How to compute the Median value of an Image

[Kadir_Kirtac]

Hello, how can I compute the median value of my image?

[alexey_stetsenko]

Hi Kadir,

find a histogram and take the central value of it (median = histogram[histogram.length /2])

sorry:

find a histogram, sort histogram index by value and take level with mean histogram index, example in c#:

*****
IppiSize roi_size = new IppiSize(Width, Height);
byte tmp_max = 0;
byte tmp_min = 0;
ippiMinMax_8u_C1R((byte*)data, Stride, roi_size, &tmp_min, &tmp_max);
min = tmp_min;
max = tmp_max;
int[] hist = new int[tmp_max - tmp_min + 1];
int[] levels = new int[hist.Length + 1];
fixed (int* phist = &hist[0], plevels = &levels[0])
{
ippiHistogramEven_8u_C1R((byte*)data, Stride, roi_size, phist, plevels, levels.Length, tmp_min, tmp_max + 1);
}

int[] tmp = Array.Sort(levels);
int median = tmp[tmp.Length / 2];

Array.Sort(levels, hist);
int median = levels[levels.Length / 2];

[Kadir_Kirtac]

wouldn't the following code work for median value calculation ?

Ipp32f *pSrcSort = ippsMalloc_32f(len);
ippsCopy_32f(pSrc, pSrcSort, len);
ippsAbs_32f_I(pSrcSort, len);
ippsSortAscend_32f_I(pSrcSort);
Ipp32f medVal = pSrcSort[len/2];

[alexey_stetsenko]

it will be working fine, but you need to allocate more memory

[alexey_stetsenko]

and another problem may be happened, if image has the aligned allocation in the memory, then it will be wrong result...

for small images like width * height < 256 your code ispreferable

[alexey_stetsenko]

the last two line of code is not correct, as i thought

Array.Sort(levels, hist);
int median = levels[levels.Length / 2];

It should be found the first index of integral amplitude which has intersection with Width*Height/2 value:

int a = 0;
int center = Width * Height / 2;
for(int i = 0; i < hist.Length; i++)
{
a += hist;
if (a >= center)
{
median = level;
break;
}
}

[Kadir_Kirtac]

what you mean by aligned allocation is probably, if the image width is not a number which is multiple of 8, than IPP adds zero values to align the image width to the nearest multiple of 8. Than I will get a wrong result in median calculation by considering also these zero values at the end of the aligned lines, since I can't specify the ROI in ipps functions, am I right ?

[alexey_stetsenko]

Ipp32f *pSrcSort = ippsMalloc_32f(len);
ippsCopy_32f(pSrc, pSrcSort, len);
ippsAbs_32f_I(pSrcSort, len);
ippsSortAscend_32f_I(pSrcSort);
Ipp32f medVal = pSrcSort[len/2];

yes, that is what i did mean, the IPP use a 32-bytes boundary, but if you change the function ippsCopy_<> to ippiCopy_<> it will be OK. Is the pixel of image the float value? Why you use the ippsAbs_32f_I(pSrcSort, len)?

[Kadir_Kirtac]

Quoting alexey.stetsenko

Ipp32f *pSrcSort = ippsMalloc_32f(len);
ippsCopy_32f(pSrc, pSrcSort, len);
ippsAbs_32f_I(pSrcSort, len);
ippsSortAscend_32f_I(pSrcSort);
Ipp32f medVal = pSrcSort[len/2];

yes, that is what i did mean, the IPP use a 32-bytes boundary, but if you change the function ippsCopy_<> to ippiCopy_<> it will be OK. Is the pixel of image the float value? Why you use the ippsAbs_32f_I(pSrcSort, len)?

Yes, image pixel values are of type float; to make the negative signed values positive, I use the ippsAbs function;

so ippiCopy may help to ignore the aligned zero values, I will try, thank you.

[alexey_stetsenko]

to make the negative signed values positive, I use the ippsAbs function ..

I don't know the background of the task, but in this case the result can be wrong. If you need to get a real median value, you shouldn't use the ippsAbs function.