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

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)
[x,cc_x,b_x,betac_x,zc_x,sig_x,t,xc_t,bx_t,betax_t] = Read_Predict('predict');

%   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.

Answer 1

[Star Strider]

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’?

[Summer]

t starts at 1.71 and ends at 5730

[Stephen23]

Then read my answer and you will know why you are getting this error.

[Summer]

Then having another function that implements loops is useless for this case? I know the values of t are the problem but unless I fix it somehow in the original function then there is no other way?

[Star Strider]

I assume ‘xc_t’ and ‘betax_t’ are arrays and not functions. Even though you index ‘t’ in your calling loop, the value of it that you pass to the function is a value, not an index. It has to be an index — probably an index that refers to an element of the actual ‘t’ vector — for it to work in the line that is throwing the error. Indices are always positive integers.

I’m having a difficult time understanding your code, so I cannot suggest a specific fix for the problem.

[Star Strider]

I’m having problems figuring out what you are doing in your code. In the function that is throwing the error, you need to pass the time vector as one argument, perhaps calling it ‘tv’. Then, refer to it (using index ‘k1’ here):

xa=(x-xc_t(tv(k1))+bx_t(tv(k1)))/(sr2*betax_t(tv(k1)));

and if you want to save ‘xa’ as a vector (assuming it is a scalar in your original code):

xa(k1) = (x-xc_t(tv(k1))+bx_t(tv(k1)))/(sr2*betax_t(tv(k1)));

This is most easily done in a for loop.

[Summer]

Values of ‘xc_t’ and ‘betax_t’ are read from the same table containing 't' and since those depend on t, I wanted their values that will be used in any of the computations to be those corresponding to those of 't'. Say for k1, t = 1.71 and values of xc and betax should be those corresponding to this time value.

"Even though you index ‘t’ in your calling loop, the value of it that you pass to the function is a value, not an index. It has to be an index — probably an index that refers to an element of the actual ‘t’ vector — for it to work in the line" That is exactly the case here!

tv(k1) would be a value, right?

[Star Strider]

Good point! (Early morning here when I wrote that.)

Should be:

xa(k1) = (x-xc_t(k1)+bx_t(k1))/(sr2*betax_t(k1));

This would correspond to ‘tv(k1)’ in time and by subscript reference.

[Stephen23]

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]

"xa(k1) = (x-xc_t(k1)+bx_t(k1))/(sr2*betax_t(k1));"

Worked this time! Thanks :)

[Star Strider]

My pleasure!

Answer 2

[Stephen23]

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.

[Summer]

but that's why I created the second function with the loops so that instead of reading t=1.71 it goes like t(1) = 1.71

[Image Analyst]

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 =
     []

[Stephen23]

True, [] simply returns an empty array. I have edited my answer to reflect this.

[Summer]

Thanks, Stephen!

Answer 3

[Luca]

clc; clear;

teta=linspace(0,pi/2,100);
R=1;
g=9.81;

v_quadro = g*R*cos(teta.*2) + R*g(2^0.5-1)-4*R^3*(1-cos(teta))*cos(teta)/((2*R-(2^0.5)*R)^2);

Where is the error, please? Thanks.

[Walter Roberson]

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]

Thanks, it works.

Answer 4

[common fernando]

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)
10. fs=50*512; %sampling freq.
11. dt =1/fs;
12. N0=fs/50; %number of samples/cycle
13. m=50; %no. of cycles
15. t = dt*(0:m*N0); %data window
16. fi=50; %Frequency test
17. ww=wgn(201,1,-40);
18. size(transpose(ww))
19. t =dt*(0:200);
20. y=sin(2*pi*fi*t + 0.3);
21. v=sin(2*pi*fi*t + 0.3)+transpose(ww)
24. tmax=1;
27. n=N0-1:-1:0;
28. f0=50;
29. f=50.88;
31. Hc=2/N0*cos(2*pi*n/N0+pi/N0);
32. Hs=-2/N0*sin(2*pi*n/N0+pi/N0);
33. t_est=[];
34. f_est=[];
35. j_max=tmax*fs;
36. for j=1:j_max+1
37. x=v(((j-1):(j+ N0-2))*(dt));
38. c(j)=x*Hc';
39. s(j)=x*Hs';
40. if(j>N0)
41. Ac(j-N0)=sqrt(sum(c(end-N0+1:end).^2)/N0);
42. As(j-N0)=sqrt(sum(s(end-N0+1:end).^2)/N0);
43. cc(j-N0)=c(end-N0+1:end)*Hc';
44. ss(j-N0)=c(end-N0+1:end)*Hs';
45. if(j>2*N0)
46. Acc(j-2*N0)=sqrt(sum(cc(end-N0+1:end).^2)/N0);
47. Ass(j-2*N0)=sqrt(sum(ss(end-N0+1:end).^2)/N0);
48. ccc(j-2*N0)=cc(end-N0+1:end)*Hc';
49. ccs(j-2*N0)=cc(end-N0+1:end)*Hs';
50. ssc(j-2*N0)=ss(end-N0+1:end)*Hc';
51. sss(j-2*N0)=ss(end-N0+1:end)*Hs';
52. 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);
53. t_est=[t_est;(j-1)*dt];
54. f_est=[f_est;ff];
55. end
56. end
57. end
59. t_est;
60. f_est
62. plot(t_est, f_est,'red')
63. hold on
64. RMSE = sqrt(mean((f_est-fi).^2))
67. xlabel('time')
68. ylabel('frequency')
69. title('3LDFT WHITE NOISE ')
70. plot (t,fi)
71. hold off

[Walter Roberson]

j starts at 1. j-1 is 0. So you are trying to index at 0

[common fernando]

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
3. fs=200; %sampling freq.
4. dt =1/fs;
5. N=fs/200%number of samples/cycle
6. m=6 %no. of cycles
7. t =dt*(0:400); %dt*(0:m*N); %data window
9. fi=50; %Frequency test
11. ww=wgn(201,1,-40);
12. size(transpose(ww))
13. t =dt*(0:200);
14. y=sin(2*pi*fi*t + 0.3);
15. v=sin(2*pi*fi*t + 0.3)+transpose(ww);
17. tmax=1;
19. % v : as function of time
20. % fs : sampling frequency (Hz)
21. % tmax : time of final estimation
22. % to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)),50*512,1)
23. % to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)+randn(1)*.1),50*512,1)
28. dt=1/fs;
29. v_old=v(1);
30. i_max=tmax*fs;
31. if(v_old==0)
32. old_cross=0;
33. else
34. old_cross=-1;
35. end
36. new_cross_detected=0;
37. t_est=[];
38. f_est=[];
39. for i=200:1:400
40. v_new=v(i.*dt);
41. if(v_old.*v_new<0)
42. new_cross=(i-1+v_old/(v_old-v_new))*dt;
43. new_cross_detected=1;
44. elseif(v_new==0)
45. new_cross=i*dt;
46. new_cross_detected=1;
47. end
48. if(new_cross_detected==1)
49. if(old_cross~=-1)
50. t_est=[t_est;i*dt];
51. f_est=[f_est;1/(2*(new_cross-old_cross))];
52. end
53. old_cross=new_cross;
54. new_cross_detected=0;
55. end
56. v_old=v_new;
57. end
58. t_est
59. f_est
60. plot(t_est,f_est,'red')
64. o=rms(fi)
65. c=rms(f_est)
66. RMSE = sqrt(mean(c - o).^2)
71. hold on
72. xlabel('time')
73. ylabel('frequency')
74. title('ZC white noise')
77. plot (t,fi)
78. hold off

[Steven Lord]

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