Error using vertcat: dimension of arrays being concenated are not consistent

Hello,

I am using xlsread (x=xlsread(filename,sheet,xlrange))

to obtain values for x from my excel file, x is a 27x1 double.

(x=

1.01

1.05

0.004

0.65

etc...)

Then I have matrix A (nx27). I try to call some of the values for A from x, for example(i present just a few rows for the idea):

A= [1 -1 0 1 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; % just some random row

0 0 0 1 -1 0 x(20) 1 0 (x(21)-x(23)) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]

A is a jacobian matrix I use , so some elements for A are calculation operations from partial differentials.

When i use this, it gives me the error using vertcat: dimensions of arrays being concatenated are not consistent

The error points to the line where the term 'A=' is. Does it have something to do with it? Anyways, i didnt find an answer after searching for 5 hours from here and trying few things.

Any idea why this happens? is it something with xlsread? I am completely lost with this. Should i use something different for obtaining values for x from excel?

BR,

Jouni

[Walter Roberson]

We need the real code that constructs A

[Jouni Salovaara]

Right, sorry about that. Here is the code, left out irrevelant stuff. I have already calculated the partial differentials on paper and just wrote the results as the matrix A to the script. The idea for me is just to bring values from x to the matrix A. The matrix A does not change in this exercise, only the values of x.

path='C:\Users\user\location\Matlab';
filename='importfile.xlsx';
fullpath= [path, '\', filename];
sheet2='values';
xlrange3= 'A2:A28';
x=xlsread(filename,sheet2,xlrange3);

A=[1 -1 0 1 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 
0 0 1 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 
x(14) -x(15) (u(4) - x(16)) 0 0 0 0 0 0 0 0 0 0 x(1) -x(2) -x(3) -u(2) 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 (-x(21)+x(23)) 0 0 0 0 0 0 0 0 0 0 0 0 0 -x(7) 0 x(7) 1 0 0 0; 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 1 1; 
x(13) -x(15) 0 x(17) x(19) -x(20) (x(21)-x(23)) 0 0 0 0 0 x(1) 0 x(2) 0 x(4) 0 x(5) -x(6) x(7) 0 -x(7) 0 0 0 0; 
0 0 0 0 0 0 (x(21)-x(22)) 0 0 0 0 0 0 0 0 0 0 0 0 0 x(7) -x(7) 0 0 -1 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 u(2) -u(2) 0 0 0 0 0 0 1 0 0;
0 0 0 0 0 0 (x(22)-u(4)) 0 0 0 0 0 0 0 0 0 0 0 0 0 x(7) 0 0 0 -1 0 0;
0 0 (x(15)-u(3)) 0 0 0 0 0 0 0 0 0 0 0 0 0 x(3) 0 0 0 0 0 0 0 0 0 1];

and here is the error:

Error using vertcat

Dimensions of arrays being concatenated are not consistent.

Error in test (line 23)

A=[1 -1 0 1 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;

Answer 1

[Guillaume]

The error is simple. At least one of the rows of your A does not have the same number of elements as the others.

Certainly,

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 1 1; 
x(13) -x(15) 0 x(17) x(19) -x(20) (x(21)-x(23)) 0 0 0 0 0 x(1) 0 x(2) 0 x(4) 0 x(5) -x(6)
x(7) 0 -x(7) 0 0 0 0; 

is clearly wrong since that last row has only 7 elements. Note that newlines in a matrix construction are equivalent to semicolons.

[Jouni Salovaara]

Oh, man! maybe I'm done for today :D

Thank you very much for the answer! All sorted out now.

BR,

Jouni