Selman 2013-4-18
Try this out to see if it solves the issue:
clear
clc
N=20; %for example
H=sym(zeros(N,4));
B=1:N;
%Concatenation
for j=1:4
for m=1:N
H(m,j)=['A',num2str(j),num2str(m)];
end
end
disp(H)
  1 个评论
Nuno
Nuno 2013-4-18
Thank you Ahmed A. Selman! To get what I want I just had to make H(j,m) instead of H(m,j).

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