Help with linspace and indexing

I have a vector of values where "roughly" the middle value is close to zero. It can be positive or negative. (these represent real locations on an fluorescent image of a regular pitched object array)

xactual=
    -30.6552
    -23.6552
    -17.6552
    -11.6552
     -5.6552
      0.3448
      6.3448
     12.3448
     18.3448
     24.3448
     30.3448
     36.3448

I want to be able to subtract this offset (0.34 in this case) off all of the values, so I tried:

xactualZero=min(abs(xactual(:)))
xactual=xactual-xactualZero;

However this doesnt handle the sign, as I said this value closest to zero can be positive or negative.

Also I want to find the index of this value (thats closest to zero) and create a linspace vector of equal length, starting with zero in the same location, but increment positively and negatively by 6 units. i.e

xreal=
      -30
      -24
      -18
      -12
       -6
        0
        6
       12
       18
       24
       30
       36

I am trying to calculate optical distortion and have the "actual locations" of my objects that I know are 6 pixels seperated I can work out the distortion as

dist=xreal-xactual.

Answer 1

[Jan]

[dummy, index] = min(abs(xactual(:)))
xactualZero    = xactual(index);
xactual        = xactual - xactualZero;

If your problem is almost solved by the min function, reading the documentation of this command will help you. If it is not solved by a specific option of the command, look at the "See also" line on the bottom. This is extremely useful.

[Jason]

Thanks Jan, thats what I was after. What about the generation of a second vector using linspace? Thanks

[Stephen23]

>> 6*((1:numel(xactual))-index)
ans =
   -30  -24  -18  -12   -6    0    6   12   18   24   30   36

Or do you particularly need to use linspace?

[Jason]

Thats perfect, thankyou. How do I accept both of your answers?

Answer 2

[Image Analyst]

You might be interested in imregionalmin() or imregionalmax().