Find returns empty with inconsistent size

I have a variable x, that is a vector with a variable number of elements. When applying the function

find(x>0), for:

find([-1 -1 -1 -1 -1 -1]>0)
ans = 
  1×0 empty double row vector

find([-1 -1]>0)
ans =
  1×0 empty double row vector

find([-1]>0)
ans =
     []
     

The problem I have is that when x has only one element, and the size of the output is not consistent. For instance, I would like to get an empty with the same size as "1×0 empty double row vector".

Any suggestions?

Answer 1

[Walter Roberson]

find([-1 -1 -1 -1 -1 -1]>0)

ans =

  1×0 empty double row vector

find([-1 -1 -1 -1 -1 -1].'>0)

ans =

  0×1 empty double column vector

find([-1 -1]>0)

ans =

  1×0 empty double row vector

find([-1 -1].'>0)

ans =

  0×1 empty double column vector

find([-1]>0)

ans =

     []

size(ans)

ans = 1×2

0 0

So when there is 1 row, the result has 0 columns. When there is 1 column, the result has 0 rows. When there is 1 row and 1 column (scalar), the result has 0 rows and 0 columns.

[Rub Ron]

How can I make that when x is a scalar, my the empty output has is 1x0 size?

[Walter Roberson]

There are no options to find() that can make that behaviour happen. You can, however, do

idx = find([-1]>0);
if isempty(idx); idx = zeros(1,0); end

Answer 2

[Awais Saeed]

This command is for vectors and matrices. find(X > 0) returns the indices of the array X where elements are greater than zero. If none is found, it returns an empty matrix. Now using this command on scalers ( [-1] in your case) does not make any sense.

However, if you want 1×0 empty double row vector then you have to check whether X is a scaler or not. If X is a scaler, then do

X = -1:-2 % this will be an empty vector and will give you 1×0 empty double row vector

Use

isscalar()

yo check if X is a scaler ot not

[Rub Ron]

are you aware of an alternative function to find, that when the output is empty, at leat it provide 1x0 or 0x1 for the empty output?

[Awais Saeed]

No there is not. To get 1x0 empty double row for scaler values, just check if they are scalers or not. If they are, then by doing

j : k % where j > k, will give you 1x0 empty double row vector

Here is how you do it

x = [-1 -1 -1 -1 -1 -1]; % check with x = 0; x = -1 etc. as well
if (isscalar(x))
    result = x : x - 1
else
    result = find(x>0)
end

What Walter Roberson suggested will also work for you.

[Walter Roberson]

x = [-1]

x = -1

if (isscalar(x))
    result = x : x - 1
else
    result = find(x>0)
end 

result =

  1×0 empty double row vector

x = [+1]

x = 1

if (isscalar(x))
    result = x : x - 1
else
    result = find(x>0)
end

result =

  1×0 empty double row vector

x = [-inf]

x = -Inf

if (isscalar(x))
    result = x : x - 1
else
    result = find(x>0)
end 

result = NaN

In that second case, result should not be empty. In the third case, result should be 1x0 not NaN. Your code x:x-1 is presuming an outcome and is using odd operators to try to enforce it.