Finding the mean and standard deviation

2012 8 30
1 个回答
3 次查看（30 天）

0 个投票

I is an rgb image
d = unique(I);
out = [];
for jj = 1:numel(d)
r = bwconncomp(I == d(jj),4);
out = [out;[repmat({d(jj)},r.NumObjects,1) r.PixelIdxList']];
end
and i get output as(below is example some part of output)
[ 44] [ 1x3 double]
[ 46] [ 1x3 double]
[ 19] [ 1x3 double]
[ 44] [ 1x3 double]
[ 20] [ 1x3 double]
[ 43] [ 1x3 double]
[ 19] [ 1x3 double]
[ 39] [ 1x3 double]
[ 41] [ 1x3 double]
[ 20] [ 1x3 double]
[ 40] [ 1x3 double]
[ 41] [ 1x3 double]
[ 21] [ 1x3 double]
[ 42] [ 1x3 double]
for each block i have to find the mean and standard deviation ,please help

请先登录，再回答此问题。

回答（1 个）

Image Analyst
Image Analyst 2012-8-30

0 个投票

Post this as a follow up to the original question at http://www.mathworks.com/matlabcentral/answers/46552-grouping-the-pixel-having-same-values I don't even know what you mean by the mean. Each "block" has the value d(jj) so that's the mean, and of course the std dev would be zero.

3 个评论
显示 1更早的评论 隐藏 1更早的评论

kash
kash 2012-8-30
I dont want o make that thread complicated ,thats posted a new question twice ,my output must be as
A rgb image as input and then divide the pixels having same value into blocks (have to display the image) and then find the meana nd standard deviation for each block
In the above for each block i tries using
meanValues = cellfun(@(x) mean(x(:)),out(:,2))
but dontknow if it is correct or not,please help
Image Analyst
Image Analyst 2012-8-30
Again, since each block is a unique color, the mean color should be that color and the SD should be zero. Isn't that right?
kash
kash 2012-8-31
for example
I=imread('rice.png');
C = mat2cell(I,[128 128],[128 128]);
meanValues = cellfun(@(x) mean(x(:)),C)
meanValues =
131.3774 129.7532
94.9357 88.9209
so for each block i get mean values ,is it possible to do the same for my problem

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Neighborhood and Block Processing 的更多信息

标签

提问:

kash
kash
2012-8-30

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by