Find Xmin and Ymin to Imcrop an image in Matlab

I want to crop my image. Cropping of the image using coding is as follow I = imcrop(I,[xmin ymin width height]). I know the width and height of image, but I donot know how to find xmin and ymin cordinates of an image from the different pixels values of an image. sum() or any() or find() function can be used, but I donot know how to apply these functions I donot the basic concept, how xmin and ymin value are chosen from thousands of values image have.

2 Comments

What do you call xmin and ymin of an image?
Thomas sir, I got your point. Thanks a lot.
To know these cordinates after imcrop command type data = clipboard('pastespecial'); you will get all the four values
Giving an example
S = imcrop(I); %Crop an image You can crop the image which will appear in the figure window,here afetr selecting the area to be cropped, right click the mouse and click to copy position and then type the following code. data = clipboard('pastespecial');% Four values will be stored here

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Answers (2)

xmin is the x coordinate of the lower left corner of the cut you want to make in the image,
ymin is the y coordinate of the lower left corner of the cut you want to make in the image,
width is the width of the cut you want to make,
height is the height of the cut you want to make.

1 Comment

I am more of a newbie in Matlab,but I would think since the spatial co-ordinates in Matlab are given from top-to-bottom for the y-axis, then that would make the ymin to be (at least visually) the y-co-ordinate of the top left corner of the cut you want to make.

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Why don't you just call imcrop() with no input arguments and interactively make your cropping box?

2 Comments

But I need coordination of the cropped image, I need to know xmin ymin ... after cropping it
Documentation says
[___,rect2] = imcrop(_) returns the cropping rectangle in rect2, a four-element position vector, in addition to the cropped image.
So, did you actually supply a second output argument, rect, when you called it? My guess is no. So why don't you do that?

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Asked:

on 30 Apr 2015

Commented:

on 12 Dec 2017

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