need help with " Error using vertcat Dimensions of matrices being concatenated are not consistent"

Can anybody help me, cant understand why do i get this error when i run it?

i have attached my code.

thanks

[Jan]

Please post the complete error message, such that the readers can focus on fixing the error instead of guessing where it happens.

[Abhilash]

Error using vertcat

Dimensions of arrays being concatenated are not consistent.

Z = table ([5:6;5],['M','F','M'],'VariableName',{'age','gender'},'RowName',{'A','B','C'})

[Stephen23]

"Dimensions of arrays being concatenated are not consistent."

[5:6;5]
 ^^^    1x2, i.e. 1 row 2 columns
     ^  1x1, i.e. 1 row 1 column

How do you expect to vertically concatenate arrays with different numbers of columns?

Also note that square brackets are a contcatenation operator, so your code

['M','F','M']

concatenates three scalar characters into one character vector, and is exactly equivalent to:

'MFM'

You probably intended to define a string column vector:

["M";"F";"M"]

[ABDO]

h= 0.3; % Rayon de lissage
dx=0.3;
N =floor(1/dx);
x =0:dx:(pi/sqrt(3)); % Positions des particules
dt = 1/4;
t = 0:dt:1;
M = length(t) - 1;
% Initialisation de la solution approchée et de la matrice A
A = zeros(N, N);
b=zeros(N,1);
uapp=zeros(N,1);
%sol et cond exacte
u_exact = @(x) cos(sqrt(3)*x);
% Construction de la matrice A et du vecteur b
for i = 1:N
    for j =N
        r_ij = abs(x(i) - x(j));
        if r_ij <= h         
            d2W_dx2 = second_derivative_monaghan_kernel(r_ij, h);
            W_value = monaghan_kernel(r_ij, h);
            A(i, j) = (u_exact(x(j))- u_exact(x(i))) * d2W_dx2;
            b(i) = -3 * u_exact(x(j)) * W_value ;
        end
    end
end 
u_app(1)=1;
u_app(N)=u_exact(pi/sqrt(3));
uapp = b\A; 
uapp = [0; uapp];
% Affichage des résultats
plot(x , uapp, 'r');
hold on;
plot(x, u_exact(x), 'b--'); 
xlabel('x');
ylabel('u');
legend('Solution approchée (SPH)', 'Solution exacte');
title('Solution approchée et exacte par SPH');
grid on;
% Fonctions du noyau de Monaghan
 function W_value =monaghan_kernel(r_ij, h)
    C= (1/ h);
    if r_ij < h
        W_value =C*((2 / 3) - (r_ij / h)^2 +(1/2 *(r_ij / h)^3));
    elseif r_ij<2*h
           W_value =C*((1/6)*(2-(r_ij / h))^3);      
    else
       W_value=0;
    end 
end
function d2W_dx2 = second_derivative_monaghan_kernel(r_ij, h)
    C = (1 / h); % Corrected the denominator
    if r_ij < h
        d2W_dx2 = -2 * C * (1 - (3 * (r_ij / h)));
    elseif r_ij < 2*h
        d2W_dx2 = 2 * C * (2 - (r_ij / h)); % Removed extra characters
    else
        d2W_dx2 = 0;
    end
end  

Ihave a problem simultaning a numerical solution

I need some

[Voss]

@ABDO: Post a new question. Use this link: https://www.mathworks.com/matlabcentral/answers/questions/new/

Answer 1

[Thorsten]

When you vertically connect matrices using ; the matrices must have the same number of columns.

 A = ones(10, 3);
 B = zeros(20, 4);
 C = [A; B]

You get

 Error using vertcat
CAT arguments dimensions are not consistent.

Instead you can write

 B = zeros(20, 3);
 C = [A; B]; % works, A and B have same number of rows

[alon stuart]

thanks for answering. i understand what you have suggested but i don't think this is the case. ive get a 3X1 vector by multiplying a 3 by 3 matrix with a 3X1 vector then i try to plot this vector by attaching a timeline vector from 0 to 5 sec. any other suggestions ?

[Thorsten]

Your t has 51 values. Next you try to use t an index into your 3x3 matrix by

 Qout_doot(t)

That does not work, because

1. t is not a positive integer value

2. t has more elements than Qout_doot.

You could use

 t = linspace(0, 5, size(Qout_doot, 2));
 plot(t, Qout_doot)

[alon stuart]

thanks thorsten but its still doesnt work

[Thorsten]

Ok, so what goes wrong? My example should work for Qout_doot of any size without a Matlab error. But the result may not be what you are looking for? Or what exactly doesn't work?

[alon stuart]

iam not sure what is the problem exactly

the first code work perfectly so why the seconed one doesnt work? (first code): "

      P0 =

       cos(t1 + t2)/2 + (4*cos(t1))/5
       sin(t1 + t2)/2 + (4*sin(t1))/5
                             4/5 - d3
                                    1
      t1=4*t; t2=t^2+3; d3=10*t;
      P=subs(P0)
      P=P(1:3);
      P = matlabFunction(P);
      t = 0:0.1:10;
      plot(t,P(t));
"
(second code):"

syms t1 t2 d3 t4  t

F=P0(1:3);

J = [diff(F(1),t1) diff(F(1),t2) diff(F(1),d3); diff(F(2),t1) diff(F(2),t2) diff(F(2),d3); diff(F(3),t1) diff(F(3),t2) diff(F(3),d3)]

Qin_doot=[4;2*t;10];

Qout_doot=simplify(J*Qin_doot)
t1=4*t; t2=t^2+3; d3=10*t; t4=0;
Qout_doot=subs(Qout_doot)
Qout_doot =matlabFunction (Qout_doot); 
t = 0:0.1:5;
plot(t,Qout_doot(t));
"

thanks again

Answer 2

[Jan]

I guess that this fails:

t = 0:0.1:10;
...
Qin_doot = [4; 2*t; 10];

the sizes of the scalars 4 and 10 do not match with the size of t.

The description as text "ive get a 3X1 vector by multiplying ..." is less useful than posting the line of code you are talking of.

[alon stuart]

Jan thanks for answering, and you are right! ill detail again:

"Qout_doot=subs(Qout_doot)

Qout_doot = matlabFunction(Qout_doot); t = 0:0.1:5; plot(t,Qout_doot(t));"

when Qout_doot is a 3X3 matrix that dependent on the variable t

[John Amakali]

hi guys , i have the same problem... i have attached my code. someone help please.

Warning: Unsuccessful read: A timeout occurred before the Terminator was reached.. 
Error using vertcat
Dimensions of matrices being concatenated are not consistent.

Error in GUI>pushbutton1_Callback (line 144)
    set(M,'Zdata',[Tdata; zeroIndex],'Cdata',[Tdata; Tdata]);

Error in gui_mainfcn (line 95)
        feval(varargin{:});

[Walter Roberson]

We do not know what size(Tdata) is or size(zeroIndex) .

The message about timeout suggests that you might be doing a serial read or read from arduino or raspberry, and that it is not returning as much data as you expect. If your Tdata is the actual received data and zeroIndex is built under the assumption that all of the requested data was received, then they could easily end up with incompatible sizes.

Answer 3

[A N]

I've got another problem with using vertcat Dimensions of matrices being concatenated are not consistent.

I'm trying to write a code to make a transmission characteristic of the Gaussian filter from this

title ,,Computional Surface and Roundness Metrology"

My code is:

lambdac=0.8; %mm
dx=0.001; %mm

alpha=sqrt(log(2)/pi);
disp(alpha);
x=(-lambdac:dx:lambdac);

S=(1/alpha*lambdac).*exp(-pi*(x/alpha*lambdac).^2);

S=S/sum(S);
figure(1)
plot(x,S)

xlabel('Odległość [mm]')

ylabel('Wzmocnienie')

xlim([-1 1])

%Przykład 5.2
m=size(S,1); %długość filtru gaussowskiego
l=8; %długość punktów pomiarowych na profilu

n=1/dx; %liczba punktów pomiarowych na profilu

   S=[zeros(n/2-floor(m/2),1); S; zeros(n/ 2-floor(m/2)-1,1)];  %copied from %previous link

Sf=fft(S); %transformata Fouriera S

j=(2:1:floor (n/2)+1); %generuje falę o długości dla danych X 

wave=n*dx./(j-1);

figure (2)

semilogx (wave, 100*abs(Sf(2:floor (n/2)+1,1)));

xlabel('Długość fali [mm]')
ylabel('Amplituda [%]')

hold off

I've tried also with

S=[zeros(n/2-floor(m/2)-1,1); S; zeros(n/ 2-floor(m/2)+1,1)];

but no result. After n=1/dx I've found:

a=zeros(n/2-floor(m/2),1);
size(a);
disp(size(a));

S;
size(S);
disp(size(S));

c=zeros(n/2-floor(m/2)+1,1);
size(c);
disp(size(c));

and result:

     500     1
             1        1601

     501     1

Any idea how to solve it?

[Stephen23]

So S has 1601 columns and you are trying to vertically concatenate it with two arrays, each of which has just one column. You need to ensure that the number of columns or rows is the same, e.g.:

R = n/2-floor(m/2);
S = [zeros(1,R-1), S, zeros(1,R+1)];

Answer 4

[A N]

Great! Works!

After it I would like to do a FFT of S, but

My code:

Sf=fft(S); %transformata Fouriera 

j=(2:1:floor(n/2)+1); %faa o danej długości

size(j);
disp(size(j));

wave=n*dx./(j-1);

size(wave);

disp(size(wave));

figure(2)

semilogx(wave,100*abs(Sf(2:floor(n/2)+1,1)))

xlabel('Długość fali [mm]')

ylabel('Amplituda [%]')

Command Windows shows as a result of those displays:

     1   500

     1   500

Index exceeds matrix dimensions.

Error in proba (line 61)
semilogx(wave,100*abs(Sf(2:floor(n/2)+1,1)));

Error in proba (line 65)
semilogx(wave,100*abs (2:floor(n/2)));"

Do you know how to solve it? Maybe my error is easy but I've been doing only spectral FFTs and would like to know more about examples like this (no Fs and spectral). In Help I found only spectral FFT. Line with semilog I copied from this book, and really hard to find out - is this wrong?

Edit:

Writing:

semilogx(wave,100*abs(Sf(2:floor(n/2)+1)));

impacts of this plot:

[Walter Roberson]

At the command line command

dbstop if error

and run the code. When it stops, look at size(Sf) and look at the value of n and calculate floor(n/2)+1 and see if it is within the number of rows of Sf.

[A N]

Dimension of

 Sf is 1 2601.

May I ,,lean" this matrix somehow or or make the others greater?

[Walter Roberson]

Okay so Sf is a row vector of length 2601. But what is n at the time that the problem occurs?

[A N]

Wrote

dbstop if error

in Editor and a screen of my Workspace (so I think that n is 500, but no sure).

[Walter Roberson]

Sf(2:floor(n/2)+1,1) asks for row 2:501, column 1, but your Sf is a column vector that is 1 x 2601 not 2601 x 1.

[A N]

Wrote:

semilogx(wave,100*abs(Sf(2:floor(n/2)+1)))

And get this plot:

[Walter Roberson]

The plot did not get attached.

Answer 5

[MANYINGA MARC]

Bonjour la communauté. J'ai un code de calcul, Dan lequel je voudrais faire varier les éléments d'une matrice À, en fonction d'une variable temps t. Par exemple

t=0:10:60 A=[1 2*t 3 3*t 5 8 5 6 3] Affiche error using vertcat Besoin d'aide

[Walter Roberson]

A=[1 2*t 3 3*t 5 8 5 6 3] would be an error if t is not a row vector.