how can i convert a matrix with decimal values into a image and then retrieve the same matrix back from the image with minimal error?? plz help

i have a matrix with decimal values both positive n negative. want to convert it into a image n then get the previous matrix from this image with minimal errors.

[tania]

i would rather correct my question by saying that i want to convert decimal values(234.67,3.6789 etc) to values in between the range 0-255 to view it as a coloured rgb image n then retrive the same decimal values.

[Guillaume]

That is obviously not possible, once you've removed information you can't magically recreate it.

Maybe you should explain why you want to do that in the first place. From our point of view, there is no point to your question as a matrix with decimal values can be displayed as a colour image just as well.

[tania]

i want that this information that is being removed from the original matrix to be stored somewhere and then use it when needed. and plz tell how can i display a matrix with decimal point values both negative and positive as a coloured image???

matrix is of the type B =

    8.9428   98.9408   26.0083   43.7976  142.7446   45.2182    4.8901   96.7147   37.7709  135.9462;
 -102.2958  -77.7892  -58.6407  -64.7477 -138.0498 -105.1648  -84.1800  -56.4408  -43.5661 -131.9481;
    6.3647  145.9462  124.7070    2.3231    7.9015   99.9508  132.2800   28.6386   92.5636  122.6641;
   10.7168   97.3487  120.5047  147.6096  110.6787   80.8690  100.3763   64.2379   39.7921   39.1092;
   78.2475  120.0496    9.0707   25.0753   40.3679  104.7158   28.5650   72.3033  123.6564   89.1534;
   14.5095   68.0697   59.8887   15.9325   63.4253   99.9792   55.3375   18.0917  147.3995    3.3769;
  122.7223   64.8587   79.0314   55.8615   82.1806   26.7199   69.1089   88.4261  109.5373   63.7889;
  122.6321  123.7971   62.5199   29.7178  141.4105   19.2022  147.2457   33.9282   51.5816   46.9078;
  108.3659   12.5205   98.5290   73.4531   62.6616  149.8621   23.4607   57.6929   87.6104   24.2227;
   22.4798   19.9757   94.1960   50.9240  147.4579   25.6682  128.3284   87.4480   16.1654   26.8149]

thanks in advance..

[Image Analyst]

This is not a color image (you don't have three independent color channels, only one), so your only option is to apply a colormap - a pseudocolor look up table - to turn the numbers into colors.

colormap(hsv(256));
colorbar;

[Walter Roberson]

To display a matrix that might have both negative and positive values as a colored image, use

imagesc(B)

The results of this will depend upon the range of data in the array. To get consistent results we would need to know the range of values that might occur in the array. Will the entries always be in the range -255 to +255 ? Could they be -256 exactly to +256 exactly? Could they be -256 + eps(256) to +256 - eps(256) which is to say (-256,256) exclusive?

Answer 1

[Walter Roberson]

sz = size(YourDoubleMatrix);
R = typecast(uint64(YourDoubleMatrix(:) * 2^56),'uint8');
R1 = reshape(R, [8 sz]);
R2 = permute(flipud(R1), [2 3 1 4]);
R3 = reshape(R2, [sz(1), sz(2)*8, sz(3)]);

Now R3 is an image the same height but 8 times as wide as the original image. The first horizontal 1/8th of R3 in the image is equivalent to uint8(floor(YourDoubleImage)) -- so for example for 234.67 and 3.6789 it would have 234 and 3. The second horizontal 1/8th of R3 is equivalent to uint8(floor((YourDoubleImage - floor(YourDoubleImage)) * 256)) -- that is, it has the first 8 bits of the fractions, such as the first 8 bits of 0.67 and 0.6789. The next horizontal 1/8th of R3 in the image corresponds to the next 8 bits, and so on, until the last 1/8th of R3 corresponds to the last 8 bits. The last section will be pure black except for the parts of YourDoubleImage that are less than 16.

This arrangement stores all of the information in the original for values that are at least 1, if I have calculated correctly. It would be possible to store all of the information completely for all values, but if that were done, no portion of the visual result would look like the result of displaying uint8(YourDoubleMatrix) which it seems likely you would like to see.

[tania]

my matrix is

B =

    [8.9428   98.9408   26.0083   43.7976  142.7446   45.2182    4.8901   96.7147   37.7709  135.9462;
  102.2958   77.7892   58.6407   64.7477  138.0498  105.1648   84.1800   56.4408   43.5661  131.9481;
    6.3647  145.9462  124.7070    2.3231    7.9015   99.9508  132.2800   28.6386   92.5636  122.6641;
   10.7168   97.3487  120.5047  147.6096  110.6787   80.8690  100.3763   64.2379   39.7921   39.1092;
   78.2475  120.0496    9.0707   25.0753   40.3679  104.7158   28.5650   72.3033  123.6564   89.1534;
   14.5095   68.0697   59.8887   15.9325   63.4253   99.9792   55.3375   18.0917  147.3995    3.3769;
  122.7223   64.8587   79.0314   55.8615   82.1806   26.7199   69.1089   88.4261  109.5373   63.7889;
  122.6321  123.7971   62.5199   29.7178  141.4105   19.2022  147.2457   33.9282   51.5816   46.9078;
  108.3659   12.5205   98.5290   73.4531   62.6616  149.8621   23.4607   57.6929   87.6104   24.2227;
   22.4798   19.9757   94.1960   50.9240  147.4579   25.6682  128.3284   87.4480   16.1654   26.8149]

THIS IS THE CODE I EXCECUTED..

A=rand(10); %random matrix(10x10) with values between 0 to 1.

B=A*150;
sz = size(B);

R = typecast(uint64(B(:) * 2^56),'uint8');

R1 = reshape(R, [8 sz]);
R2 = permute(flipud(R1), [2 3 1 4]);

R3 = reshape(R2, [sz(1), sz(2)*8, sz(3)]);

BUT THERE IS ERROR ..

Error using flipud (line 19)
X must be a 2-D matrix.

Error in HELP (line 6)
R2 = permute(flipud(R1), [2 3 1 4]);

THANKS FOR UR ASSISSTANCE..

[Image Analyst]

What do you want to do? If you have an RGB image in the range of 0-150 like that and just want to display it, then simply round and call imshow():

imshow(uint8(B));
% Apply colormap, if desired
colormap(hsv(256));
colorbar;

[Walter Roberson]

You must be using R2013b or earlier. Use

R2 = permute(flipdim(R1,1), [2 3 1 4]);

Note: the code was designed for 3 dimensional floating point arrays, not for 2 dimensional floating point arrays.

Also the code was designed for non-negative values.

[tania]

thanks

[tania]

how can i retrieve the the values like (0.67 and 0.6789) from the resultant matrix R3???

[Walter Roberson]

Forward:

sz = size(B);

R = typecast(uint64(B(:) * 2^56),'uint8');

R1 = reshape(R, [8 sz]);
R2 = permute(flipdim(R1,1), [2 3 1 4]);

R3 = reshape(R2, [sz(1), sz(2)*8, sz(3)]);

Reverse:

sz = size(R3);
U3 = reshape(R3, [sz(1), sz(2)/8, 8, sz(3)]);
U2 = flipdim(permute(U3, [3 1 2 4]),1);
UB = reshape(double(typecast(U2(:), 'uint64')) / 2^56, [sz(1), sz(2)/8, sz(3)]);

[Image Analyst]

I just don't understand why you need to go through all this flipping, permuting, reshaping, casting, normalizing, resizing, etc. What do you want to do tania? Just work with floating point images? I do it all the time and don't have to go through all that. Just work with them as is - floating point or not. If you want to save them in a standard image file format, convert to uint8 or uint16. If you want to save as floating point, save into a .mat file. I think it should be as simple as that. Am I wrong?

[Walter Roberson]

"i want to convert decimal values(234.67,3.6789 etc) to values in between the range 0-255 to view it as a coloured rgb image n then retrive the same decimal values." together with "i want that this information that is being removed from the original matrix to be stored somewhere and then use it when needed."

So the poster needs all of the bits stored somewhere but the image to still be in RGB uint8 format.

I could have achieved this by typecast()'ing the double array to uint8 and doing a trivial reshape back to 3D, but none of the result would have been understandable. I choose to do the permutes and reshapes so that at least one portion of the image (the left-most 1/8th) would show up the same as if you had used uint8(floor(TheImage)) and thus would look the same as if you had not stored all the extra information, but at the same time for the extra information to be recoverable from the image.

I might have pointed one of my previous answers that showed how to represent single precision float as TIFF images, a process that could be generalized to double precision if the right tag name is found. But that would have violated the constraint that the image had to be 0-255.

So the poster wants an image as if they had uint8()'d the data but with the extra bits retrievable. Why that is important is unclear to me, but the permute and flip and so on is one way of achieving it.

[Image Analyst]

Well I could be wrong because I don't have the Mind Reading Toolbox like you have :-) But I think this might be a case of the poster asking for "x" but not telling us that they really want to get at "z", whereas if they told us in advance that they want "z" we'd say that approach "x" is not the right approach like they thought, and they should do approach "y". That kind of thing happens all the time.

The comment the poster added later, below the original post, said that what she really needed to do was to have a pseudcolored version of a gray scale matrix (the badly-named "B") for display (which we told her how to do), as well as saving the original floating point matrix for later use (which can be done just by using the original B floating point variable if it's used in the same instance of the program, or by saving to a mat file if it's needed in a later instance of the program or by a different program).

tania should clarify. Or maybe she already has because she accepted this answer.

Answer 2

[Image Analyst]

Your matrix already is an image. Why do you think it's not? Images can be floating point - they don't have to be uint16 or uint8 you know.

[Walter Roberson]

TIFF might be the only image file format that allows you 64 bits of floating point per pixel. I was able to find the TIFF tag for 32 bits of floating point per pixel but not for 64, other than to confirm that it exists.

[Image Analyst]

She didn't say she needed to write it out to disk - make a round trip out to disk in an industry standard format. If that is what is needed, then I suggest she just use a .mat file - unless for some reason the disk file also needs to be read in and used by some program other than MATLAB.

Answer 3

[amreena najeeb]

good afternoon sir

i am doing image steganography btt i have problem in inverse cat map i am using the following code for encryption but feel problem in decryption

function X = three_d_catmap(Y)

p = size(Y,1); % get the number of pixels on each side

X = zeros(size(Y)); % make space for X (all zeros to start)

for i = 1:p % loop through all the pixels

for j = 1:p

newi = mod(((i-1) + (j-1)),p) + 1; % get new i coord (a+b) modulas p

newj = mod(((i-1) + 2*(j-1)),p) + 1; % get new j coord (a+2b) modulas p

X(newi,newj,:) = mod((4*(i-1)*1*(j-1)+Y(i,j,:)),p) + 1;

end

end%first applied this