Dividing an image into blocks of 64*64

Question

NC 2018-6-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/405917-dividing-an-image-into-blocks-of-64-64

评论： NC 2018-6-17

I want to divide an image of size into blocks of 64*64 non-overlapping blocks and then apply a user defined function to calculate the ratio between highest pixel value and total pixel values. So how to divide the image into non-overlapping blocks ?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2018-6-16

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/405917-dividing-an-image-into-blocks-of-64-64#answer_324902

在 MATLAB Online 中打开

aux_fun = @(block_info) YourFunction(block_info.block);
result = blockproc(aux_fun, [64 64], 'TrimBorder', false);

YourFunction should take in a 64 x 64 block and return the ratio or a vector of information.

Note: more work is needed if the image is not an exact multiple of 64 on each side.

8 个评论
显示 6更早的评论隐藏 6更早的评论

Walter Roberson 2018-6-16

在 MATLAB Online 中打开

512 and 768 are exact multiples of 64 on each side, so you do not need to do anything extra.

More generally, blockproc often gets invoked for images that are not exact multiples. For example, an image that was 500 by 700 would not be an exact multiple of 64 on either side. When that happens, blockproc ends up processing partial blocks, such as a block that has 64-12 = 52 rows. When that is a possibility, you need to decide how you want to handle the situation. One approach is to have blockproc pad those blocks (several padding methods are available.) Another approach, the default, is that your function will get invoked on a block that is not the size you expected, such as 52 x 64, and your routine takes care of the details.

For example you talk about the ratio between the highest pixel value and the total pixel values, which would be like

max(block(:)) ./ sum(block(:))

which does not depend upon the block size. But is it a good formula to continue to use if the block might be smaller because you are near an edge? Like if you had one pixel in the block that was 255 and the rest were 1, then with 64 x 64, that would be 255 ./ (255 + 4095) or about 0.0586 . But if you were near an edge and were dealing with a 32 x 64 partial block, then it would be 255 ./ (255 + 2047) or about 0.1108 . Is it reasonable for the purposes of your calculation that a half-sized block would come out with a value close to twice as large? Perhaps, perhaps not.

Walter Roberson 2018-6-17

在 MATLAB Online 中打开

I = imread('cameraman.tif');   %for example
aux_fun = @(block_info) YourFunction(block_info.data);
result = blockproc(I, [64 64], aux_fun, 'TrimBorder', false)

with

function result = YourFunction(block_of_image)
   result = double(max(block_of_image(:))) ./ sum(double(block_of_image(:)));

NC 2018-6-17

Thank you very much. The problem solved.

请先登录，再进行评论。

Dividing an image into blocks of 64*64

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

8 个评论
显示 6更早的评论隐藏 6更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

Community Treasure Hunt

Dividing an image into blocks of 64*64

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

8 个评论 显示 6更早的评论隐藏 6更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

Community Treasure Hunt

WeChat

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

8 个评论
显示 6更早的评论隐藏 6更早的评论