Create a matrix with a for loop

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

[per isakson]

Replace

isreal(s) == 1

by

isreal(s)

[John BG]

this doesn't solve the problem

[per isakson]

No but it's still a relevant comment.

[John BG]

the initial s is as follows:

s =
  Column 1
  0.000000000000000 + 0.288771407778945i
  Column 2
  0.000000000000000 + 0.288979906876575i
  Column 3
  0.000000000000000 + 0.289642223181746i
  Column 4
  0.000000000000000 + 0.291767011359047i
  Column 5
  0.000000000000000 + 0.298807152333598i
  Column 6
  0.000000000000000 + 0.324953285195147i
  Column 7
                NaN + 0.000000000000000i
  Column 8
  0.274222586190323 + 0.000000000000000i
  Column 9
  0.121267812518166 + 0.000000000000000i
  Column 10
  0.064418039894926 + 0.000000000000000i
  Column 11
  0.035623524993955 + 0.000000000000000i

if you decide to apply isreal(s), look what happens:

   isreal(s)
  ans =
    logical
     0

despite some of the values of s are clearly real, applying isreal() does not return those with null imaginary.

Instead, look for those components of s that have null imag(s), please have a look at my answer, thanks in advance

John BG

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Answer 1

[John BG]

Hi Colin

this is John BG [e-mail]

1.

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));

2.

NaNs to 1

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

3.

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

4.

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

6.

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?

7.

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)]

or

   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:[e-mail] [e-mail]>

[Walter Roberson]

Most of those find() are not needed. For example,

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

can be

       s(isnan(s)) = 1;

Also, be careful with the order of the statements:

>> abs(3-5i) > 1
ans =
  logical
   1

but 3-5i should probably be replaced by 0 under the ~isreal() rule.

[John BG]

No, why should 3-5i be zeroed?

the rule regarding real is

elseif isreal(s) == 1
        s = 0;

or as you have pointed out

elseif isreal(s)
        s = 0;

I recommend to instead replace this with

elseif imag(s) ==0
        s = 0;

or

elseif imag(s) < err1
        s = 0;

Walter, let's comment the rules supplied in the question, shall we?

1.

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

this to me means if s<0 do nothing, which as you said, since s may or may not be complex, doesn't help much.

Perhaps Colins means

if (real(s)<0) && (abs(s)<=1) 
        s = s;

but so far no comment from the question originator, and in any case,

s=s

same as do nothing.

2.

    elseif s > 1 || (isnan(s)==1)
          s = 1;

Again, let me start with the NaN

NaN+5i or -3+NaN*1j is same as NaN, as long as there's a NaN, doesn't matter whether in real or imaginary parts, the value is NaN, so all NaNs to 1.

I assumed

abs(s)>1

3.

      elseif isreal(s) == 1
          s = 0;

to me this is the same as

if imag(s)==0

then don't care about real(s) and zero it.

4.

    else
          s = 0;

this last else bring everything down to zero, shouldn't be here, why starting any calculation if it all ends up being zeroed?

therefore I encourage the question originator Mr Lynch to add some comments to help decide what answer does qualify to accepted answer.

Answer 2

[Walter Roberson]

Change

      RTIlogk{i} = [i k n s];

to

      RTIlogk{2*i+7} = [i k n s];

Or, better yet, learn this pattern:

ivals = -3:.5:2;
num_i = length(ivals);
RTIlogk = cell(num_i,1);
for i_idx = 1 : num_i
  i = ivals(i_idx);
  ...
  RTIlogk{i_idx} = ...
end

This pattern does not require that the values be equally spaced or even real valued.

[Walter Roberson]

And be careful: the < operator ignores any complex component for the comparison. You should always check for complex before using the relative comparison operators.

[John BG]

And

1.

assert(3>4)
Assertion failed.

as expected throws error if condition false.

2.

assert does nothing if condition true:

assert(3<4)

4.

as Walter mentions above, in the following line operators < > do not compute imaginary numbers, returning false for the real parts only, and in turn assert returns error

assert(1j*3<1j*4)
Assertion failed.

assert(0+1j*3<0+1j*4)
Assertion failed.

assert(3i<4i)
Assertion failed.

5.

However, to further warn about the use of operators < > with figures that have non-null imaginary parts

assert(3i<4)

now assert does nothing, which implies that

3i<4

returns true

and

assert(3i>4)
Assertion failed.

6.

As Walter points out,

avoid using operators

<
>

with numbers that have non null imaginary parts.

apply abs() real() or any other function before using < > to precisely avoid href = ""</a> dealing with complex figures.

Regards

John BG

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