I want to find the height and width of a letter in generalize form but I dont know what algorithm should I apply so that I can find the height and width of any letter
I am using MATLAB as well as openCv. Anyone can suggest me anything how to approach
this image is my test image
my main task is to find the height and width of words say for eg "Football"
You read the image with imread. You find the first instance where pixels are not white (255). You can sum rows and columns to get it quick, but you have to hide the add in the corner for this to work. You can then use the difference between y and x to get the width and height.
img = imread('unKDO.jpg');
% hide lettergenerator add
img(565:end,448:end) = 255;
% see when pixels are less than white
y(1) = find(mean(img) < 255,1);
y(2) = find(mean(img) < 255,1,'last');
x(1) = find(mean(img,2) < 255,1);
x(2) = find(mean(img,2) < 255,1,'last');
figure;
imshow(img)
hold on
plot(y(1),x(1),'*r')
plot(y(2),x(2),'*g')
Related
I have a black and white image as shown below:
I want to separate the white components of this image and then save them as separate image. This image has four white parts. I want to separate them and save four new images; each containing a white part of the image.
To achieve this, I wrote the following code:
BW=imread('img11_Inp.jpg');
imshow(BW);
BW=imbinarize(BW);
[L, num] = bwlabel(BW);
for k = 1 : num
thisBlob = ismember(L, k);
h = int2str(k);
filname = strcat(h,'_Out.jpg');
imwrite(thisBlob,filname);
figure
imshow(thisBlob, []);
end
Problem
This code separates the white parts and saves them but the size of the white part saved in the new image is same as in the original image. See the output images below:
Output images
Desired Output images
I want the output images to contain increased size of the white part of the original image. Following images are the ones I want:
Question
How can I modify the above code so that I can get the desired output images ?
Steps:
Find the boundary of the white portion.
To include the black portion, subtract a constant from the top left corner.If it is less than or equal to zero, it means we have reached or exceeded the left corner of the actual image, so set it 1. If it is greater than zero then all is fine.
Make similar adjustments for the right bottom corner.
Crop to the desired size.
Code:
%Finding the boundary of the white
[~, c1] = find(thisBlob, 1); [~, r1] = find(thisBlob.', 1);
[~, c2] = find(thisBlob, 1, 'last'); [~, r2] = find(thisBlob.', 1, 'last');
%Making adjustments to include the black portion
k = 10; %constant defining max number of black pixels
mxlim = size(X); %to be used to confirm that we don't exceed the boundary of the image
r1 = r1-10; r1(r1<=0)=1; c1 = c1-10; c1(c1<=0)=1;
r2 = r2+10; r2(r2>mxlims(1)) = mxlim(1); c2 = c2+10; c2(c2>mxlim(2)) = mxlims(2);
%Extracting the desired portion
thisBlob = thisBlob(r1:r2, c1:c2);
Output for the provided images:
You can change the number of black pixels by changing the constant k in the code.
Test Case when the white portion is on the Edge:
To verify if it also works if the white portion is on the edge like this image:
The code gives the following output for the above image:
Actually, what you want to perform is a crop with a little bit of span around the object. This can be easily achieved using imcrop that you must call providing the rectangle you want to keep.
In order to identify the rectangle:
Find the minimum an maximum rows that contain a white pixel (y-axis);
Find the minimum an maximum columns that contain a white pixel (x-axis);
Calculate width and height of the rectangle using maximum - minimum.
Since you want to crop using a little margin (in my example I set its value to 10 but you have full control over it), you must subtract that margin to minimum values and adding it to maximum values, but paying attention not to go out of the boundaries of the image (that's where the little min-max game comes into play).
Here is the full working code:
img = imread('img11_Inp.jpg');
imshow(img);
img_bin = imbinarize(img);
[lab,num] = bwlabel(img_bin);
span = 10;
for k = 1:num
file = [num2str(k) '_Out.jpg'];
blob = ismember(lab,k);
blob_size = size(blob);
col_idx = find(any(blob == true,1));
x1 = max([1 (min(col_idx) - span)]);
x2 = min([blob_size(2) (max(col_idx) + span)]);
width = x2 - x1;
row_idx = find(any(blob == true,2));
y1 = max([1 (min(row_idx) - span)]);
y2 = min([blob_size(1) (max(row_idx) + span)]);
height = y2 - y1;
blob_crop = imcrop(blob,[x1 y1 width height]);
imwrite(blob_crop,file);
figure();
imshow(blob_crop,[]);
end
Also, don't use int2str(k) in order to obtain a string representation of your index. Your index is actually a double so you are forcing a double (no pun intended) cast: double -> int and then int -> char array. Just use num2str.
Result:
I have an image (white background with 1-5 black dots) that is called main.jpg (main image).
I am trying to place another image (secondary.jpg) in every black dot that is found in main image.
In order to do that:
I found the black pixels in main image
resize the secondary image to specific size that I want
plot the image in every coordinate that I found in step one. (the black pixel should be the center coordinates of the secondary image)
Unfortunately, I don't know how to do the third step.
for example:
main image is:
secondary image is:
output:
(The dots are behind the chairs. They are the image center points)
This is my code:
mainImage=imread('main.jpg')
secondaryImage=imread('secondary.jpg')
secondaryImageResized = resizeImage(secondaryImage)
[m n]=size(mainImage)
for i=1:n
for j=1:m
% if it's black pixel
if (mainImage(i,j)==1)
outputImage = plotImageInCoordinates(secondaryImageResized, i, j)
% save this image
imwrite(outputImage,map,'clown.bmp')
end
end
end
% resize the image to (250,350) width, height
function [ Image ] = resizeImage(img)
image = imresize(img, [250 350]);
end
function [outputImage] = plotImageInCoordinates(image, x, y)
% Do something
end
Any help appreciated!
Here's an alternative without convolution. One intricacy that you must take into account is that if you want to place each image at the centre of each dot, you must determine where the top left corner is and index into your output image so that you draw the desired object from the top left corner to the bottom right corner. You can do this by taking each black dot location and subtracting by half the width horizontally and half the height vertically.
Now onto your actual problem. It's much more efficient if you loop through the set of points that are black, not the entire image. You can do this by using the find command to determine the row and column locations that are 0. Once you do this, loop through each pair of row and column coordinates, do the subtraction of the coordinates and then place it on the output image.
I will impose an additional requirement where the objects may overlap. To accommodate for this, I will accumulate pixels, then find the average of the non-zero locations.
Your code modified to accommodate for this is as follows. Take note that because you are using JPEG compression, you will have compression artifacts so regions that are 0 may not necessarily be 0. I will threshold with an intensity of 128 to ensure that zero regions are actually zero. You will also have the situation where objects may go outside the boundaries of the image. Therefore to accommodate for this, pad the image sufficiently with twice of half the width horizontally and twice of half the height vertically then crop it after you're done placing the objects.
mainImage=imread('https://i.stack.imgur.com/gbhWJ.png');
secondaryImage=imread('https://i.stack.imgur.com/P0meM.png');
secondaryImageResized = imresize(secondaryImage, [250 300]);
% Find half height and width
rows = size(secondaryImageResized, 1);
cols = size(secondaryImageResized, 2);
halfHeight = floor(rows / 2);
halfWidth = floor(cols / 2);
% Create a padded image that contains our main image. Pad with white
% pixels.
rowsMain = size(mainImage, 1);
colsMain = size(mainImage, 2);
outputImage = 255*ones([2*halfHeight + rowsMain, 2*halfWidth + colsMain, size(mainImage, 3)], class(mainImage));
outputImage(halfHeight + 1 : halfHeight + rowsMain, ...
halfWidth + 1 : halfWidth + colsMain, :) = mainImage;
% Find a mask of the black pixels
mask = outputImage(:,:,1) < 128;
% Obtain black pixel locations
[row, col] = find(mask);
% Reset the output image so that they're all zeros now. We use this
% to output our final image. Also cast to ensure accumulation is proper.
outputImage(:) = 0;
outputImage = double(outputImage);
% Keeps track of how many times each pixel was hit by the object
% This is so that we can find the average at each location.
counts = zeros([size(mask), size(mainImage, 3)]);
% For each row and column location in the image
for i = 1 : numel(row)
% Get the row and column locations
r = row(i); c = col(i);
% Offset to get the top left corner
r = r - halfHeight;
c = c - halfWidth;
% Place onto final image
outputImage(r:r+rows-1, c:c+cols-1, :) = outputImage(r:r+rows-1, c:c+cols-1, :) + double(secondaryImageResized);
% Accumulate the counts
counts(r:r+rows-1,c:c+cols-1,:) = counts(r:r+rows-1,c:c+cols-1,:) + 1;
end
% Find average - Any values that were not hit, change to white
outputImage = outputImage ./ counts;
outputImage(counts == 0) = 255;
outputImage = uint8(outputImage);
% Now crop and show
outputImage = outputImage(halfHeight + 1 : halfHeight + rowsMain, ...
halfWidth + 1 : halfWidth + colsMain, :);
close all; imshow(outputImage);
% Write the final output
imwrite(outputImage, 'finalimage.jpg', 'Quality', 100);
We get:
Edit
I wasn't told that your images had transparency. Therefore what you need to do is use imread but ensure that you read in the alpha channel. We then check to see if one exists and if one does, we will ensure that the background of any values with no transparency are set to white. You can do that with the following code. Ensure this gets placed at the very top of your code, replacing the images being loaded in:
mainImage=imread('https://i.stack.imgur.com/gbhWJ.png');
% Change - to accommodate for transparency
[secondaryImage, ~, alpha] = imread('https://i.imgur.com/qYJSzEZ.png');
if ~isempty(alpha)
m = alpha == 0;
for i = 1 : size(secondaryImage,3)
m2 = secondaryImage(:,:,i);
m2(m) = 255;
secondaryImage(:,:,i) = m2;
end
end
secondaryImageResized = imresize(secondaryImage, [250 300]);
% Rest of your code follows...
% ...
The code above has been modified to read in the basketball image. The rest of the code remains the same and we thus get:
You can use convolution to achieve the desired effect. This will place a copy of im everywhere there is a black dot in imz.
% load secondary image
im = double(imread('secondary.jpg'))/255.0;
% create some artificial image with black indicators
imz = ones(500,500,3);
imz(50,50,:) = 0;
imz(400,200,:) = 0;
imz(200,400,:) = 0;
% create output image
imout = zeros(size(imz));
imout(:,:,1) = conv2(1-imz(:,:,1),1-im(:,:,1),'same');
imout(:,:,2) = conv2(1-imz(:,:,2),1-im(:,:,2),'same');
imout(:,:,3) = conv2(1-imz(:,:,3),1-im(:,:,3),'same');
imout = 1-imout;
% output
imshow(imout);
Also, you probably want to avoid saving main.jpg as a .jpg since it results in lossy compression and will likely cause issues with any method that relies on exact pixel values. I would recommend using .png which is lossless and will also likely compress better than .jpg for synthetic images where the same colors repeat many times.
I wish to crop this original image
to a new image which will contain only the bag with minimum white pixels (essentially reducing the size image to the bag borders)
.
Therefore I decided to first convert it to a binary image
but I do not know how to find the bag corner coordinates [xmin ymin width height] in order to use them with imcrop(I,rect).
Any help will be great.
The script:
clc;
close all;
url='http://oi65.tinypic.com/i19md1.jpg' ;
rgbImage = imread(url);
grayImage = rgb2gray(rgbImage);
binaryImage = grayImage < 250;
imshow(binaryImage);
That's a very easy task to perform. Since binaryImage contains a mask that you want to use to crop the image, you can find the top-left corner (xmin,ymin) of where you want to crop by finding the smallest column and row coordinate respectively that is non-zero in the mask, then to find width and height, find the bottom-right corner that is non-zero, then subtract the two x coordinates for the width and the two y coordinates for the height. You'll need to add 1 to each difference to account for self-distances (i.e. if you had a width that was 1 pixel wide, you should get a width of 1, not 0). You can use find to help you find the row and column locations that are non-zero. However, imcrop requires that the x coordinates reflect horizontal behaviour and y coordinates reflect vertical behaviour where find returns row and column locations respectively. That's why you'll have to flip them when you call find:
[y,x] = find(binaryImage); %// Find row and column locations that are non-zero
%// Find top left corner
xmin = min(x(:));
ymin = min(y(:));
%// Find bottom right corner
xmax = max(x(:));
ymax = max(y(:));
%// Find width and height
width = xmax - xmin + 1;
height = ymax - ymin + 1;
You can now go ahead and crop the image:
out = imcrop(rgbImage, [xmin ymin width height]);
imshow(out);
I get this for your cropped image:
I'm trying to make a barcode scanner in matlab. In a barcode every white bar is 1 and every black bar is 0. i'm trying to get these bars . But this is the problem:
as you can see the bars are not the same width one time they are 3 pixels ... then 2 pixels etc ... And to make it even worse they differ in images too. So my question is . How can i get the values of these bars without knowing the width of 1 bar. Or how do i give them all the same width. (2 of the same bars can be next to eachother). It's not possible to detect the transition between bars because a transition is possible after a certain amount of pixels ... and then there can be another bar or the same bar. But because it's not possible to know this certain amount of pixels it's not possible to detect a transition. It's also not possible to work with some kind of window because the bars have no standard width. So how can i normalize this ?
A barcode :
thx in advance !
Let's assume that the bars are strictly vertical (as in your example). Here is a possible workflow:
%# read the file
filename = 'CW4li.jpg';
x = imread(filename);
%# convert to grayscale
x = rgb2gray(x);
%# get only the bars area
xend = find(diff(sum(x,2)),1);
x(xend:end,:) = [];
%# sum intensities along the bars
xsum = sum(x);
%# threshold the image by half of all pixels intensities
th = ( max(xsum)-min(xsum) ) / 2;
xth = xsum > th;
%# find widths
xstart = find(diff(xth)>0);
xstop = find(diff(xth)<0);
if xstart(1) > xstop(1)
xstart = [1 xstart];
end
if xstart(end) > xstop(end)
xstop = [xstop numel(xth)];
end
xwidth = xstop-xstart;
%# look at the histogram
hist(xwidth,1:12)
%# it's clear that single bar has 2 pixels (can be automated), so
barwidth = xwidth / 2;
UPDATE
To get relative bar width we can devide width in pixels to minimum bar width:
barwidth = xwidth ./ min(xwidth);
I believe it's good assumption that there always will be a bar on width 1.
If you won't get integer value (due to noise, for example), try to round the numbers to closest integer and get residuals. You can summarize those residuals to get quality assessment of recognition.
Some clustering algorithm (like k-mean clustering) might also work well.
I have an image in MATLAB:
im = rgb2gray(imread('some_image.jpg');
% normalize the image to be between 0 and 1
im = im/max(max(im));
And I've done some processing that resulted in a number of points that I want to highlight:
points = some_processing(im);
Where points is a matrix the same size as im with ones in the interesting points.
Now I want to draw a circle on the image in all the places where points is 1.
Is there any function in MATLAB that does this? The best I can come up with is:
[x_p, y_p] = find (points);
[x, y] = meshgrid(1:size(im,1), 1:size(im,2))
r = 5;
circles = zeros(size(im));
for k = 1:length(x_p)
circles = circles + (floor((x - x_p(k)).^2 + (y - y_p(k)).^2) == r);
end
% normalize circles
circles = circles/max(max(circles));
output = im + circles;
imshow(output)
This seems more than somewhat inelegant. Is there a way to draw circles similar to the line function?
You could use the normal PLOT command with a circular marker point:
[x_p,y_p] = find(points);
imshow(im); %# Display your image
hold on; %# Add subsequent plots to the image
plot(y_p,x_p,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
You can also adjust these other properties of the plot marker: MarkerEdgeColor, MarkerFaceColor, MarkerSize.
If you then want to save the new image with the markers plotted on it, you can look at this answer I gave to a question about maintaining image dimensions when saving images from figures.
NOTE: When plotting image data with IMSHOW (or IMAGE, etc.), the normal interpretation of rows and columns essentially becomes flipped. Normally the first dimension of data (i.e. rows) is thought of as the data that would lie on the x-axis, and is probably why you use x_p as the first set of values returned by the FIND function. However, IMSHOW displays the first dimension of the image data along the y-axis, so the first value returned by FIND ends up being the y-coordinate value in this case.
This file by Zhenhai Wang from Matlab Central's File Exchange does the trick.
%----------------------------------------------------------------
% H=CIRCLE(CENTER,RADIUS,NOP,STYLE)
% This routine draws a circle with center defined as
% a vector CENTER, radius as a scaler RADIS. NOP is
% the number of points on the circle. As to STYLE,
% use it the same way as you use the rountine PLOT.
% Since the handle of the object is returned, you
% use routine SET to get the best result.
%
% Usage Examples,
%
% circle([1,3],3,1000,':');
% circle([2,4],2,1000,'--');
%
% Zhenhai Wang <zhenhai#ieee.org>
% Version 1.00
% December, 2002
%----------------------------------------------------------------
Funny! There are 6 answers here, none give the obvious solution: the rectangle function.
From the documentation:
Draw a circle by setting the Curvature property to [1 1]. Draw the circle so that it fills the rectangular area between the points (2,4) and (4,6). The Position property defines the smallest rectangle that contains the circle.
pos = [2 4 2 2];
rectangle('Position',pos,'Curvature',[1 1])
axis equal
So in your case:
imshow(im)
hold on
[y, x] = find(points);
for ii=1:length(x)
pos = [x(ii),y(ii)];
pos = [pos-0.5,1,1];
rectangle('position',pos,'curvature',[1 1])
end
As opposed to the accepted answer, these circles will scale with the image, you can zoom in an they will always mark the whole pixel.
Hmm I had to re-switch them in this call:
k = convhull(x,y);
figure;
imshow(image); %# Display your image
hold on; %# Add subsequent plots to the image
plot(x,y,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
In reply to the comments:
x and y are created using the following code:
temp_hull = stats_single_object(k).ConvexHull;
for k2 = 1:length(temp_hull)
i = i+1;
[x(i,1)] = temp_hull(k2,1);
[y(i,1)] = temp_hull(k2,2);
end;
it might be that the ConvexHull is the other way around and therefore the plot is different. Or that I made a mistake and it should be
[x(i,1)] = temp_hull(k2,2);
[y(i,1)] = temp_hull(k2,1);
However the documentation is not clear about which colum = x OR y:
Quote: "Each row of the matrix contains the x- and y-coordinates of one vertex of the polygon. "
I read this as x is the first column and y is the second colum.
In newer versions of MATLAB (I have 2013b) the Computer Vision System Toolbox contains the vision.ShapeInserter System object which can be used to draw shapes on images. Here is an example of drawing yellow circles from the documentation:
yellow = uint8([255 255 0]); %// [R G B]; class of yellow must match class of I
shapeInserter = vision.ShapeInserter('Shape','Circles','BorderColor','Custom','CustomBorderColor',yellow);
I = imread('cameraman.tif');
circles = int32([30 30 20; 80 80 25]); %// [x1 y1 radius1;x2 y2 radius2]
RGB = repmat(I,[1,1,3]); %// convert I to an RGB image
J = step(shapeInserter, RGB, circles);
imshow(J);
With MATLAB and Image Processing Toolbox R2012a or newer, you can use the viscircles function to easily overlay circles over an image. Here is an example:
% Plot 5 circles at random locations
X = rand(5,1);
Y = rand(5,1);
% Keep the radius 0.1 for all of them
R = 0.1*ones(5,1);
% Make them blue
viscircles([X,Y],R,'EdgeColor','b');
Also, check out the imfindcircles function which implements the Hough circular transform. The online documentation for both functions (links above) have examples that show how to find circles in an image and how to display the detected circles over the image.
For example:
% Read the image into the workspace and display it.
A = imread('coins.png');
imshow(A)
% Find all the circles with radius r such that 15 ≤ r ≤ 30.
[centers, radii, metric] = imfindcircles(A,[15 30]);
% Retain the five strongest circles according to the metric values.
centersStrong5 = centers(1:5,:);
radiiStrong5 = radii(1:5);
metricStrong5 = metric(1:5);
% Draw the five strongest circle perimeters.
viscircles(centersStrong5, radiiStrong5,'EdgeColor','b');
Here's the method I think you need:
[x_p, y_p] = find (points);
% convert the subscripts to indicies, but transposed into a row vector
a = sub2ind(size(im), x_p, y_p)';
% assign all the values in the image that correspond to the points to a value of zero
im([a]) = 0;
% show the new image
imshow(im)