Size of an image as a single value

I have the ff info from an image using the size function; 3456x 4608x 3 uint8, however i need the size of the image as a single value to compute for spatial frequency in cycles/degree of visual angle. any idea how i can go about this. thank you

[Dyuman Joshi]

" however i need the size of the image as a single value ..."

What would that value be? You need to define the value first.

[Naa Adjeley]

I mean i need the size as a single computed number not in different dimensions. And Please, which unit are the dimensions (3456x 4608x 3), produced from the size function: size (image);

[Dyuman Joshi]

There are no units to the output of size. The function size() returns the dimensions of an array, which are just plain numbers.

However, for an array representing an image, each element is considered a pixel. Given that the image you have is a 3D array, it means that you are working with a RGB image, and it can be considered 3456 pixels tall, 4608 pixels wide with the 3 pages corresponding to the Red, Green and Blue intensities/values.

If by physical size you mean the number of bytes it takes in storage, you can calculate it manually as @FPDodds has shown below or you can call whos on the array like this -

y = rand(5,6,7);
whos y

  Name      Size             Bytes  Class     Attributes

  y         5x6x7             1680  double              

[Stephen23]

"And Please, which unit are the dimensions (3456x 4608x 3), produced from the size function: size (image);"

Probably: number of vertical pixels x number of horizontal pixels x number of color channels

https://www.mathworks.com/help/matlab/creating_plots/image-types.html

Answer 1

[Image Analyst]

"I actually meant the physical size when displyed though." That depends on the figure size. You can resize the figure to any size up to the size of your monitor. You can put a ruler up to your screen and measure it. Why do you need to know this?

[DGM]

That would also require knowing the viewing distance from the monitor, since the units are in visual angle. Whether either display size and viewing distance are enforced is unknown, but certainly viewing distance isn't something that can be programmatically determined.

[Naa Adjeley]

I got some codes to compute the radially average spectrum and the spatial frequency in cycles per degree of visual angle of some pictures, but it seems i need for find the image width of the pictures i want to run the analysis on, as at now i only have the viewing distance at which each picture was taken, but no idea on the physical size of the pictures or image width. Pictures were taken with Xiaomi, Redmi Note 3.

[Image Analyst]

You could estimate it from the thin lens equation if you know the focal length that was used, and the size of the image sensor in the camera.

[Naa Adjeley]

any ideas how i can get those values, also i recieved the ff codes from a friend for calculation of the image width in cm, the value i get is 165.27cm after running the code, but thinking that is too much for an image width or?

% Read the image

info = imfinfo('case1.jpg');

% Find the resolution

resolution_dpi_x = info.XResolution;

resolution_dpi_y = info.YResolution;

% Display the resolution

disp(['Resolution (DPI): ', num2str(resolution_dpi_x), num2str(resolution_dpi_y)]);

DPI= num2str(resolution_dpi_x)

% Convert pixels to centimeters

width_pixels= 4608;

width_cm = (width_pixels/DPI) * 2.54;

% Display the width in centimeters

disp(['Image Width: ', num2str(width_cm), ' cm']);

[Image Analyst]

Are you sure the units of info.XResolution are DPI? If so, the code looks right but I'd get the width from the image itself

theImage = imread('case1.jpg');
[heightInPixels, width_pixels, numberOfColorChannels] = size(theImage)
DPI = resolution_dpi_x
width_cm = (width_pixels/DPI) * 2.54;
% Display the width in centimeters
fprintf('Image Width: %f cm.\n', width_cm);

[Naa Adjeley]

yes pls, the DPI value produced from the "XResolution" was the same as the DPI displayed on the image properties. thank you for the codes, i did run it and had an image width of 165.27cm. thank you.

[Image Analyst]

If this Answer solves your original question, then could you please click the "Accept this answer" link to award the answerer with "reputation points" for their efforts in helping you? They'd appreciate it. Thanks in advance. 🙂 Note: you can only accept one answer (so pick the best one) but you can click the "Vote" icon for as many Answers as you want. Voting for an answer will also award reputation points.

[Naa Adjeley]

okay, well noted. I just did that but for some answers, I am unable to see the vote icon.

[Image Analyst]

Only official "Answers" will have accept and vote icons. Comments below those answers will not. Thanks for accepting! 🙂

Answer 2

[FPDodds]

Hi, the units in the dimensions you provide are pixels. I.e. your image is 3456 pixels tall and 4608 pixels wide. The 3 are the layers of the image, presumably RGB. Depending on bit depth of the image, there overall size will be different. You say uint8 so there are 8 bits per pixel for each colour. This gives you an overall image size of 382,205,952 bits = 47MB.

If you mean physical size when displayed, then you want to look at your screen resolution and work out how many pixels per inch you're getting.

[Naa Adjeley]

oh okay, thank you. I actually meant the physical size when displyed though.