Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.
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Buenas tardes en el codigo que estoy estudiando cuando lo trnascribi sale el error:
Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.
Error in ejercicio_15_3_4 (line 83)
indices(i)= find(coordenadas_ascendentes(i) == coordenadas_x);
el código es el siguiente:
clc
clear;
x=0:0.01:40;
k=(14-1.2*x)/2.25;
g=(8-x)/1.1;
h=9-2.5*x;
z= @(X,Y) 1.75*X + 1.25*Y;
plot(x,k,x,g,x,h);
hold on;
grid;
title('Maximización')
xlabel x;
ylabel y(x);
%Generación de al intersección en el eje y
ordenada= [ k(1) g(1) h(1)];
miny=min(ordenada);
plot(x(1),miny, 'o');
%Generar intersección del eje x
x1=find(k==0);
x2=find(g==0);
x3=find(h==0);
ejex=[x(x1) x(x2) x(x3)];
minx=min(ejex);
plot(min(ejex),0,'o');
xlim([0, minx+3]);
ylim([0, miny+3]);
%Secuencia para graficar la intersección entre las funciones
f(1,:)=k;
f(2,:)=g;
f(3,:)=h;
combinaciones=nchoosek(1:3,2);
cant_comb=size(combinaciones);
cant_comb=cant_comb(1);
c=1;
for i=1:cant_comb
CUMPLE=1;
comb_temp=combinaciones(i,:);
f_uno=f(comb_temp(1),:);
f_dos=f(comb_temp(2),:);
if isempty(find(f_uno==f_dos))
intersecciones(i,:)=find(f_uno==max(f_uno));
else
intersecciones(i,:)=find((f_uno==f_dos));
end
inter_temp=intersecciones(i);
for j=1:3
if f_uno(inter_temp)>f(j,inter_temp)
CUMPLE=0;
end
end
if CUMPLE==1
inter_validas(c,1)=inter_temp;
inter_validas(c,2)=comb_temp(1);
c=c+1;
end
end
%Intersecciones válidas
cant_inter_validas= size(inter_validas);
cant_inter_validas=cant_inter_validas(1);
for i=1:cant_inter_validas
coordenadas(i,:)=[x(inter_validas(i,1)),f(inter_validas(i,2),inter_validas(i,1))];
plot(coordenadas(i,1), coordenadas(i,2),'o');
end
coordenadas_x=coordenadas(:,1);
coordenadas_ascendentes=sort(coordenadas_x);
coordenadas_finales(:,1)=coordenadas_ascendentes;
for i=1:length(coordenadas_ascendentes)
indices(i)= find(coordenadas_ascendentes(i) == coordenadas_x);
coordenadas_finales(i,2)= coordenadas(indices(i),2);
end
%Polígono
PX=[0,x(1), (coordenadas_finales(:,1))',minx];
PY=[0,miny, (coordenadas_finales(:,2))',0];
%Gráfico del polígono
fill(PX,PY,'r');
z1=z;
for i=1:length(PX)
X=PX(i);
Y=PY(i);
valorz(i)=z1(X,Y);
end
indicezmax=find(valorz==max(valorz));
coordenadamaxz=[PX(indicezmax) PY(indicezmax)];
plot (PX(indicezmax), PY(indicezmax),'o','MarkerSize',10);
str=strcat('El óptimo se alcanza en :',num2str(PX(indicezmax)),',',num2str(PY(indicezmax)));
text1=text(minx-3,miny-3,str);
str=strcat('Y tiene un valor de:',num2str(max(valorz)));
text2=text(minx-3,miny-3,miny-1,str);
2 Comments
Dyuman Joshi
on 23 Nov 2022
find() returns all the indices that satisfy the condition tested.
y=[1 3 2 0 2 1 4 2 0 2];
i=find(y==y(3))
When a condition is satisfied more than once, it will return multiple (corresponding) indices.
You can not assign multiple values to a single element of an array.
You can, however, get the n index/indices from first or last that satisfies the condition -
%1 from first
i1=find(y==y(3),1)
%2 from last
il=find(y==y(3),2,'last')
Voss
on 23 Nov 2022
indices(i)= find(coordenadas_ascendentes(i) == coordenadas_x);
returns anything other than a single index, you'll get an error.
In particular, it may be that no element of coordenadas_x is equal to coordenadas_ascendentes(i), in which case find returns an empty array.
Answers (1)
Sanjana
on 7 Jun 2023
Hi Marcos,
I understand that there is a mismatch in the dimensions of the variables in the assignment operation. The problem here is at this line,
indices(i)= find(coordenadas_ascendentes(i) == coordenadas_x);
For “==” operation, the arrays should have same dimensions, please modify the below lines,
for i=1:length(coordenadas_ascendentes)
indices(i)= find(coordenadas_ascendentes(i) == coordenadas_x);
coordenadas_finales(i,2)= coordenadas(indices(i),2);
end
to,
indices= find(coordenadas_ascendentes == coordenadas_x);
coordenadas_finales(:,2) = coordenadas(indices,2);
Hope this helps!
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