MATLAB: Create a matrix with a for loop

matrix

Hey there!

I am attempting to create a matrix which has n, s, k, and i as columns. However, I keep getting this error no matter how I try to make the matrix:

Subscript indices must either be real positive integers or logicals.

Is there a way around this problem considering that I need to make a list that comprises of numbers that aren't positive integers?

 for i = -3:.5:2 
    k = 10^i;
    n = (100 / k);
    s = sqrt((-.5^2 .* ((100 / k)-(.5.*(100 / k)))./(100-(100 / k)))./((((100 / k)-(.5.*(100 / k)))./(100-(100 / k)))-1));
    if (0<s) && (s<=1) 
        s = s;
    elseif s > 1 || (isnan(s)==1)
        s = 1;
    elseif isreal(s) == 1
        s = 0;
    else
        s = 0;
    end
    RTIlogk{i} = [i k n s]; 
end

Best Answer

Yes. Create a vector from your original ‘i’, then refer to it by subscript.

This works:

iv = -3:.5:2;
for i = 1:numel(iv)
    k = 10^iv(i);
    n = (100 / k);
    s = sqrt((-.5^2 .* ((100 / k)-(.5.*(100 / k)))./(100-(100 / k)))./((((100 / k)-(.5.*(100 / k)))./(100-(100 / k)))-1));
    if (0<s) && (s<=1)
        s = s;
    elseif s > 1 || (isnan(s)==1)
        s = 1;
    elseif isreal(s) == 1
        s = 0;
    else
        s = 0;
    end
    RTIlogk{i} = [i k n s];
end

Related Solutions

MATLAB: Turning imaginary numbers into real numbers

To get rid of that, your code can be improved as follows:

   for k = 1:1:100
      n = (100 / k);
      s =  sqrt((((.25*(n-(.5*(100-n))))/(n))-.25)/(-((n - .5*(100-n)))/(n)));
      if (0<s) && (s<=1) 
          s = s;
      elseif s > 1 
          s = 1;
      elseif (real(s)>=0) && imag(s)~=0
          s = 1;
      end
      RTDlineark{k} = [k]; 
      RTDlinearkn{k} = [n];
      RTDlinearks{k} = [s];
  end

MATLAB: Create a matrix with a for loop

Hi Colin

this is John BG jgb2012@sky.com

is there need for the for loop?

i = -3:.5:2;
k = 10.^i;
n = (100 ./ k);
s = sqrt((-.5^2 .* ((100 ./ k)-(.5.*(100 ./ k)))./(100-(100 ./ k)))./((((100 ./ k)-(.5.*(100 ./ k)))./(100-(100 ./ k)))-1));

NaNs to 1

     s(find(isnan(s)))=1

abs(s)>1 to 1, instead of

if s >1

get the indices of s meeting constrain abs(s)>1 with command find

     s(find(s(abs(s)>1)))=1

if you want the previous 2 constraints combined then, instead of

      if s > 1  % || (isnan(s)==1)
      s = 1;

use

        intersect(find(s(abs(s)>1), find(isnan(abs(s)))

5. instead of

          isreal(s)

use

     s(find(imag(s)==0))=0

instead of

    if (s>0) && (s<=1) 
          s = s;

use

    intersect( s(abs(s)>0),  s(abs(s)<=1))

but wouldn't it be just fine to leave these remaining as they are?

the last

else
      s = 0;

would zero useful data, omitted

so all together

      i = -3:.5:2;
  k = 10.^i;
  n = (100 ./ k);
  s = sqrt((-.5^2 .* ((100 ./ k)-(.5.*(100 ./ k)))./(100-(100 ./ k)))./((((100 ./ k)-(.5.*(100 ./ k)))./(100-(100 ./ k)))-1));
     s(find(isnan(s)))=1         % NaNs to 1
     s(find(s(abs(s)>1)))=1   % abs(s)>1
     s(find(imag(s)==0))=0  % zero reals

for the generation of RT do you want to use

        [i k n abs(s)]

   RTIlogk=   [i' k' n' abs(s)']
     RTIlogk =
     1.0e+05 *
    Columns 1 through 3
    -0.000030000000000   0.000000010000000   1.000000000000000
    -0.000025000000000   0.000000031622777   0.316227766016838
    -0.000020000000000   0.000000100000000   0.100000000000000
    -0.000015000000000   0.000000316227766   0.031622776601684
    -0.000010000000000   0.000001000000000   0.010000000000000
    -0.000005000000000   0.000003162277660   0.003162277660168
                     0   0.000010000000000   0.001000000000000
     0.000005000000000   0.000031622776602   0.000316227766017
     0.000010000000000   0.000100000000000   0.000100000000000
     0.000015000000000   0.000316227766017   0.000031622776602
     0.000020000000000   0.001000000000000   0.000010000000000
    Column 4
     0.000002887714078
     0.000002889799069
     0.000002896422232
     0.000002917670114
     0.000002988071523
     0.000003249532852
                     0
                     0
                     0
                     0
                     0

if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:jgb2012@sky.com jgb2012@sky.com>

Best Answer

Related Solutions

MATLAB: Turning imaginary numbers into real numbers

MATLAB: Create a matrix with a for loop

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