how to convert double to uint8

Question

meenu v 2020-9-17

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/595522-how-to-convert-double-to-uint8

编辑： DGM 2024-2-21

for example

if G is 2052*831 double

i wanna convert it to 2052*831*3 uint8

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the cyclist 2020-9-17

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 2020-9-17

hi, meant an RGB image

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Answer 1

Image Analyst 2020-9-17

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/595522-how-to-convert-double-to-uint8#answer_496324

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);

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Quan Zheng 2024-2-21

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 2024-2-21

编辑：DGM 2024-2-21

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.

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