ind2sub doesn't work

Check it out:

>> A = zeros(3);
>> A(2,3) = 1 
A =
     0     0     0
     0     0     1
     0     0     0
>> tmp = A'
tmp =
     0     0     0
     0     0     0
     0     1     0
>> [maxVal, maxInd] = max(tmp(:))  
maxVal =
     1
maxInd =
     6

And now the WTF comes:

>> ind2sub(size(tmp), maxInd)
ans =
     6
>> ind2sub(maxInd, size(tmp))
ans =
     3     3

Seems like function is dead.

Answer 1

[Alfonso Nieto-Castanon]

You are funny.

ind2sub reshapes the siz first argument to match the number of output variables used (in your case a single output argument), so that the resulting subscripts suffice to fully index the original elements. In your case:

 ind2sub(size(tmp), maxInd)

is just the same as:

 ind2sub(numel(tmp), maxInd)

[Egor]

thanks! I am really funny :)

Answer 2

[Matt J]

>> [i,j]=ind2sub(size(tmp), maxInd)

i =

       3

j =

       2

[Egor]

thanks!

Answer 3

[James Tursa]

The typical use of ind2sub in your case is to request two outputs:

[I J] = ind2sub(size(tmp), maxInd)
I =
     3
J =
     2

Which is the correct row & column for the index passed to the function.

What you have done is this:

ind2sub(size(tmp), maxInd)  -->  ind2sub([3 3], 6)

So you have asked a function that is typically expecting to produce two outputs (the number of elements in the first argument) to instead produce only one output. It did the only reasonable thing given the way you called it (less than the expected number of outputs) ... it returned the single linear index into your single output variable.

The next thing you did was this:

ind2sub(maxInd, size(tmp))  -->  ind2sub(6,[3 3])

So you have told the function the dimension of the matrix is 1x6 and you have two indexes you want to convert to subscripts ... a 3 and a 3. So it did the correct conversion. Three elements into a 1x6 matrix yields the subscript 3. That's again a correct result.

You should probably re-read the doc on ind2sub, particularly the section on "Effects of Returning Fewer Outputs".

[Egor]

thank you! I read through 'help' function, so expected it always to return 2D value.