Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

7 visualizaciones (últimos 30 días)
Dear community, help with this.
Voc=Clean(:,7);
a=Voc(1:3);
j=1;k=9;
t=length(Voc);
for i=1:t;
v= Voc(j:k,1);
a=cellfun(@mean,v);
Voc_p(i,1)=a;
j=j+9;
k=k+9;
end
A datatable was convert into a array previously and appears this error:
Unable to perform assignment because the indices on the left side are not compatible with the size of the
right side.
Error in Prueba_1 (line 29)
Voc_p(i,1)=a;
How can I fix this?
Thank u

Respuesta aceptada

Walter Roberson
Walter Roberson el 10 de Jun. de 2019
v= Voc(j:k,1);
That is creating a non-scalar cell array
a=cellfun(@mean,v);
That is going through each cell member and taking the mean of each. The fact that you did not get an error on this line tells us that each cell member was a scalar or a vector, not empty and not 2 D or more. The end result will be a vector of length k-j+1 = 9.
Voc_p(i,1)=a;
You are trying to store that vector of length 9 into a single location, Voc_p(i,1)
  13 comentarios
Walter Roberson
Walter Roberson el 11 de Jun. de 2019
You moved the *9 to inside the floor() when it needs to be outside.
lastidx = floor(nrow/9) * 9;
Jonathan Bijman
Jonathan Bijman el 11 de Jun. de 2019
It works!
Thank U so much Walter for all the troubles I caused u.
I hope can I count for your help in the future, for this amateur like me.
Thank u so much =)

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by