I am a beginner in Matlab who is working on medical image processing of retinal OCT images. My aim is to align all the images to 1 height value. I want to find the maximum height of the layer in the eye.
For example, if input :
the output: returns this height:
I have tried this approach as outlined in Hand_height but it returns the height of the complete image.
Iterate over X and find the first peak (blue point) using findpeaks in the vertical direction (Y) to generate the first layer (blue line),
and then determine the peak with the smallest index in the Y-direction.
Please see the image!
In order to find maximum high you should find the top border of a retina.Here you have an example of how to find it.
Related
I have to resize image i.e if its dimension is 3456x5184 to 700X700 as my code needs image with less number of pixels otherwise it takes too much time to give results.So, when I use imresize command it changes the dimensions of image but at the same time it changes the shape of image i.e the circle in image which I also need to detect looks like oval instead of being cirle. I need your suggestions to resolve this problem. I am really grateful to you people.
Resizing images is done by either subsampling (to get smaller images) or some kind of interpolation (to get larger images)
Input is either a factor or a final dimension for width and height.
The only way to fit a rectangle into a square by simply resizing it is to use different scales for width and height. Which of course will yield in a distorted image.
To achieve what you want you can either crop a 700x700 region from your image or resize image using the same factor for with and height. Then you can fit the larger dimension into 700 and fill the rest around the other dimension with black or whatever you prefer.
I have a small problem with finding the pixel size of an image. I am to find size of nano and micro particles on my BW image. I used regionprops to get the area - then the diameter. Now i know the value in pixels. How do i convert to micro or nano meter scale? Do I take into account the sensor size(6.5umx6.5um) of my camera?
I use MATLAB for image processing.
Thank you
there is a function called imfinfo which will return a struct. In this struct you will maybe find three fields (it depends on the coder that you used for the image format) called XResolution, YResolution and ResolutionUnit. Using this 3 fields you can easily get pixel size, for example if XResolution=10, YResolution=10 and ResolutionUnit='meter' then you have a 100cm2 pixels (its a bit unreal i know :))
I hope this helps and that your image file contains the XResolution and YResolution information in your header.
Using ImageMagick's convert to barrel-distort a photo to correct a strongly visible pincushion distortion, I provide positive a, b or c values (from a database for my lens + focal length). This results in an image that is corrected, has the original width and height, but includes a non-rectangular, bent/distorted border, as the image is corrected towards its center. Simplified example:
convert rose: -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0' out.png
How can I automatically crop the black, bent border to the largest possible rectangle in the original aspect ratio within the rose?
The ImageMagick website says, that a parameter "d" is automatically calculated, that could do this (resulting in linear distortion effectively zooming into the image and pushing the bent border right outside the image bounds), but the imagemagick-calculated value seems to aim for something different (v6.6.9 on ubuntu 12.04). If I guess and manually specify a "d", I can get the intended result:
convert rose: -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0 +0.6' out.png
The given formular a+b+c+d=1 does not seem to be a proper d for my cropping case. Also, d seems to depend on the aspect ratio of the image and not only on a/b/c. How do I make ImageMagick crop the image, or, how to I calculate a proper d?
Update
I found Fred's ImageMagick script innercrop (http://www.fmwconcepts.com/imagemagick/innercrop/index.php) that does a bit what I need, but has drawbacks and is no solution for me. It asumes arbitrary outer areas, so it takes long to find the cropping rectangle. It does not work within Unix pipes, and it does not keep the original aspect ratio.
Update 2
Contemplating on the problem makes me think that calculating a "d" is not the solution, as changing d introduces more or less bending and seems to do more than just zoom. The d=1-(a+b+c) that is calculated by imagemagick results in the bent image touching the upper/lower bounds (for landscape images) or the left/right bounds (for portrait images). So I think the proper solution would be to calculate where one of the new 4 corners will be given a/b/c/d, and then crop to those new corners.
The way I understand the docs, you do not use commas to separate the parameters for the barrel-distort operator.
Here is an example image, alongside the output of the two commands you gave:
convert o.png -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0' out0.png
convert o.png -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0 +0.6' out1.png
I created the example image in order to better visualize what you possibly want to achieve.
However, I do not see the point you stated about the automatically calculated parameter 'd', and I do not see the effect you stated about using 'd=+0.6'...
I'm not sure I understand your wanted result correctly, so I'm assuming you want the area marked by the yellow rectangle cropped.
The image on the left is out0.png as created by the first command above.
In order to guess the required coordinates, we have to determine the image dimensions first:
identify out0.png
out0.png PNG 700x700 700x700+0+0 8-bit sRGB 36KB 0.000u 0:00.000
The image in the center is marked up with the white rectangle. The rectangle is there so you can look at it and tell me if that is the region you want cropped. The image on the right is the cropped image (without scaling it back to the original size).
Is this what you want? If yes, I can possibly update the answer in order to automatically determine the required coordinates of the cropping. (For now I've done it based on guessing.)
Update
I think you may have mis-understood the purpose of the barrel-distortion operation. It is meant for correcting a barrel (slight) distortion, as is produced by camera lenses. The 3 parameters a, b and c to be used for any specific combination of camera, lens and current zoom could possibly be stated in your photo's EXIF data. The formula were a+b+c+d = 1 is meant to be used when the new, distortion-corrected image should have the same dimensions as the original (distorted) image.
So to imitate the barrel-correction, we should probably use the second image from the last row above as our input:
convert out3.png -virtual-pixel gray -distort barrel '0 -0.2 0' corrected.png
Result:
I have a binary image with two white vertical segments separated by a small gap. I would like to calculate the distance between the two segments. Or better the gap.
My first attempt: find the profile of the two segments (using bwboundary and bwtraceboundary) and then find the intersection between this profile with horizontal line scanning the whole image. The number of lines without intersection represents the distance between the two segments.
I would like to find this gap without detecting the profile. Is there a way?
Thank you.
You can use measuretool from the MATLAB File Exchange by Jan Neggers to retrieve geometrical information of images.
I am creating a Microsoft Word document using the OpenXml library. Most of what I need is already working correctly. However, I can't for the life of me find the following bit of information.
I'm displaying an image in an anchor, which causes text to wrap around the image. I used WrapSquare but this seems to affect the last line of the previous paragraph as shown in the image below. The image is anchored to the second paragraph but causes the last line of the first paragraph to also indent around the image.
Word Screenshot http://www.softcircuits.com/Client/Word.jpg
Experimenting within Word, I can make the text wrap how I want by changing the wrapping to WrapTight. However, this requires a WrapPolygon with several coordinates. And I can't find any way to determine the polygon coordinates so that they match the size of the image, which is in pixels.
The documentation doesn't even seem to indicate what units are used for these coordinates, let alone how to calculate them from pixels. I can only assume the calculation would involve a DPI value, but I have no idea how to determine what DPI will be used when the user eventually loads the document into Word.
I would also be satisfied if someone can explain why the issues described above is happening in the first place. I can shift the image down and the previous paragraph is no longer affected. But why is this necessary? (The Distance from text setting for both Left and Top is 0".)
The WrapPolygon element has two possible child elements of LineTo and StartPoint that each take a x and y coordinate. According to 2.1.1331 Part 1 Section 20.4.2.9, lineTo (Wrapping Polygon Line End Position) and 2.1.1334 Part 1 Section 20.4.2.14, start (Wrapping Polygon Start) found in the [MS-OI29500: Microsoft Office Implementation Information for ISO/IEC-29500 Standard Compliance]:
The standard states that the x and y attributes are represented in
EMUs. Office interprets the x and y attributes in a fixed coordinate
space of 21600x21600.
As far as converting pixels to EMUs (English Metric Units), take a look at this blog post for an example.
I finally resolved this. Despite what the standard says, the WrapPolygon coordinates are not EMUs (English Metric Units). The coordinates are relative to the fixed coordinate space (21600 x 21600, as mentioned in the quote provided by amurra).
More specifically, this means 0,0 is at the top, left corner of the image, and 21600,21600 is at the bottom, right corner of the image. This is the case no matter what the size of the image is. Coordinates greater than 21600 extend outside the image.
According to this article, "The 21600 value is a legacy artifact from the drawing layer of early versions of the Microsoft Office."