How do I convert a 3D greyscale image into a 2D greyscale image?

Hi all,

I have 140 x 160 x 16 greyscale image of a single object. Each image along its z-axis (length is 16) is slightly different as it traverses the depth of the object. How do I convert this 3D image into a 2D image? Please note, that I would like obtain an image as if the original image have been compacted to a single slice (i.e. z-axis length is 1) by pressing down on the x-y plane of the original image.

Thank you!

All best

Answer 1

[Image Analyst]

There are several things you can do that are typically done. One is to get a mean projection along the slices, where you take the mean value along each vertical column of pixels.

meanSliceImage = mean(grayImage3d, 3);

The other is to get a max projection, where you take the max along each vertical column of pixels.

maxProjectionImage = max(grayImage3d, [], 3);

Or you can simply extract a slice. For example to get slice #9 you could do

slice9 = grayImage3d(:, :, 9);

[D Overduin]

Thank you for your time!

Could you elaborate on your former two proposed solutions? Would every pixel in that given column be set to same mean or max intensity value of that column?

Thanks!

[Walter Roberson]

No. It is more like

P(I, J) = max(Img(I, J, :))

This would model a fully transparent light emitting object where for any particular pixel what you perceive is the brightest pixel in that direction.

[Walter Roberson]

And not, for example, a fully transparent light emitting object where for any particular pixel what you perceive is the total light in that direction.

[Walter Roberson]

You have to take into account: is it a light emitting object, in which the different spots are different inherent brightness? Is it a light emitting object in which the different spots are all the same inherent brightness but the material might have variable opaqueness? Is it a light reflecting object in which the source lights are different brightness but the surface is consistent reflectively? A light reflecting object in which the surface is variable reflectivity (which is the situation for most coloured solids)? If reflecting is the reflection only from the surface, or is it semi-transparent and the light is being absorbed and scattered back internally?

Your output would be very different if you are modelling a block of concrete than if you are modelling mica or if you are modelling a Snow Globe.

[Image Analyst]

If you have a volumetric image, imagine it as a rectangular block, and you stomped on the top of it to crush it into a 2-D image.

Each column of the 3-D image, not to be confused with a column from the 2-D image, collapses into a single pixel in the 2-D image, and it's value would be either the max value in that vertical column, or the mean value, depending on what you chose.

Or if you chose a single slice, it would be like taking two horizontal slices through your block and pulling out the slice in between the cuts.

But rather than asking "What can be done?" you should already know what you need to do and should be asking "How can I obtain that?" Don't just take some arbitrary random suggestion someone makes because it probably won't solve the problem that you ultimately need to solve.

Answer 2

[Walter Roberson]

There isn't just one solution.

If you think of the data as a solid opaque object being viewed from parallel to an axes, then the task becomes a simple matter of indexing the first or last pane along x, y, or z.

But perhaps you are viewing it from iindefinitely far, but from an angle, then you need to apply a rotation matrix to the coordinates and then find the closest part of the object in a given direction to retain and drop the rest in that direction. (See also Projection Matrix)

You use the same basic technique if you are viewing from a point instead of from a direction.

Then there is also the possibility that the object should not be considered completely solid. In that case, you need the rotation matrix type approach but you also need Alpha Blending to calculate the net colour (brightness) perceived.

A different approach that is used, especially for more complex scenes, is Ray Tracing.

[D Overduin]

Thank you for your help!

In my desired converted from 3D to 2D image, I am viewing in parallel to the z-axis. The object in the image is not transparent. The unoccupied space (filled, for example, with air) is transparent, however. Therefore, one can see a given part of the object that is located, for example, at slice 14 (i.e. grayImage3d(:, :, 14) if it is not blocked from view by any other part of the object located at any of the 13 previous slices (i.e. grayImage3d(:, :, 1:13)).

Thanks!

[Walter Roberson]

Okay, so what value or range indicates transparent?

[Walter Roberson]

zc = sum( cumprod( grayImage3d~=0, 3), 3);

zc is now the count, 0 to 16, of the number of 0's before the first non-zero entry. 0 would mean that grayImage3d(I, J, 1) is nonzero, and 16 would mean that all of the pixels at (I, J, :) are zero. And this means that if the count is not 16, then (I, J, count+1) is the content of the first nonzero pixel (and if it is 16 then the pixel should be transparent)

Now that you know which layer is the first nonzero, you can extract those pixels into an array.

[Image Analyst]

What do you mean by "The unoccupied space (filled, for example, with air) is opaque"? This is not usual. Usually, like in CT or MRI images, air is transparent (low signal), and tissue is opaque with a much higher signal (gray level).

Do you want it to be like you're looking straight down the z axis and the color in your 2-D image is the color of the first, non-air voxel you encounter? Kind of like you were looking at the "thing" as you held it in your hand or stood over it lying on the ground and looked down on it.

[D Overduin]

@Walter RobersonI have applied your suggestion to my image.

zc = sum( cumprod( grayImage3d~=0, 3), 3);
unique(zc)
ans = 16

@Image Analyst You are correct with your suggestion. I corrected it. Concerning your description, you are spot-on.

[Image Analyst]

So did either of us suggest something that ended up working (doing what you wanted)? If so, please accept that Answer (you can accept only one). Thanks in advance.