how to filling a matrix

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dakhli mohamed
dakhli mohamed 2018 年 12 月 28 日
コメント済み: dakhli mohamed 2018 年 12 月 29 日
how to fill a matrix of size 180 * 180 by results obtained by another program knowing that I will go through the matrix in a way 4 * 4 example
The first box must be filled by 3 values of 1 and a 0 by the following four boxes by 2 values of 1 and two values of 0, the following four values a value of 1 and a value of 0, and so on. at the end of the matrix 180 * 180
note: the number of values of 1 and 0 is the output of another program and it is the program entry that I am developing it
Resulat=[1,0,1,1,0.0,.........
1,1,0,0.0.1,..........]
  4 件のコメント
dakhli mohamed
dakhli mohamed 2018 年 12 月 28 日
dakhli mohamed
dakhli mohamed 2018 年 12 月 28 日
thank you for you KSSV

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採用された回答

Image Analyst
Image Analyst 2018 年 12 月 28 日
Well you could use blockproc() to march along the data one element at a time and replacing the output block with the number of zeros specified in the input block in random locations. This requires the Image Processing Toolbox. Essentially it's
outputMatrix = blockproc(outputMatrix,[windowSize windowSize], myFilterHandle)
A full demo is attached.
0000 Screenshot.png
A =
4 2 1 4 1
4 4 4 2 4
1 3 3 4 2
1 4 4 4 2
outputMatrix =
1 1 1 1 0 0 1 1 0 0
1 1 0 0 0 1 1 1 0 1
1 1 1 1 1 1 0 1 1 1
1 1 1 1 1 1 1 0 1 1
1 0 1 1 1 1 1 1 1 1
0 0 1 0 0 1 1 1 0 0
0 0 1 1 1 1 1 1 0 1
1 0 1 1 1 1 1 1 1 0
Note how each 2x2 block of the output array has as many 1's in it as the corresponding number in the input array A.
  6 件のコメント
Image Analyst
Image Analyst 2018 年 12 月 28 日
It already IS saved in a variable called outputMatrix.
If you want to store it to a file on disk that another program can read in, you can use save():
fileName = fullfile(pwd, 'outputMatrix.mat')
save(fileName, 'outputMatrix');
then, in the other program use load():
s = load(fileName);
outputMatrix = s.outputMatrix;
dakhli mohamed
dakhli mohamed 2018 年 12 月 29 日
thank you so much

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その他の回答 (1 件)

Omer Yasin Birey
Omer Yasin Birey 2018 年 12 月 28 日
編集済み: Omer Yasin Birey 2018 年 12 月 28 日
Hi dahkli, try this
result = eye(180);
for i = 1:numel(result)
temp = rand>(0.25*mod(i,3));
result(i) = temp;
end
  2 件のコメント
dakhli mohamed
dakhli mohamed 2018 年 12 月 28 日
I have an array of size 95 * 95 pixel each pixel contains a value either (1 or 2 or 3 or 4)
I will eventually create another size array (190 * 190)
if the pixel contains a value of 1 so the four subpixel new array containing a 1 and 3 zeros
if the pixel contains a value of 2 so the new array contains 2 valeues of 1 and 2 values of 0
if the pixel contains the value 3 so the new array contains 3 value of 1 and a value of 0
if the pixel contains the value 4 so the new tab contains 4 values of 1 and zeros value of 0
exempl:
A=[1,3
2.4]
Result new array=[1.0 1,1
0,0 1,0
1.0 1,1
1,0 1,1]
the four sub-pixel to the left of the second array represents the first array pixel (A)
and in my ca the table A is of size 95 * 95
Help please !!!!!!!
dakhli mohamed
dakhli mohamed 2018 年 12 月 28 日
thank you for you Omer Yasin Birey

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