Why am I getting the error "Array indices must be positive integers or logical values"?
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Yogesh Pai
el 5 de Dic. de 2018
Comentada: Walter Roberson
el 3 de Abr. de 2025
I want to compare the exponential of matrix "A" obtained by the summation formula and the one with the eigen values. However, I'm getting the following error when exponential is computed with the diagonalised form (E here).
Array indices must be positive integers or logical values.
A = [1,2,1,3; 3,1,2,0; 2,2,1,4; 1,3,2,1];
[V,D] = eig(A);
exp_A = 0;
for k = 0:100
exp_A = exp_A + (A^k)/factorial(k);
end
disp(exp_A);
304.3054 340.7344 263.9243 320.0369
275.4796 308.2690 238.7711 289.0775
389.2552 436.7654 339.1998 410.7063
309.9581 347.7556 269.7020 326.5840
E = V*(exp(D))\V;
disp(E);
0.0008 + 0.0000i -0.0001 - 0.0000i -0.0001 + 0.0000i -0.0007 - 0.0000i
-0.0001 - 0.0000i -0.8364 - 0.1567i 0.1039 + 0.0000i 0.8393 + 0.1795i
-0.0001 + 0.0000i 0.1039 + 0.0000i -0.8364 + 0.1567i 0.8393 - 0.1795i
-0.0007 - 0.0000i 0.8393 + 0.1795i 0.8393 - 0.1795i -0.7766 - 0.0000i
5 comentarios
Respuesta aceptada
Kelly Kearney
el 21 de Oct. de 2024
Editada: MathWorks Support Team
el 21 de Oct. de 2024
MATLAB uses 1-based indexing, so the first element is accessed with index 1, not 0. This means array indices must be positive integers. For example, access the first element of a vector "A".
A = [10, 9, 8, 7, 6]; Aval = A(1) % This returns 10. Meanwhile, A(0) returns an error.
MATLAB also accepts logical indexing, which can be useful when working with conditional statements. For example, say you want to know the values of "A" that is larger than 7. Use the > operator to return a logical array whose elements are logical 1 when an element in "A" is larger than 7; and logical 0 otherwise.
ind = A > 7
Now that you know the locations of the elements meeting the condition, you can access the individual values that are larger than 7 by using "ind" as the index array.
Avals = A(ind)
For more information, see Array Indexing.
As others have mentioned, without providing the full code that caused the error, it is hard to pinpoint why this error occurs. For example, you can have a variable named "sum = 2" sitting in your workspace. So, when you try to call the built-in "sum" function to find the sum of array elements, such as "sum(0.1:0.1:1)", the existing variable is overshadowing it. The error occurs because MATLAB cannot extract the element with index 0.1, 0.2, and so on. To avoid this error, you should not use a variable name that conflicts with MATLAB built-in functions.
2 comentarios
Más respuestas (11)
Roya Kakar
el 19 de En. de 2022
Please answer my quesion. I get the following error after running this line of code
sm =sum(table2array(X));
Array indices must be positive integers or logical values.
henry espinoza
el 19 de Jun. de 2020
Editada: Walter Roberson
el 21 de Nov. de 2024
Array indices must be positive integers or logical values.
Error in Tarea3_HE (line 15)
Xc1 = (-1 /(PRM(u)*(w)))*j
1 comentario
Walter Roberson
el 21 de Nov. de 2024
Your variable u contains zeros or negative values or fractions.
You might have thought that PRM is a function, but at that point in the code it is a variable.
laura celis
el 12 de Mzo. de 2022
Editada: Walter Roberson
el 21 de Nov. de 2024
tengo este error
Array indices must be positive integers or logical values.
Error in recortedesenales (line 38)
z1= z(A:B);
1 comentario
Walter Roberson
el 21 de Nov. de 2024
Your variable A(1,1) is negative, or zero, or contains fractions, and B(1,1) is greater than A(1,1) .
B(1,1) is not necessarily an integer and is not necessarily an integer-value more than A(1,1). For example, 3:5.1 is a valid indexing vector because it expands to [3 4 5]
Roshan Siwakoti
el 21 de Oct. de 2022
a(i) =((delta_p2)/(2*tau(j)*(t_traj(j+1)-t_traj(j))))-((delta_p1)/(2*tau(j)*(t_traj(j)-t_traj(j-1))))
Array indices must be positive integers or logical values
3 comentarios
Kelly Kearney
el 21 de Oct. de 2022
Possible guess... Is a(i) supposed to be a(j)? If you haven't otherwise set i to be an integer, it defaults to sqrt(-1), which cannot be used as an index. Same goes for j; check that it's an integer. Best practices in Matlab recommend avoiding i or j as indices for this very reason.
Walter Roberson
el 21 de Oct. de 2022
I predict that your j is 1 so j-1 is 0 which is not a valid index.
Ferlyn
el 14 de Nov. de 2022
Editada: Walter Roberson
el 21 de Nov. de 2024
% c. x(t) = 3*exp(-2*t)u(t)
subplot (4,2,3)
y = 3*exp(-2*t).*(u(t));
plot (t,y,'-g','Linewidth', 2);
axis ([-1 5 -1 5]);
grid on;
xlabel ('t in sec');
ylabel ('x(t)');
title ('plot of (c)');
Here's my code and it says "Array indices must be positive integers or logical values". What should I do to fix it? Thank you.
1 comentario
Walter Roberson
el 21 de Nov. de 2024
I predict that u is a vector and t is not positive integers.
It is common for u(t) to intend to refer to the unit step function applied to t -- but it is also common for u to be defined as a vector. For example you might have
u = t>=0;
and if you had that and then tried u(t) it would be an attempt to index the vector u at locations given by t, but t probably starts with 0.
Nerea Espinosa Giralt
el 19 de Nov. de 2022
Hola!
Tengo este error." Array indices must be positive integers or logical values."
Atenuacion=(k*R_001^alfa*u(p))*(R_001)^(0.38/Ls-0.25/(1+n(p/0.01)^(-0.36)*longitud^m)*Ls);
¿podriais ayudarme por favor?
Gracias,
Saludos
3 comentarios
Image Analyst
el 19 de Nov. de 2022
You have u(p) which is u(0.05). There is no 0.05'th element of a matrix. Only the 1st element, 2nd element, 3rd element, etc.
Steven Lord
el 19 de Nov. de 2022
The section of code with n(p/0.01) also looks like it might be missing a multiplication operator. If it isn't and you intended p/0.01 to give you an integer value you can use as an index into n, you may encounter difficulties due to floating point arithmetic. Even if that expression did happen to give an integer value (or you called round to force it to give an integer value) as written unless that integer value is 1 you're asking for an element of n that doesn't exist.
x = 1;
y = x(2) % There is no second element
Fariha
el 12 de Ag. de 2023
Editada: Walter Roberson
el 21 de Nov. de 2024
Hello, my code:
yoy(j,1)=100*(log(Real_Price(ii,1))-log(Real_Price(ii-4,1)));
every time I run it says '
Array indices must be positive integers or logical values.
'
can you help me understand my problem?
1 comentario
DGM
el 12 de Ag. de 2023
MATLAB sees that you're trying to index into an array. Array indices must be real positive (nonzero) integers (or logicals). So:
Is the index jj a real, positive integer?
Is the index ii a real, positive integer?
Is ii>4 so that ii-4 is always positive?
Is log() still interpreted as a function, or have you created a variable with the same name?
Maximilian
el 18 de Ag. de 2023
Hello, why am I getting the "Array indices must be positive integers or logical values." error for the code below. I cleared the workspace so that shouldn't be a problem.
Vi = 0;
Vf = 1;
R = 1000;
C = 10 * 10^-6;
t = linspace(0,0.1,1001);
r = R*C;
V(t) = Vi + (Vf-Vi).*(1 - (2.72 .^(-t/r)));
Plot(t,V(t))
2 comentarios
Walter Roberson
el 19 de Ag. de 2023
It is common to get confused between a formula and an array
A line such as
V(t) = Vi + (Vf-Vi).*(1 - (2.72 .^(-t/r)));
is typically intended to be a formula -- the left and right t are intended to refer to values
MATLAB does support creating formulas with that syntax, but only using the Symbolic Toolbox, such as
syms t
V(t) = Vi + (Vf-Vi).*(1 - (2.72 .^(-t/r)));
MATLAB would then refer to V as a "symbolic function".
MATLAB main support for creating formulas is for numeric formulas (that do not typically involve the Symbolic Toolbox). Those use a different syntax called "anonymous functions" that would look like
V = @(t) Vi + (Vf-Vi).*(1 - (2.72 .^(-t/r)));
In both the case of the "symbolic function" and this "anonymous function", V would become an object describing calculations that would be done if you pass values to the object, such as
V(3.2)
In both the symbolic function and the anonymous function, the t would refer to some indefinite value of t to be given value later, and the symbolic or anonymous function describes how to calculate the output in that future time when you finally get passed a specific value in place of t
But if you are not working with symbolic functions or anonymous functions (or some Object Oriented classes) then () is either for passing parameters to a function, or else for indexing. In the original code
t = linspace(0,0.1,1001);
V(t) = Vi + (Vf-Vi).*(1 - (2.72 .^(-t/r)));
you are working with definite numeric values for t not with a symbolic variable, so the right hand side is calculating definite numeric values, and the left side V(t) would be considered indexing.
Almost any time you have definite numeric values for t, V(t) on the left side of an assignment would be interpreted as an attempt to index. (There are some potential exceptions for Object Oriented classes.)
Lokesh
el 4 de Jun. de 2024
Editada: Walter Roberson
el 21 de Nov. de 2024
Array indices must be positive integers or logical values.
Error in p3 (line 38)
f_triple_prime(i) = (f(i+1) - 3*f(i) + 3*f(i-1) - f(i-2)) / (deta^3);
2 comentarios
Elián
el 21 de Nov. de 2024
Editada: Walter Roberson
el 21 de Nov. de 2024
clear,
close all;
A=[1,9,1];
B=[2,4,7];
C=[10,-1,5];
paso=1;
retardo=0.01;
cont=1;
t=0:0.1:1;
x1=t*B(1)+(1-t)*A(1);
y1=t*B(2)+(1-t)*A(2);
z1=t*B(3)+(1-t)*A(3);
plot3(x1(1),y1(1),z1(1),'*r')
xlabel('eje x')
ylabel('eje y')
zlabel('eje z')
axis([-5 5 -5 5 -20 20])
title ('Trayectoria de una particula en el espacio (Grafica Elian Sayas)')
hold on
grid on
H=scatter3(x1(1),y1(1),z1(1),200, 'MarkerEdgeColor', 'k', 'LineWidth' ,3);
tiempo=['t=' ,num2str(t(1)), 's'];
Ht=text (A(1)+1, A(2)+1, A(3)+1, tiempo, 'Color', 'b', 'FontSize', 12);
for i= 2:paso:length(t)
delete(H),
delete(Ht),
T=plot3([x1 (i-1) x1(1)], [y1(1-1) y1(i)], [z1(i-1) z1(i)], 'b', 'LineWidth',1.2);
H=scatter3(x1(1) ,y1(i),z1(i),200, 'MarkerEdgeColor' , 'k', 'LineWidth',3);
tiempo=['t=',num2str(t(i)),'s'];
Ht=text(A(1)+1,A(1)+1,A(1)+1, tiempo, 'Color', 'b', 'FontSize',12);
F(cont)=getframe(gcf);
cont=cont+1;
pause(retardo)
end
hold on
t2=1:0.1:2;
x2=(2-t2) *B(1)+(t2-1) *C(1) ; y2=(2-t2) *B(2)+(t2-1) *C(2);
z2= (2-t2) *B(3)+(t2-1)*C(3) ;
plot3(x2(1), y2(1), z2(1), '*y')
for i= 2: paso: length(t2)
delete (H) , delete(Ht) ,
T=plot3([x2(i-1) x2(i)],[y2(i-1) y2(1)],[z2(1-1) z2(i)],'b' ,'LineWidth',1.
H=scatter3(x2(i),y2(i),z2(i),200,'MarkerEdgeColor','k','LineWidth' ,3);
tiempo=['t2=',num2str(t2(i)),'s'];
Ht=text(A(1)+1,A(1)+1,A(1)+1,tiempo,'Color','b','FontSize',12);
F(cont)=getframe(gcf) ; cont=cont+1;
pause(retardo)
end
hold on
t3=2:0.1:20;
x3=cos(t3-2)-1+C(1);
y3=sin(t3-2)+C(2);
z3=t3-2+C(3);
plot3(x3(1),y3(1),z3(1),'*m');
for i= 2: paso: length(t3)
delete(H),
delete(Ht),
T=plot3([x3(i-1) x3(1)],[y3(1-1) y3(1)],[z3(i-1) z3(1)],'b', 'LineWidth',1.2);
end
video-Videowriter('ANIMACION_TRAYECTORIA', 'Uncompressed AVI');
open(video)
writeVideo(video,F)
close(video)
1 comentario
Walter Roberson
el 21 de Nov. de 2024
Your line
T=plot3([x2(i-1) x2(i)],[y2(i-1) y2(1)],[z2(1-1) z2(i)],'b' ,'LineWidth',1.
is missing a ")" at least. it was probably intended to be
T=plot3([x2(i-1) x2(i)],[y2(i-1) y2(1)],[z2(1-1) z2(i)],'b' ,'LineWidth',1.2);
Astrid
el 3 de Abr. de 2025
Editada: Walter Roberson
el 3 de Abr. de 2025
%apartir de codiciones actualizar los valores
h2b=zeros(1,length(nh));
for i=1:length(nh)
if i==3 % cuando n=0
h2b = 0.9975;
elseif i==4% cuando n=1
h2b(i)=-0.9975+0.995.*h2b(i-1);
elseif i<4
h2b(i)=0.995.*h2b(i-1);
end
end
1 comentario
Walter Roberson
el 3 de Abr. de 2025
Consider the case of i=1 then i==3 is false and i==4 is false, but i<4 is true. So the line
h2b(i)=0.995.*h2b(i-1);
is to be executed. But when i=1 then that would be
h2b(1)=0.995.*h2b(1-1);
which would be
h2b(1)=0.995.*h2b(0);
which is a problem because indices must be positive integers and cannot be 0.
Notice by the way that the for loop has no work to do if i is greater than 4. In such a case you might as well code it without a loop, such as
L = length(nh);
if L >= 1
h2b(1) = something;
end
if L >= 2
h2b(2)=0.995.*h2b(1);
end
if L >= 3
h2b(3) = 0.9975;
end
if L >= 4
h2b(4)=-0.9975+0.995.*h2b(3);
end
This can potentially be reduced a fair bit if it is certain that nh will be at least a certain number of elements. For example if it is certain that length(nh) will be at least 4 then the code reduces to
h2b(1) = something;
h2b(2)=0.995.*h2b(1);
h2b(3) = 0.9975;
h2b(4)=-0.9975+0.995.*h2b(3);
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