How to Convert data into Image using Matlab

조회 수: 64(최근 30일)
Stephen john
Stephen john 2022년 9월 8일
댓글: Walter Roberson 2022년 10월 3일
Hello Everyone, I Hope you are doing well. I have the following data, I have written a code to convert the data into Image. But when i round the values the Image does not same as the plot, I am doing round because each value represent a pixel value.
The Dataplot.jpg shows the original data, which is also attached in dataScan.mat. The output image after the code is imagefromdataset.jpg.
Is there is any way to get the same shape as original data. Due to rounding the value the shape changes.
How can i modified the code to get the same shape in image.
%% create grayscale shapes that resemble the data
[numImages, lenImage] = size(dataset);
imSz = 1000; % assuming images are 1000x1000
imbg = false(imSz); % background "color"
imfg = ~imbg(1,1); % forground "color"
imSizeOut=[1000 1000]; % ImageSize
for imNum = 1:numImages
imData = round(dataset(imNum,:)); % get pattern
[~,Y] = meshgrid(1:imSz); % make a grid
% black and white image
BW = imbg;
% resize (from 1000x1000)
% convert to uint8 (0 255)
im = im2uint8(BW);
%im = BW;
im = flipud(BW);
  댓글 수: 2
Stephen john
Stephen john 2022년 9월 8일
@Rik I want to create a binary Image thats why i am doing this , is there is any other method Which is use to convert data into image? The wrong is when i round decimal values the shape is change

댓글을 달려면 로그인하십시오.


Rik 2022년 9월 8일
The shape did not actually change. You just expanded the y from [-2 18] to [0 1000].
plot(dataset,'o'),ylim([0 1000])
If you want to reproduce this data yourself, one way to that is to create the coordinate grid explicitly:
x=1:numel(dataset);%this is what you did implictly
X=linspace(min(x), max(x) ,resolution);
Y=linspace(min(y), max(y) ,resolution);
% determine index in X of each x, likewise for y (note that the coordinates
% are flipped for the y-direction in images).
ind_x=interp1(X, 1:numel(X) ,x,'nearest');
%compute linear index
ind=sub2ind([resolution resolution],ind_y,ind_x);
% create the image
% Since you apparently want a dilation, apply it before displaying the
% image.
  댓글 수: 8
Stephen john
Stephen john 2022년 10월 3일
the following error
Error using matlab.internal.math.interp1
Sample points must be unique.
Error in interp1 (line 188)
VqLite = matlab.internal.math.interp1(X,V,method,method,Xqcol);

댓글을 달려면 로그인하십시오.

Walter Roberson
Walter Roberson 2022년 10월 3일
yeah it has only one values means we have a straight white line in the image.
Y=linspace(min(y), max(y) ,resolution);
With your y values all being the same, min(y) and max(y) are going to be equal, so you would be generating a constant vector with resolution elements in it.
Your Y values (the independent variable for interpolation purposes) are all the same, but interp1() requries that the values of the independent variables are all different.
I need a code which works on this data as well.
Well you cannot use interp1() in that case.
Your situation is asking interp1([5 5 5], [3 2 1], [5 5 5 5], 'nearest') . Yes, 5 (in the query values) appear in the independent vector [5 5 5], but do you say that they should be considered to be come from location 3, or location 2, or location 1 ?
You will need to detect this case and treat it differently.
  댓글 수: 8
Walter Roberson
Walter Roberson 2022년 10월 3일
What error message are you getting?
Scaled_x = [1:200, 200, 200:-1:150, 150, 150:175];
Scaled_y = [50*ones(1,200), 150, 151:151+51-1, 83, 84:-1:59];
Coords = [Scaled_x(:), Scaled_y(:)];
M = insertShape(M, 'line', Coords, 'Color', 'white'); %outputs RGB
M = M(:,:,1) ~= 0; %convert RGB to binary

댓글을 달려면 로그인하십시오.

Community Treasure Hunt

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

Start Hunting!

Translated by