# "Subscript indices must either be real positive integers or logicals." again! See codes

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Summer on 10 Jan 2015
Commented: Steven Lord on 13 Jun 2020
Hi,
I have created two functions in Matlab. One is supposed to produce a concentration profile for a given location and time, but the current code gives this error "Subscript indices must either be real positive integers or logicals.". The second function, called testing, recalls the first function to do the calculation and uses loops instead, but still the same error is appearing.
1. Concentration profile code
function [c,t] = Disp_Conc(x,y,z,t)
%This function calculates the concentration of a an agent at a particular time and location
% Data generated through Dispersion model (This is a 1x61 vector)
% Time averaged volume concentration: concentration contour parameters
%erf = error function
sr2 = sqrt(2.0);
xa=(x-xc_t(t)+bx_t(t))/(sr2*betax_t(t));
xb = (x-xc_t(t)-bx_t(t))/(sr2*betax_t(t));
ya = (y+b_x(x))/(sr2*betac_x(x));
yb = (y-b_x(x))/(sr2*betac_x(x));
za = (z-zc_x(x))/(sr2*sig_x(x));
zb = (z+zc_x(x))/(sr2*sig_x(x));
c = (cc_x(x) * (erf(xa)-erf(xb)) * (erf(ya)-erf(yb)) * (exp(-za*za)+exp(-zb*zb)));
time=tm(t);
end
**Running this function, I get an error saying "Subscript indices must either be real positive integers or logicals." due to the fact that values of t and x are both fractional (both x and t are read from a 1x61 vector). To overcome this, I created another .m file that recalls the concentration function and uses loops as below:
2. Function "testing":
function [c, t] = testing( x,y,z,t )
%The values of x and t are already being read from a saved 1x61 vector. z and y can vary and must be specified by the user for a given location. The value of z should be set equal to zero for all cases and doesn't change.
for ts=1:length(t)
for xs=1:length(x)
z=0;
for y=0:100; %(k index?)
X=x(ts);
T=t(xs);
[c(ts),time(ts)]=Disp_Conc(X,y,z,T);
end
end
end
end
**Running "testing" for a particular parameters say testing(5,0,0,8.79), I get the below errors:
"Subscript indices must either be real positive integers or logicals.
Error in Disp_Conc (line 12)
xa=(x-xc_t(t)+bx_t(t))/(sr2*betax_t(t));
Error in testing (line 10)
[c(ts),time(ts)]=Disp_Conc(X,y,z,T);"
I don't know what is wrong this time? The original problem was because of indexing and now I still get the same error with the new function.
Any help would be greatly appreciated. Thx.

Star Strider on 10 Jan 2015
In the line that throws the error:
xa=(x-xc_t(t)+bx_t(t))/(sr2*betax_t(t));
what is the value of ‘t’?

Stephen Cobeldick on 10 Jan 2015
Typically there are two methods for looping over non-integer values in MATLAB:
1. derive the values within each loop, using a function or calculating the values from the loop variables.
2. use the loop variables to look up the values in a predefined array.
Simple examples for each of these, where we want the non-integer value 1/k in each iteration:
for k = 1:4
disp(1/k)
end
A = 1./(1:4);
for k = 1:4
disp(A(k))
end
While the second method looks more complicated, it actually has many advantages in MATLAB: it helps memory management , and makes it easier to write vectorized code. Note how the second example uses indexing, and the index value is an integer!
As you have arrays of data, you should consider using the look-up method. You will probably need these function to help you: size, numel.
Summer on 10 Jan 2015
"xa(k1) = (x-xc_t(k1)+bx_t(k1))/(sr2*betax_t(k1));"
Worked this time! Thanks :)
Star Strider on 10 Jan 2015
My pleasure!

Stephen Cobeldick on 10 Jan 2015
Edited: Stephen Cobeldick on 10 Jan 2015
In MATLAB all array indices must be logical or positive numeric integers. This means that the following is permitted:
>> A = [123,456,789];
>> A(2) % 2 is an integer
ans = 456
But the following produces an error:
>> A(3.14159) % 3.14159 is not an integer
Subscript indices must either be real positive integers or logicals.
Note that negative integers and zero are not permitted indices.
You need to review your code and check all of the array indexing. In particular it seems that you are using some data values (e.g. t, x) as indices.

Show 1 older comment
Image Analyst on 10 Jan 2015
Stephen, very good explanation (+ a vote). Since we answer this at least once or twice a day, I don't know why it's not on the FAQ yet. So I used your excellent explanation as the basis for a new FAQ entry: http://matlab.wikia.com/wiki/FAQ#How_do_I_fix_the_error_.22Subscript_indices_must_either_be_real_positive_integers_or_logicals..22.3F
By the way, empty seems to be okay
>> emptyvector = []
emptyvector =
[]
>> A(emptyvector)
ans =
[]
Stephen Cobeldick on 10 Jan 2015
True, [] simply returns an empty array. I have edited my answer to reflect this.
Summer on 10 Jan 2015
Thanks, Stephen!

Luca on 21 Aug 2017
Edited: Walter Roberson on 21 Aug 2017
clc; clear;
teta=linspace(0,pi/2,100);
R=1;
g=9.81;
Where is the error, please? Thanks.

Walter Roberson on 21 Aug 2017
Your g is not a vector, but you are trying to index it at (2^0.5-1).
MATLAB does not have any implicit multiplications. You need to add ".*" or "*" in g(2^0.5-1)
Luca on 22 Aug 2017
Thanks, it works.

christina fayez on 11 Jun 2020
hi guys I do the code for white noise of 3LDFT algorithm but give me error
Subscript indices must either be real positive integers or logicals.
Error in threeDFT (line 37)
x=v(((j-1):(j+ N0-2))*(dt));
1. function [t_est,f_est]=threeDFT(v,fs,tmax,N0)
2. % v : volt as function of time
3. % fs : sampling frequency (Hz)
4. % tmax : time of final estimation
5. % N0 : number of samples in the window
6. % to test: [t,f]=threeDFT(@(t)(220*sin(2*pi*50.1*t+pi/2)),50*512,1,512)
7. % to test: [t,f]=threeDFT(@(t)(220*sin(2*pi*50.1*t+pi/2)+randn(size(t))*.1),50*512,1,512)
8. fs=50*512; %sampling freq.
9. dt =1/fs;
10. N0=fs/50; %number of samples/cycle
11. m=50; %no. of cycles
12. t = dt*(0:m*N0); %data window
13. fi=50; %Frequency test
14. ww=wgn(201,1,-40);
15. size(transpose(ww))
16. t =dt*(0:200);
17. y=sin(2*pi*fi*t + 0.3);
18. v=sin(2*pi*fi*t + 0.3)+transpose(ww)
19. tmax=1;
20. n=N0-1:-1:0;
21. f0=50;
22. f=50.88;
23. Hc=2/N0*cos(2*pi*n/N0+pi/N0);
24. Hs=-2/N0*sin(2*pi*n/N0+pi/N0);
25. t_est=[];
26. f_est=[];
27. j_max=tmax*fs;
28. for j=1:j_max+1
29. x=v(((j-1):(j+ N0-2))*(dt));
30. c(j)=x*Hc';
31. s(j)=x*Hs';
32. if(j>N0)
33. Ac(j-N0)=sqrt(sum(c(end-N0+1:end).^2)/N0);
34. As(j-N0)=sqrt(sum(s(end-N0+1:end).^2)/N0);
35. cc(j-N0)=c(end-N0+1:end)*Hc';
36. ss(j-N0)=c(end-N0+1:end)*Hs';
37. if(j>2*N0)
38. Acc(j-2*N0)=sqrt(sum(cc(end-N0+1:end).^2)/N0);
39. Ass(j-2*N0)=sqrt(sum(ss(end-N0+1:end).^2)/N0);
40. ccc(j-2*N0)=cc(end-N0+1:end)*Hc';
41. ccs(j-2*N0)=cc(end-N0+1:end)*Hs';
42. ssc(j-2*N0)=ss(end-N0+1:end)*Hc';
43. sss(j-2*N0)=ss(end-N0+1:end)*Hs';
44. ff=f0*N0/pi*atan(tan(pi/N0)*((ccc(j-2*N0).^2+ccs(j-2*N0).^2)./(ssc(j-2*N0).^2+sss(j-2*N0).^2)).^.25);
45. t_est=[t_est;(j-1)*dt];
46. f_est=[f_est;ff];
47. end
48. end
49. end
50. t_est;
51. f_est
52. plot(t_est, f_est,'red')
53. hold on
54. RMSE = sqrt(mean((f_est-fi).^2))
55. xlabel('time')
56. ylabel('frequency')
57. title('3LDFT WHITE NOISE ')
58. plot (t,fi)
59. hold off

Walter Roberson on 11 Jun 2020
j starts at 1. j-1 is 0. So you are trying to index at 0
christina fayez on 12 Jun 2020
I have another error
clear all; close all; clc
Attempted to access v(1.005); index must be a positive integer or logical.
Error in code (line 41)
v_new=v(i.*dt);
>>
1. %frequency and amplitude
2. fs=200; %sampling freq.
3. dt =1/fs;
4. N=fs/200%number of samples/cycle
5. m=6 %no. of cycles
6. t =dt*(0:400); %dt*(0:m*N); %data window
7. fi=50; %Frequency test
8. ww=wgn(201,1,-40);
9. size(transpose(ww))
10. t =dt*(0:200);
11. y=sin(2*pi*fi*t + 0.3);
12. v=sin(2*pi*fi*t + 0.3)+transpose(ww);
13. tmax=1;
14. % v : as function of time
15. % fs : sampling frequency (Hz)
16. % tmax : time of final estimation
17. % to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)),50*512,1)
18. % to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)+randn(1)*.1),50*512,1)
19. dt=1/fs;
20. v_old=v(1);
21. i_max=tmax*fs;
22. if(v_old==0)
23. old_cross=0;
24. else
25. old_cross=-1;
26. end
27. new_cross_detected=0;
28. t_est=[];
29. f_est=[];
30. for i=200:1:400
31. v_new=v(i.*dt);
32. if(v_old.*v_new<0)
33. new_cross=(i-1+v_old/(v_old-v_new))*dt;
34. new_cross_detected=1;
35. elseif(v_new==0)
36. new_cross=i*dt;
37. new_cross_detected=1;
38. end
39. if(new_cross_detected==1)
40. if(old_cross~=-1)
41. t_est=[t_est;i*dt];
42. f_est=[f_est;1/(2*(new_cross-old_cross))];
43. end
44. old_cross=new_cross;
45. new_cross_detected=0;
46. end
47. v_old=v_new;
48. end
49. t_est
50. f_est
51. plot(t_est,f_est,'red')
52. o=rms(fi)
53. c=rms(f_est)
54. RMSE = sqrt(mean(c - o).^2)
55. hold on
56. xlabel('time')
57. ylabel('frequency')
58. title('ZC white noise')
59. plot (t,fi)
60. hold off
Steven Lord on 13 Jun 2020
There's no such thing as element 1.005 of an array in MATLAB. If as line 19 states you want v to be a function of t rather than a vector created using a vector of values t you should probably define it as an anonymous function.
v = @(t) sin(2*pi*fi*t + 0.3)+transpose(ww);
If you do this, v(1.005) will not be an attempt to index into a vector but will be an attempt to evaluate the anonymous function when t is 1.005.
As a simpler example that shows the technique:
x = 1:10;
y1 = @(x) x.^2;
y1(5.5) % works, returns (5.5)^2 = 30.25
y2 = x.^2;
y2(5.5) % will not work