How to find median value of a columns of a matrix ?

I have a zero-padded 8*8 matrix, and I want to find the median value of 3*3 block and then change the value of the middle element to this median value, then go to next block and do the same. how can I do this?

Here is my code:

 A=[44 25 22 55 21 11;35 43 2 23 45 66;12 22 19 12 76 88; 21 23 43 12 65 75;...
    10 86 56 55 23 97;23 75 88 5 55 71];
A_padded = padarray(A, [1 1]); % my 'zero-padding'

Thanks in advance for the help!

Answer 1

[Walter Roberson]

 A=[44 25 22 55 21 11;35 43 2 23 45 66;12 22 19 12 76 88; 21 23 43 12 65 75;...
    10 86 56 55 23 97;23 75 88 5 55 71];
A_padded = padarray(A, [1 1]); % my 'zero-padding'
A_padded

A_padded = 8×8

0 0 0 0 0 0 0 0 0 44 25 22 55 21 11 0 0 35 43 2 23 45 66 0 0 12 22 19 12 76 88 0 0 21 23 43 12 65 75 0 0 10 86 56 55 23 97 0 0 23 75 88 5 55 71 0 0 0 0 0 0 0 0 0

simpliest_way = medfilt2(A_padded)

simpliest_way = 8×8

0 0 0 0 0 0 0 0 0 0 22 22 21 21 0 0 0 22 22 22 22 45 21 0 0 21 22 22 23 65 65 0 0 12 22 23 43 65 65 0 0 21 43 55 55 55 55 0 0 0 23 55 23 23 0 0 0 0 0 0 0 0 0 0

more_complicated = blockproc(A_padded, [1 1], @(B) median(B.data(:)), 'BorderSize', [1 1], 'TrimBorder', false)

more_complicated = 8×8

0 0 0 0 0 0 0 0 0 0 22 22 21 21 0 0 0 22 22 22 22 45 21 0 0 21 22 22 23 65 65 0 0 12 22 23 43 65 65 0 0 21 43 55 55 55 55 0 0 0 23 55 23 23 0 0 0 0 0 0 0 0 0 0

[majid]

sorry, I have a question about your answer.

is your answer median of each middle column of block?

if i want to find median of row of each block,how is your answer change?

furthermore, I frgot to ask in question that also I want to save the median value of each block in a variable.

how can I do this?

[Walter Roberson]

is your answer median of each middle column of block?

No. As you requested, it is the median of each 3 x 3 block. And then it "slides" one position further over and takes that 3 x 3 block.

So for example,

A B C D

E F G H

I J K L

First it would take the median of

A B C

E F G

I J K

and then it would "slide over" one block to take median of.

B C D

F G H

J K L

Median of each 1 x 3 "sliding" block:

 A=[44 25 22 55 21 11;35 43 2 23 45 66;12 22 19 12 76 88; 21 23 43 12 65 75;...
    10 86 56 55 23 97;23 75 88 5 55 71];
A_padded = padarray(A, [1 1]); % my 'zero-padding'
A_padded

A_padded = 8×8

0 0 0 0 0 0 0 0 0 44 25 22 55 21 11 0 0 35 43 2 23 45 66 0 0 12 22 19 12 76 88 0 0 21 23 43 12 65 75 0 0 10 86 56 55 23 97 0 0 23 75 88 5 55 71 0 0 0 0 0 0 0 0 0

more_complicated = blockproc(A_padded, [1 1], @(B) median(B.data(:)), 'BorderSize', [0 1], 'TrimBorder', false)

more_complicated = 8×8

0 0 0 0 0 0 0 0 0 25 25 25 22 21 11 0 0 35 35 23 23 45 45 0 0 12 19 19 19 76 76 0 0 21 23 23 43 65 65 0 0 10 56 56 55 55 23 0 0 23 75 75 55 55 55 0 0 0 0 0 0 0 0 0

If you only want discrete blocks instead of sliding blocks, then that is possible, but it is a bit awkward to write as an anonymous function. It isn't terrible in the 1 x 3 case, though:

even_more_complicated = blockproc(A_padded, [1 3], @(B) [B.data(1), median(B.data), B.data(2)])

even_more_complicated = 8×9

0 0 0 0 0 0 0 0 0 0 25.0000 44.0000 22.0000 22.0000 55.0000 11.0000 5.5000 0 0 35.0000 35.0000 2.0000 23.0000 23.0000 66.0000 33.0000 0 0 12.0000 12.0000 19.0000 19.0000 12.0000 88.0000 44.0000 0 0 21.0000 21.0000 43.0000 43.0000 12.0000 75.0000 37.5000 0 0 10.0000 10.0000 56.0000 55.0000 55.0000 97.0000 48.5000 0 0 23.0000 23.0000 88.0000 55.0000 5.0000 71.0000 35.5000 0 0 0 0 0 0 0 0 0 0

[majid]

thank you very much!

I want to slide blocks.

even_more_complicated = blockproc(A_padded, [1 3], ....
      @(B) [B.data(1), median(B.data), B.data(2)])

can you help me about How can I do the same for the middle column of each sliding block ?

[Walter Roberson]

What size of output are you expecting?

For example,

A B C D

E F G H

I J K L

First block would be [A B C] and you said you want to replace the middle with the median, which would give you [A M1 C] where M1 is a median. Then you slide over to [B C D] and you want to replace the middle with the median, which would give you [B M2 D] . Now you want to put those two together on output... and you would want the first output row to look like

[A M1 C B M2 D]

??? So if there were N columns of input, then you would want the output to have (N-2)*3 columns ??

Or do you want to output only the medians, so the output for the first row would be [M1 M2] ?

If you want to output only the medians, and it is for the 1 x 3 sliding, then the code is what I posted earlier,

more_complicated = blockproc(A_padded, [1 1], @(B) median(B.data(:)), 'BorderSize', [0 1], 'TrimBorder', false)

[majid]

I need the seond case.

fo matrix

A B C D

E F G H

I J K L

we first derive median of middle column (variable X1)and middle row (variable X2)

A B C

E F G

I J K

Then we go 1 element forward now we want to derive this matrix median

B C D

F G H

J K L

and so on!

finally we derive:

median of each rows : X1=[M1 M2 M3 ... Mn]

median of each column: X2=[N1 N2 N3 ... Nn]

do I explain clear?