So I have loaded the .mat file:
load Yale_32x32.mat;
X = fea';
I then can view the image using:
imshow(reshape(X(:,1),32,32),[])
There are 165 images, so the second dimension can be any number in the range from 1 to 165. Suppose, I want to add 'salt-pepper' noise to one of the images. If I try to do:
J = imnoise(reshape(X(:,1),32,32),'salt & pepper', 0.05);
and then:
imshow(J,[]);
... it will display to me a noise on the purely white background. What am I doing wrong?
EDIT
X(:,1) gives me:
70
68
49
53
50
50
37
33
26
13
17
61
69
109
....
etc.
After applying J = imnoise(X(:,1),'salt & pepper', 0.05);, I get:
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
1
etc.
EDIT
Fixed: I had to normalize my image. Thank you.
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I have a square Matrix N x M, odd dimensions, and I want to put a diamond of zeroes, for example, for a 5 x 5 matrix:
1 3 2 4 2
5 7 8 9 5
3 2 4 6 3
6 8 2 1 3
3 3 3 3 3
Is transform to:
1 3 0 4 2
5 0 8 0 5
0 2 4 6 0
6 0 2 0 3
3 3 0 3 3
How can this be done efficiently?
I'll bite, here is one approach:
% NxN matrix
N = 5;
assert(N>1 && mod(N,2)==1);
A = magic(N);
% diamond mask
N2 = fix(N/2);
[I,J] = meshgrid(-N2:N2);
mask = (abs(I) + abs(J)) == N2;
% fill with zeros
A(mask) = 0;
The result:
>> A
A =
17 24 0 8 15
23 0 7 0 16
0 6 13 20 0
10 0 19 0 3
11 18 0 2 9
I also had some time to play around. For my solution there are no limits concerning A being odd or even or larger than 1. Every integer is fine (even 0 works, though it does not make sense).
% NxN matrix
N = 7;
A = magic(N);
half = ceil( N/2 );
mask = ones( half );
mask( 1 : half+1 : half*half ) = 0;
mask = [ fliplr( mask ) mask ];
mask = [ mask; flipud( mask ) ];
if( mod(N,2) == 1 )
mask(half, :) = []
mask(:, half) = []
end
A( ~mask ) = 0;
A
I am first creating a square sub-matrix mask of "quarter" size (half the number of columns and half the number of rows, ceil() to get one more in the case N is odd).
Example for N=7 -> half=4.
mask =
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
I then set it's diagonal values to zero:
mask =
0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0
Mirror the mask horizontally:
mask =
1 1 1 0 0 1 1 1
1 1 0 1 1 0 1 1
1 0 1 1 1 1 0 1
0 1 1 1 1 1 1 0
Then mirror it vertically:
mask =
1 1 1 0 0 1 1 1
1 1 0 1 1 0 1 1
1 0 1 1 1 1 0 1
0 1 1 1 1 1 1 0
0 1 1 1 1 1 1 0
1 0 1 1 1 1 0 1
1 1 0 1 1 0 1 1
1 1 1 0 0 1 1 1
As N is odd we got a redundant row and redundant column that are then removed:
mask =
1 1 1 0 1 1 1
1 1 0 1 0 1 1
1 0 1 1 1 0 1
0 1 1 1 1 1 0
1 0 1 1 1 0 1
1 1 0 1 0 1 1
1 1 1 0 1 1 1
The logical not is then used as a mask to select the values in the original matrix that are set to 0.
Probably not as efficient as #Amro's solution, but it works. :D
My solution:
looking at the first left half of the matrix
in the first row 0 is in the middle column (let's call it mc)
in the second row the 0is in column mc-1
and so on while the rows increase
when you reach column 1 the sequence continue but with mc+1 but the rows decrease
In a similar way for the right half of the matrix
n=7
a=randi([20 30],n,n)
% Centre of the matrix
p=ceil(n/2)
% Identify the column sequence
col=[p:-1:1 2:p p+1:n n-1:-1:p]
% Identify the row sequence
row=[1:n n-1:-1:1]
% Transorm the row and column index in linear index
idx=sub2ind(size(a),row,col)
% Set the 0'
a(idx)=0
a =
22 29 23 27 27 21 23
29 29 21 27 24 26 24
30 28 21 27 29 28 25
28 22 24 20 27 24 25
23 26 21 20 30 20 29
26 20 26 23 25 22 25
21 24 25 25 23 21 30
a =
22 29 23 0 27 21 23
29 29 0 27 0 26 24
30 0 21 27 29 0 25
0 22 24 20 27 24 0
23 0 21 20 30 0 29
26 20 0 23 0 22 25
21 24 25 0 23 21 30
Hope this helps.
Qapla'
Using indexing (only works when N is odd):
N = 7;
% Random matrix
A = randi(100, N);
idx = [N-1:-2:1; 2:2:N];
A(cumsum([ceil(N/2) idx(:)' idx(end-1:-1:1)])) = 0
A =
60 77 74 0 54 83 9
8 48 0 76 0 28 67
6 0 32 78 83 0 10
0 27 25 5 11 39 0
76 0 49 43 67 0 16
79 7 0 86 0 70 78
57 28 85 0 81 44 81
I have a vector test2 that includes NaN 0 and 1 in random order (we cannot make any assumption).
test2 = [NaN 1 1 1 0 0 0 NaN NaN NaN 0 0 0 1 1 1 0 1 1 1 ];
I would like to group the elements containing consecutive 1 and to have in the separte vectors start and finish the first and last index of the groups.
In this case start and finish should be:
start = [2 14 18];
finish = [4 16 20];
I tried to adapt the code provided here coming up with this solution that is not working...could you help me with the right solution and tell me why the one I tried doesn't work?
a = (test2 ==1);
d = diff(a);
start = find([a(1) d]==1); % Start index of each group
finish = find([d - a(end)]==-1); % Last index of each group
start =
2 14 18
finish =
2 3 5 6 7 8 9 10 11 12 14 15 18 19
I am using MATLAB R2013b running on Windows.
I tried also using MATLAB R2013a running on ubuntu.
a = (test2 ==1)
d=diff([0 a 0])
start=find(d==1)
finish=find(d==-1)-1
Padding a zero at the beginning and end is the easiest possibility. Then the special cases where a group starts at index 1 or ends at last index don't cause problems.
Full output:
>> test2 = [NaN 1 1 1 0 0 0 NaN NaN NaN 0 0 0 1 1 1 0 1 1 1 ]
test2 =
Columns 1 through 16
NaN 1 1 1 0 0 0 NaN NaN NaN 0 0 0 1 1 1
Columns 17 through 20
0 1 1 1
>> a = (test2 ==1)
a =
Columns 1 through 16
0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1
Columns 17 through 20
0 1 1 1
>> d=diff([0 a 0])
d =
Columns 1 through 16
0 1 0 0 -1 0 0 0 0 0 0 0 0 1 0 0
Columns 17 through 21
-1 1 0 0 -1
>> start=find(d==1)
start =
2 14 18
>> finish=find(d==-1)-1
finish =
4 16 20
>>
The problem is the line finish = find([d - a(end)]==-1);, in particular that a(end) == 1. There are two steps to correcting this. First, change the problem line to finish = find(d==-1); This tells MATLAB, "Look for the elements where the difference between adjacent elements is -1". In other words, the vector shifts from 1 to 0 or NaN. If you run the code, you'll get
start = 2 14 18
finish = 4 16
Now, you'll notice the last element isn't detected (i.e. we should get finish(3) == 20. This is because the length of d is one less than the length of test2; the function diff cannot calculate the difference between the last element and the non-existant last+1 element!
To remedy this, we should modify a:
a = [(test2 == 1) 0];
And you will get the right output for start and finish.
I am trying to replace the first and second column of an edgelist matrix (edgenumber x 3) by specific node numbers as such:
5 1 1
1 38 1
2 1 1
28 17 1
18 1 1
25 1 1
that the node numbers (connection from node 5 to node 1) are replaced by the corresponding values from a vector. The edgelist is generated from an unweighted 40x40 adjacency matrix.
The vector degree_list of size 40x1 contains the "real"node numbers of this edgelist which i want to add to a larger 321x321 adjacency matrix. (if there's an easier way to do that then by concatenating the edgelists, that would also be great).
degree_list=[183,150,151,39,184,185,152,...];
So of the above edgelist I would like to replace all 1s in coll 1 and 2 by 183, all 2s by 150, etc.
Then I need to keep this new edgelist which I the add to the larger edgelist, transform it back to an adjacency matrix and have my new correct bigger adjM.
I have tried to find a solution here and on other websites but not been successful. Thank you very much for your help,
Chris
Code
a1 = [
5 1 1
1 38 1
2 1 1
28 17 1
18 1 1
25 1 1]
degree_list=[183,150,151,39,184,185,152];
col12 = a1(:,[1 2])
col12_uniq = unique(col12)
degree_list = [degree_list numel(degree_list)+1:max(col12_uniq)]
uniq_dim3 = bsxfun(#eq,col12,permute(repmat(col12_uniq,1,2),[3 2 1]))
match_dim3 = bsxfun(#times,uniq_dim3,permute(degree_list(col12_uniq),[3 1 2]))
a1_out = [sum(match_dim3,3) a1(:,3)]
Output
a1 =
5 1 1
1 38 1
2 1 1
28 17 1
18 1 1
25 1 1
a1_out =
184 183 1
183 38 1
150 183 1
28 17 1
18 183 1
25 183 1
I want to implement the JPEG compression by using MATLAB. Well at the point where the symbols' probabilities (Huffman coding) are calculated i can see some NEGATIVE values. I am sure that this is not correct!!! if someone can give some help or directions i would really appreciate it. Thank all of you in advance. I use MATLAB R2012b. Here is the code:
clc;
clear all;
a = imread('test.png');
b = rgb2gray(a);
b = imresize(b, [256 256]);
b = double(b);
final = zeros(256, 256);
mask = [1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 0
1 1 1 1 1 1 0 0
1 1 1 1 1 0 0 0
1 1 1 1 0 0 0 0
1 1 1 0 0 0 0 0
1 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0];
qv1 = [ 16 11 10 16 24 40 51 61
12 12 14 19 26 58 60 55
14 13 16 24 40 57 69 56
14 17 22 29 51 87 80 62
18 22 37 56 68 109 103 77
24 35 55 64 81 104 113 92
49 64 78 87 103 121 120 101
72 92 95 98 112 100 103 99];
t = dctmtx(8);
DCT2D = #(block_struct) t*block_struct.data*t';
msk = #(block_struct) mask.*block_struct.data;
for row = 1:8:256
for column = 1:8:256
x = (b(row:row+7, column:column+7));
xf = blockproc(x, [8 8], DCT2D);
xf1 = blockproc(xf, [8 8], msk);
xf1 = round(xf1./qv1).*qv1;
final(row:row+7, column:column+7) = xf1;
end
end
[symbols,p] = hist(final,unique(final));
bar(p, symbols);
p = p/sum(p); %NEGATIVE VALUES????
I think you might have the outputs of hist (symbols and p) swapped. The probability should be calculated from the bin counts, which is the first output of hist.
[nelements,centers] = hist(data,xvalues) returns an additional row vector, centers, indicating the location of each bin center on the x-axis. To plot the histogram, you can use bar(centers,nelements).
In other words, instead of your current line,
[symbols,p] = hist(final,unique(final));
just use,
[p,symbols] = hist(final,unique(final));
Also, final is a matrix rather than a vector, so nelements will be a matrix:
If data is a matrix, then a histogram is created separately for each column. Each histogram plot is displayed on the same figure with a different color.
My question is simple. I have an rgb image and a logical matrix. I want to set the pixel which is true in the corresponding element of logical matrix to (150,160,170).
For example:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0
r= 1 1 1 1 1 g= 1 1 1 1 1 b=1 1 1 1 1 logical_mat =1 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0
I want it results in
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
r= 150 1 1 1 1 g= 160 1 1 1 1 b=170 1 1 1 1
150 150 1 1 1 160 160 1 1 1 170 170 1 1 1
150 150 150 1 1 160 160 160 1 1 170 170 170 1 1
I have tried logical index, if set pixel into same color is easy
lm = repmat(logical_mat,[1 1 3]);
rgb(lm) = 150;
But I dont know how to set the value channel by channel.
Thanks in advance.
You're already creating the right logical matrix:
lm = repmat(logical_mat,[1 1 3]);
You need to create a 3-channel color matrix the same size.
cm = repmat(cat(3,150,160,170), size(lm,1), size(lm,2))
Then, index into the color matrix with lm:
rgb(lm) = cm(lm);