How to find the size of the handwritten word in an image?

When a word is given I want to find the size of it (not the width) i.e., if a word like 'hello' is given its approximate size should be found out considering all the upperzone letters like 'l' and middle zone letters like 'e'.I have tried the bounding box but it is not efficient because it will give the maximum possible height, but i want the average height. Please help me with the code.

Answers (2)

To get the average height, you'll have to scan across the image column by column using find to find the first black pixel and the last black pixel
for col = 1 : size(grayImage, 2)
topRow(col) = find(grayImage < someThreshold, 1, 'first');
bottomRow(col) = find(grayImage < someThreshold, 1, 'last');
height(col) = bottomRow(col) - topRow(col);
end
meanHeight = mean(height);

5 Comments

what value should be used in someThreshold?
sir, I used this application to find the threshold 'Thresholding an image' from the link u provided. But for all the images i am getting the max threshold of 255 for integer and 10400 for floating point. Is it correct?? and what is the meaning of integer image and floating point image?? And which value should i use?
The app will work with either floating point images or integer images. The max threshold on the right is the highest gray level in the image I believe, or it may be the greatest value possible for integer images (I don't remember).
Integer images are typically what you read from standard format image files. You might get a floating point image if you did something like did a conversion from RGB to LAB or HSV color space, or if you summed a bunch of images and cast them to double before summing so that you would not get clipping. You should just use whatever image you have that you want to threshold.
I don't know the threshold values (low and high) that you should use. Often times it's a judgment call and that's why I built this interactive program. Sometimes you don't have a nice bimodal histogram that you can find the valley of. Sometimes it's an ugly histogram with no obvious place to threshold it at. That's why sometimes you need a human to do the thresholding at a place that is reasonably good. You might have some false blobs, but then perhaps you can go and get rid of those some other way, such as by size or shape filtering.

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may be this code will help you
clear all
close all
warning off
clc
img1=imread('word.png');
[m n x]=size(img1);
figure(1)
imshow(img1)
drawnow
img2=img1(:,:,1)<80;
img2=bwareaopen(img2,50);
figure(2)
imshow(img2)
[lab,num]=bwlabel(img2);
figure(1)
imshow(img1)
hold on
for i=1:num
[r,c]=find(lab==i);
plot([mean(c) mean(c)],[min(r) max(r)],'r');
text(mean(c)+5,m,num2str(max(r)-min(r)));
hold on
end

7 Comments

sir,should we take the average of those numbers which is on the image(i.e 20,44 etc?)if so,how to take the values displayed on the image and compute the average ?
Of course. Didn't you see my answer above. I do exactly that -- get the mean over every column.
use this for averaging
clear all
close all
warning off
clc
img1=imread('word.png');
[m n x]=size(img1);
figure(1)
imshow(img1)
drawnow
img2=img1(:,:,1)<80;
img2=bwareaopen(img2,50);
figure(2)
imshow(img2)
[lab,num]=bwlabel(img2);
figure(1)
imshow(img1)
hold on
a=0;
for i=1:num
[r,c]=find(lab==i);
plot([mean(c) mean(c)],[min(r) max(r)],'r');
text(mean(c)+5,m,num2str(max(r)-min(r)));
a=a + (max(r)-min(r));
hold on
end
a=a/num;
display(strcat('average:',num2str(a)));
average:37.2857 is the answer.is it the pixel count?no,if i'm correct.then what is the unit for the average sir?
@image analyst.Yes sir i did see your answer :) we got the average pixel count too :)
yes meghashree its the pixel count
oh ok sir .thanks a lot :) @image analyst ,sir thank you for you also :)

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Asked:

on 19 Dec 2015

Commented:

on 23 Dec 2015

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