[Image-SIG] Locate The Center of WhiteDot from a Image

[Alec Bennett]

I need to do this too. Could you possibly keep the list updated with your
findings?



On Fri, Jan 7, 2011 at 8:26 PM, Narendra Sisodiya <
narendra at narendrasisodiya.com> wrote:

>
>
> On Sat, Jan 8, 2011 at 1:44 AM, Christopher Barker <Chris.Barker at noaa.gov>wrote:
>
>> On 1/7/11 10:14 AM, Narendra Sisodiya wrote:
>>
>>> This Code Look for bounding box which I am calculating by scanning each
>>> row and column one by one. If I am finding any white pixel, I am
>>> including it into my bounding box.
>>>
>>
>>  1) there may a change that 2 or more WhiteDOT may present. One is Big
>>> whiedot and other are very smaller whitedot(just 2-3 pixel) near to this
>>> Big whitedot.
>>>
>>
>> Not sure about this one off the top of myhead, but I"d suspect some
>> smoothing may take care of it.
>>
>>
>>  I am applying this
>>> algorithm on every frame taken from camera at 2fps. So I want a high
>>> speed algorithm. PIL must be having some function or trick to do it.
>>>
>>
>> The Image..getbbox() method should do it.
>>
>>
> Thanks, I tried getbbox on wrong image. Now I am using getbbox and I am
> getting proper result that too very fast.
>
>
>> If you need more math, numpy can help. Somethign like:
>>
>> a = np.asarray(PIL_image)
>> background_color = 0
>> rows, cols = np.where(a <> background_color) # background color a uint32
>> BB = (rows.min(), rows.max(), cols.min(), cols.max())
>>
>
>
> I am unable to get what that code means ?
> May you explain how I can use above code with getbbox ? Or the above code
> using NumPy is alternate of getbbox method ?
> IF yes, then which will be the faster ?
>
> PS: I have looked at centroid algorithm, It is again going at pixel level
> and that will be slow. My application will become better with faster
> centroid algorithm.
>
>
> Also I have another question !
> I am grabbing image from camera using opencv-python. How we can compare
> python PIL with OpenCV ? My guess that PIL will is faster then opencv
> library ?
>
>
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