# Why do I get the error 'Subscript indices must either be real positive integers or logicals. ' ?

2,605 views (last 30 days)
Why do I get the following error message:
ERROR: Subscript indices must either be real positive integers or logicals.

MathWorks Support Team on 17 Feb 2021
Edited: MathWorks Support Team on 17 Feb 2021
This error occurs when you attempt to index into an array using indices that are not positive integers or logical values. Here are some tips for common situations that cause this error message:
1) Double check that your indices are positive integers. Indices in MATLAB cannot be 0, and by default, start with 1.
2) If you are using logical indexing to index into an array, be sure that your index array is of type 'logical', and not a 'double' array of 1s and 0s. You can convert a 'double' array to a logical array before attempting to use logical indexing. For example:
A = [1 2 3 4; 5 6 7 8];
ind_double = [0 1 0 1; 0 1 0 1];
ind_logical = logical(ind_double);
A(ind_logical)
For an index array 'ind', you can check its data type using the 'whos' function:
whos ind
3) If you use floating-point arithmetic to compute an index array, then the array values may not be exact integers. The 'round' function is handy when you know you have an index value that is nearly the integer index that you want. For example,
A = [1 2 3 4; 5 6 7 8];
ind_float = 2.00001;
ind_int = round(ind_float);
A(ind_float)
Below is a way to check if an index array 'ind'' contains exact integer values. This command returns a 'logical' array, where 1 indicates the index value is an exact integer, and 0 indicates it is not.
ind == round(ind)
4) If you assign a variable to the same name as a built-in function in MATLAB, then you will overwrite that function and encounter the error when you attempt to call it. For example,
max = rand(5);
A = rand(5);
max(A)
In this event, rename your variable and clear the old one to proceed:
B = max;
clear max max(A)
For more information on indexing in MATLAB, see the following documentation page:
Walter Roberson on 25 Sep 2018
"Has this been changed in a recent update of Matlab? (I updated from 2013 to 2017). Before that I never had to convert from double to logical."
R2013a:
>> A=rand(1,3), A([1 0 0])
A =
0.8147 0.9058 0.1270
Subscript indices must either be real positive integers or logicals.
Same with R2010bSP1, which is the oldest MATLAB I happen to have installed at the moment.
If we examine the R14 documentation, https://www.mathworks.com/help/releases/R14/techdoc/matlab.html in the document for " Special Characters [ ] ( ) {} = ' . ... , ; : % ! @ " we see,
"The components of V must be integers to be used as subscripts. An error occurs if any such subscript is less than 1 or greater than the size of X."
This is not a new restriction, and No, there is no setting in MATLAB that could change this.

Antoine Pichot on 19 Nov 2015
Iaredi Sabinas's comment should be a valid answer.
It may happen when you have a variable named after an existing matlab function, like min or diff. Matlab thinks you are using the variable instead of the builtin function.
Stephen Cobeldick on 19 Nov 2015
It is already part of the answer: "Another common cause is that a variable has overwritten a function name and is thus shadowing the function. For example:..." and it then proceeds to give an example of how this can happen.

Schrecklich Er on 18 Mar 2017
Edited: Walter Roberson on 4 Apr 2017
I got this message while I was trying process a image using 'imread', i had the following structure:
for i = 1 : f
for j = 1 : c
B(i,j)=([A1(i,j)*A2(i,j)]);
B1(i,j)=(A2(i,j)/A1(i,j));
end
end
i got the same error message so until 30 minutes of research I just switch the letter 'i' for a 'k' and the error message disapeared, I think that the error was there because the letter 'i' is used for imaginary numbers.
just a little hint! Hope it is useful.

Farah Nadiah on 22 Apr 2016
how about this. the word that have star...how can i declare..if i run this prgoram it get error..thnx a lot
for i=0:1: maxrow-1
for j=0:1: maxcol-1
% inv([i, j]) = 255 - image([i, j]);
for k = 0: 1
*sto([i, j, k]) = image([i, j]);*
end
end
end
Walter Roberson on 22 Apr 2016
MATLAB indexing starts at 1, not at 0. You need to add 1 to all of your indices.
Also remember that sto([i, j, k]) is indexing sto at 3 locations, sto(i), sto(j), sto(k). It is not an index into a 3 dimensional array: that would be sto(i, j, k)

Pratyush Lohumi on 18 Mar 2017
Edited: Walter Roberson on 18 Mar 2017
for n = 0:ns
s(n) = (ns-n)/ns; %slip
Tmech(n) = ph*V1eq^2*R2/((s(n)*omegas)*((R1eq + R2/s(n))^2+(X1+X2)^2)); %Electromechanical torque
end %End of slip loop
Error: Subscript indices must either be real positive integers or logicals. (Line 2)
Query: Couldn't seem to rectify the mistake?
If someone could provide a valid explanation or a corrected code for this loop, that would really help my project.
Imen BOUGRINE on 4 Apr 2017
for m = 1:num_pulse_int
%Update sensor and target positions
[sensorpos,sensorvel] = sensormotion(1/prf);
[tgtpos,tgtvel] = tgtmotion(1/prf);

padmini kutturu on 26 Apr 2017
Edited: Walter Roberson on 26 Apr 2017
for j=1:n
y(j)= (T(j)-Ts)/(Tb-Ts);
h(j)=(1/427)*(-0.0717)*L*(Tb-Ts)*(y(j+1)-y(j-1)/2*deltas);
r(j)=(e*sigma*L^2*P*Ts^3)/(kAc)*(((T(j)/Ts)^3)+((T(j)/Ts)^2)+(T(j)/Ts)+1);
c(j)=1-(h(j)*deltas/2);
a(j)=-2+(r(j)*deltas^2);
b(j)=1+(h(j)*deltas/2);
end
Walter Roberson on 25 Sep 2018
Ram: you had a k-5 subscript with k starting at 1. You need to start k at 6 or more.

Somenone can help me. I got this erro when using this code below:
W2=[];%will contain watermark signal extracted from the image
for t=1:wmsz
W2 = [W2(D_w(IND(t,1),IND(t,2))/D(IND(t,1),IND(t,2)))*10] %watermark extraction
end
Walter Roberson on 4 Oct 2017
Let us match brackets. The number will be the count of open brackets "after" the character aligned with:
W2 = [W2(D_w(IND(t,1),IND(t,2))/D(IND(t,1),IND(t,2)))*10]
1 2 3 4 3 4 32 3 4 3 4 321 0
We see from this that W2 is being indexed by
D_w(IND(t,1),IND(t,2)) / D(IND(t,1),IND(t,2))
1 2 1 2 10 1 2 1 2 10
which contains a division. So for the index into W2 to be an integer, D_w(IND(t,1),IND(t,2)) would have to be an non-zero integer exact integer multiple of D(IND(t,1),IND(t,2)) . That condition is not impossible, but it is something I would tend to doubt.

Yago Veloso on 6 Oct 2017
Edited: Walter Roberson on 6 Oct 2017
Hi everyone! I get the same error in my code, I'm trying to connect a function to my main code, where the function supply my code with all variables of my system equation solution.
Below is part of my main code where I found this error:
[Y1des]= ANN (Z1,e1,s1,D);
[ ac, ao, at_coluna, kp, r1, r2, L, vj, ro_ar, visc_ar, k_ar, RA, P, t_ar, Tar_e, T_ar_ext, Urel_e, Urel, Uabs_e, cps, ro_p_ap, hparede, cal_lat, cpar, cpl, cpv, ro_p, dp, Ubs, u, Rep1, Nu, hp1, St, Rep2, hw1, e, ae1, aet, X1, t_seg, X, ae, hp, hw, d_xp, tt,z]= variaveis (h, Y1des, hi, uar, t_min);
[G ,u, Pvse, Pve, UAEsat, UAE, He]= prop_ar (ro_ar,uar,ac, ao, at_coluna,t_ar, Tar_e, Urel_e, P);
[vl, mg, mss]= prop_leito (ro_p_ap,ro_p,ro_ar,ac,hi,vj);
%%%Initial condition
Tp(1)= 298.15;
Tar_s(1)= Tar_e - ((1-exp(-X(1)))*(Tar_e-Tp(1)));
tar_s(1) = Tar_s(1)-273.15;
Here is the error message
Subscript indices must either be real positive
integers or logicals.
Error in Dif_finitas_plus_ANN (line 44)
Tar_s(1)= Tar_e -
((1-exp(-X(1)))*(Tar_e-Tp(1)));
Thanks for the help !!
Yago Veloso on 6 Oct 2017
Walter Roberson, thanks for the help! That's what was causing the error.

MarkusP on 5 Feb 2018
Edited: MarkusP on 5 Feb 2018
hello guys, maybe someone of you is able to help me...I tried to fix my problem with the solution above but it wasn`t possible.
I get the same error message if I`m runnung the following skript. The skript is for solving a problem with methods of characteristic.
l=1;
d=0.01;
rho=1000;
f=0.1;
a=1;
p0=2e3;
v0=(p0*2*d/(l*rho*f))^0.5;
n=101;
h=l/(n-1);
v(1:n)=v0;
p(1)=p0;
for i=2:n
p(i)=p(i-1)-f*rho*v0^2*h/(2*d);
end
dt=h/a;
tmax=3;
itmax=tmax/dt;
fhr=f*h/(2*a*d);
for it=1:itmax
t=it*dt;
for i=2:n-1
pa=p(i-1);
pb=p(i+1);
va=v(i-1);
vb=v(i+1);
pc(i)=a/2*((pa+pb)/a+rho*(va-vb)+fhr*(vb*abs(vb)-va*abs(va)));
vc(i)=0.5*((pa-pb)/(a*rho)+va+vb-fhr*(vb*abs(vb)+va*abs(va)));
end
pc(1)=p0;
vb=v(2);
pb=p(2);
vc(1)=vb+(pc(1)-pb)/(a*rho)-fhr*vb*abs(vb);
vc(n)=v0*valve(t);
va=v(n-1);
pa=p(n-1);
pc(n)=pa-rho*a(vc(n)-va)+f*h*rho/(2*d)*va*abs(va); %%error is in this line
vres(it,1:n)=vc(1:n);
pres(it,1:n)=pc(1:n);
p=pc;
v=vc;
end
My valve(t) function looks like this:
function vrel = valve(t)
if t<.1
vrel=1;
else
vrel=exp(-10*(t-.1));
end
Here is the error message:
Subscript indices must either be real positive integers or logicals.
Error in MOCwaterhammer (line 45)
pc(n)=pa-rho*a(vc(n)-va)+f*h*rho/(2*d)*va*abs(va);
Thanks for you help!

Torsten on 5 Feb 2018
Edited: Torsten on 5 Feb 2018
You forgot a multiplication sign:
pc(n)=pa-rho*a->*<-here(vc(n)-va)+f*h*rho/(2*d)*va*abs(va);
Best wishes
Torsten.
MarkusP on 5 Feb 2018
Oh the solution can be so simple...
Thank you so much!

Francisca Belart on 30 Aug 2018
just using round()solved it for me.Thanks sooo much!

Hamza saeed khan on 21 Dec 2018
I think it may help. I myself remove (t) from c(t) and m(t) as follows:
clc
close all
clear all
%let values we asigned as followed
A1=4
A2=2
f1=5
f2=6
t=0.1:0.001:1
c=20*A1*cos(2*pi)*f1
m=12*A2*cos(2*pi)*f2
y=c*m
plot(y,t,'r')
xlabel('c(t)')
ylabel('m(t)')
grid on
Best wishes Hamza Pakistani lol

¿How I put a range in for loop that starts in a negative number?
Something like that
for i=-10000:100:10000
delta_g(i,1)=((4*R1^3*sp*G)/3)*(z1/(i^2+z1^2)^(3/2));
end
Salu2
Stephen Cobeldick on 8 Oct 2019
Loop over indices, rather than looping over data.

Jamal Nasir on 31 Mar 2020
this error when a negative number or real number used as indecies of matrix

Sofia Santos on 8 May 2020
Hello, I tried using round() and convert double to logical, bue when I type "whos maximum" or "whos minimum" nothing is show. I want to now the time in minimum and the time in maximum for calculate de rise time.
Can you look ate my code and see if you can help? Thank you very much!!
data_org = importdata(sinais,'\t',7); %text file with 7 columms
EDA=data_org.data;
EDA= EDA(:,6); %I just want the values in the 6th column
T=N/fs;
t = (1:N)/fs;
minimum= min(EDAuS_filter);
maximum= max(EDAuS_filter);
amp=maximo-minimo; %signal amplitude
tmin=t(minimo);
tmax=t(maximo);
rise_time= tmax-tmin;
Sofia Santos on 8 May 2020
I'm not executing this in the context of a function. The idea is to create a script that reads the six column of each text file and then filter that signal and get the amplitude and time rise. But I'm getting troubles to find the corresponding t.

Seahawkgo on 27 May 2020
Edited: Walter Roberson on 27 May 2020
"Subscript indices must either be real positive integers or logicals."
Error in Chanprojectpart32 (line 36)
E(Zr)=mean(Zr);
=====================
for j=1:N-1
s1 = 0;
s2 = 0;
for k=1:No+1
Phi=(pi-(-pi))*rand-pi;%Phi-k
Fk = Fd.*(cos((2*pi*k)/M)); %Fk
Bk = pi/(No+1)*k; %k(pi/No+1)
..........
.........
y6 = cos(2*pi*Fd*t + Phi);
s1 = s1 + y1*y2;
s2 = s2 + y5*y2;
end
Zr = NN*s1 + MM*(y3*y6);
Zi = NN*s2 + MM*(y4*y6);
end
E(Zr)=mean(Zr);
E(Zi)=mean(Zi);
##### 2 CommentsShowHide 1 older comment
Seahawkgo on 30 May 2020
Thanks

Viktoriia Buliuk on 3 Aug 2020
Hello! I try to perform the transforn of obtained data from .csv file. Could you, please, help me. I have the following errors:
Subscript indices must either be real positive integers or logicals.
Error in walvet>@(tau)hh(tau)*(2/(3^(1/2)*pi^(1/4))*exp(-((tau-ttt(j))/a(i))^2/2)*(1-((tau-ttt(j))/a(i))^2))
Error in integralCalc/iterateArrayValued (line 156)
fxj = FUN(t(1)).*w(1);
[q,errbnd] = iterateArrayValued(u,tinterval,pathlen);
Error in integralCalc (line 103)
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in walvet (line 26)
Q = integral(F,-Inf,Inf, 'ArrayValued', true);
And the code is follofing:
type wet.csv
Test = importdata('wet.csv');
t = Test(:, 1);
h = Test(:, 2);
tt = max(t);
tr = min(t);
ttt = tr:4.0003*10^-11:tt;
hh = interp1(t, h, ttt);
figure(2);
plot(t, h, 'ro');
hold on;
plot(ttt, hh, 'g');
a = 0.3*10^-10:10^-11:10^-9;
for i = 1:length(a)
for j = 1:length(ttt)
F = @(tau) hh(tau)*(2/(3^(1/2)*pi^(1/4))*exp(-((tau-ttt(j))/a(i))^2/2)*(1 - ((tau-ttt(j))/a(i))^2));
Q = integral(F,-Inf,Inf, 'ArrayValued', true);
K = (1/(a(i)^(1/2)))*Q;
S = real(K);
end
end
contour (ttt, hh, a, S, 500)
Thank you very much!!!
##### 2 CommentsShowHide 1 older comment
Viktoriia Buliuk on 11 Aug 2020
Thanks a lot!
I defined hh = @(t_per) interp1(t, h, t_per)

Destaw Masresha on 13 Dec 2020
Edited: Walter Roberson on 13 Dec 2020
A=sort(fc);
B=A((c*d-n+1):c*d); % selete the top n of the variance
Subscript indices must either be real positive integers or logicals
Walter Roberson on 13 Dec 2020
What are c and d ? What is n ?
I speculate that you constructed c and d such that you expect c*d to always be an integer, but that it is not always exactly an integer. For example if c were 0.1:0.1:1 and d were 10, then you might expect that c*d is always an integer, but that would be false:
format long g
c = 0.1 : 0.1 : 1;
mat2str(c)
ans = '[0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1]'
d = 10;
mat2str(c * d)
ans = '[1 2 3 4 5 6 7 8 9 10]'
mat2str(c * d - (1:10))
ans = '[0 0 4.44089209850063e-16 0 0 0 0 0 0 0]'
c(3) - 0.3
ans =
5.55111512312578e-17
Lessons:
• 0.1 is not exactly representable in double precision
• multiplying a double precision number that is "mathematically" a fraction, by something that "mathematically" give you back an integer... doesn't always give you back an integer. 0.3 * 10 does not give 3
• numbers calculated by fractional increments on colon operators, such as the third value of 0.1:0.1:1, are not always exactly the same as the corresponding literal value you would expect. 0.1+0.1+0.1 does not give you the same value as you would get from writing 0.3

common fernando on 5 Apr 2021
Edited: Walter Roberson on 5 Apr 2021
hello guys, maybe someone of you is able to help me...I tried to fix my problem with the solution above but it wasn`t possible.
Why do I get the following error message:
Subscript indices must either be real positive integers or logicals.
Error in threeDFT (line 33)
x=v((j-1:j+N0-2)*dt);
code:
fs=50*512;
dt =1/fs;
N0=fs/50
tmax=1;
j_max=tmax*fs;
for j=1:j_max+1
x=v((j-1:j+N0-2)*dt);
Walter Roberson on 7 Apr 2021
The code you posted before was three months ago; I would think that you have developed the code further since that time.

### Tags

No tags entered yet.

### Community Treasure Hunt

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

Start Hunting!