how to convert double to uint8

for example

if G is 2052*831 double

i wanna convert it to 2052*831*3 uint8

[the cyclist]

Just to clarify, since I think I misunderstood your question at first ...

Are you asking how to capture the full precision of a (64-bit) double, in three uint8 variables?

(Specifically, you are not trying to simply convert a double to a uint8, possibly losing precision in the process, right?)

[the cyclist]

Based on the answers I am seeing, people are guessing different things about what you mean by that third dimension (or ignoring it completely). You need to clarify what you want, so that we are not guessing.

[meenu v]

hi, meant an RGB image

Answer 1

[Image Analyst]

Try this:

% First convert G to a 3-D variable.
G = cat(3, G, G, G);
% Now convert it to uint8.  It will clip values outside the range 0-255, and will round values in the range.
G = uint8(G);

[meenu v]

sir, i meant for a RGB image

2052*831 double to 2052*831*3 uint8

[Image Analyst]

If G is already 2052*831*3 then you can either...

If G is in the range 0-255:

G = uint8(G);

If G is in some big range, like -3000 to +100000

G = uint8(255 * mat2gray(G));

Or if G is in the range 0-1, you can do

G = uint8(255 * G);

There are other ways of scaling, it just depends on what you want to do and what you're starting with.

[Quan Zheng]

Hi, I have ran into a issue a bit tougher than this. Now I have a 'double' file contains water saturations with 4 decimal points for different labels, and the number of some data are quite close to each other. Consequently, the similar numbers have been recognized as one integer after multiplying 255 and converting to uint8 and the quantity of water saturation data decreased from 220 to 120. May I ask if there are any methods which can fix this issue? I am appreciate for your time. Thanks!

[DGM]

uint8 and other integer formats can only represent integers over a fixed range. 8-bit integer formats can only represent 256 uniformly-spaced unique values. If you need more precision than that, you either should use a wider integer class (e.g. uint16) or floating point (e.g. double). What's appropriate or acceptable depends on what the end goal is. If it's required to represent it as an image which has broad decoder support, float probably won't be an option, but uint16 may be. In any case, don't use JPG.

That said, there are always ways that one could cause unnecessarily severe data loss due to truncation or rounding depending on the particular ways the conversion is handled. I have to mention that, since I haven't seen the code you used.

Also consider that if only part of your data requires higher resolution (e.g values nearest zero), you might consider storing a nonlinear transformation of the data (e.g. sqrt(mydata)).

% unit-scale float data with a bunch of closely spaced values near zero
data = [0:0.3:5 100:10:150].'/255; 

% store the data by scaling linearly
udata1 = im2uint8(data); 
fdata1 = im2double(udata1) % uniqueness gets lost due to rounding

fdata1 = 23×1

0 0 0.0039 0.0039 0.0039 0.0078 0.0078 0.0078 0.0078 0.0118

% store nonlinearly-scaled data
g = 0.5;
udata2 = im2uint8(data.^g); 
fdata2 = im2double(udata2).^(1/g) % uniqueness is preserved

fdata2 = 23×1

0 0.0012 0.0022 0.0035 0.0044 0.0062 0.0068 0.0081 0.0096 0.0104

Of course, that might not help much with the total rounding error, but it at least helps preserve uniqueness. This should probably shouldn't be considered unless simpler options are exhausted.

Answer 2

[Samiu Haque]

Use cast() function.

cast(G,'uint8');

[the cyclist]

See my comment, and note that OP says they are converting

2052*831

double to

2052*831*3

uint8.