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

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majid 2021-9-5

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https://ww2.mathworks.cn/matlabcentral/answers/1446949-how-to-find-median-value-of-a-columns-of-a-matrix

评论： majid 2021-9-5

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!

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Answer 1

Walter Roberson 2021-9-5

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https://ww2.mathworks.cn/matlabcentral/answers/1446949-how-to-find-median-value-of-a-columns-of-a-matrix#answer_781049

编辑：Walter Roberson 2021-9-5

在 MATLAB Online 中打开

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

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Walter Roberson 2021-9-5

在 MATLAB Online 中打开

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

Walter Roberson 2021-9-5

编辑：Walter Roberson 2021-9-5

在 MATLAB Online 中打开

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 2021-9-5

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?

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