Is it possible to generate a large sparse image in Julia?
The (excellent) library Images.jl by Tim Holy looks like it is meant for full image matrices where one defines every pixel.
I would like to make a bifurcation diagram, which is rather sparse. Ideally I'd like to make one that is poster-size, but the corresponding full image matrix would be way too large to generate.
Thank you for your time.
An Image can be created from any AbstractArray, which includes a sparse matrix:
julia> using Images
julia> S = sprand(10^4, 10^4, 0.01);
julia> img = grayim(S)
Gray Images.Image with:
data: 10000x10000 Base.SparseMatrix.SparseMatrixCSC{Float64,Int64}
properties:
colorspace: Gray
spatialorder: x y
Now, what you do with this image will determine how happy you'll be with this. But nothing prevents you from defining it.
UPDATE: Images now treats any AbstractArray as an image, there is no Images type anymore.
Related
I am new to the field of medical imaging - and trying to solve this (potentially basic problem). For a machine learning purpose, I am trying to standardize and normalize a library of DICOM images, to ensure that all images have the same rotation and are at the same scale (e.g. in mm). I have been playing around with the Mango viewer, and understand that one can create transformation matrices that might be helpful in this regard. I have however the following basic questions:
I would have thought that a scaling of the image would have changed the pixel spacing in the image header. Does this tag not provide the distance between pixels, and should this not change as a result of scaling?
What is the easiest way to standardize a library of images (ideally in python)? Is it possible and should one extract a mean pixel spacing across all images, and then scaling all images to match that mean? or is there a smarter way to ensure consistency in scaling and rotation?
Many thanks in advance, W
Does this tag not provide the distance between pixels, and should this
not change as a result of scaling?
Think of the image voxels as fixed units of space, which are sampling your image. When you apply your transform, you are translating/rotating/scaling your image around within these fixed units of space. That is, the size and shape of the voxels doesn't change. They just sample different parts of your image.
You can resample your image by making your voxels bigger or smaller or changing their shape (pixel spacing), but this can be independent of the transform you are applying to the image.
What is the easiest way to standardize a library of images (ideally in
python)?
One option is FSL-FLIRT, although it only accepts data in NIFTI format, so you'd have to convert your DICOMs to NIFTI. There is also this Python interface to FSL.
Is it possible and should one extract a mean pixel spacing across all
images, and then scaling all images to match that mean? or is there a
smarter way to ensure consistency in scaling and rotation?
I think you'd just to have pick a reference image to register all your other images too. There's no right answer: picking the highest resolution image/voxel dimensions or an average or some resampling into some other set of dimensions all sound reasonable.
I am trying to enlarge a point cloud data set. Suppose I have a point cloud data set consisting of 100 points & I want to enlarge it to say 5 times. Actually I am studying some specific structure which is very small, so I want to zoom in & do some computations. I want something like imresize() in Matlab.
Is there any function to do this? What does resize() function do in PCL? Any idea about how can I do it?
Why would you need this? Points are just numbers, regardless whether they are 1 or 100, until all of them are on the same scale and in the same coordinate system. Their size on the screen is just a visual representation, you can zoom in and out as you wish.
You want them to be a thousandth of their original value (eg. millimeters -> meters change)? Divide them by 1000.
You want them spread out in a 5 times larger space in that particular coordinate system? Multiply their coordinates with 5. But even so, their visual representations will look exactly the same on the screen. The data remains basically the same, they will not be resized per se, they numeric representation will change a bit. It is the simplest affine transform, just a single multiplication.
You want to have finer or coarser resolution of your numeric representation? Or have different range? Change your data type accordingly.
That is, if you deal with a single set.
If you deal with different sets, say, recorded with different kinds of sensors and the numeric representations differ a bit (there are angles between the coordinate systems, mm vs cm scale, etc.) you just have to find the transformation from one coordinate system to the other one and apply it to the first one.
Since you want to increase the number of points while preserving shape/structure of the cloud, I think you want to do something like 'upsampling'.
Here is another SO question on this.
The PCL offers a class for bilateral upsampling.
And as always google gives you a lot of hints on this topic.
Beside (what Ziker mentioned) increasing allocated memory (that's not what you want, right?) or zooming in in visualization you could just rescale your point cloud.
This can be done by multiplying each points dimensions with a constant factor or using an affine transformation. So you can e.g switch from mm to m.
If i understand your question correctly
If you have defined your cloud like this
pcl::PointCloud<pcl::PointXYZ>::Ptr cloud (new pcl::PointCloud<pcl::PointXYZ>);
in fact you can do resize
cloud->points.resize (cloud->width * cloud->height);
Note that doing resize does nothing more than allocate more memory for variable thus after resizing original data remain in cloud. So if you want to have empty resized cloud dont forget to add cloud->clear();
If you just want zoom some pcd for visual puposes(i.e you cant see what is shape of cloud because its too small) why dont you use PCL Visualization and zoom by scrolling up/down
While I use R quite a bit, just started an image analysis project and I am using the EBImage package. I need to collect a lot of data from circular/elliptical images. The built-in function computeFeatures gives the maximum and minimum radius. But I need all of the radii it computes.
Here is the code. I have read the image, thresholded and filled.
actual.image = readImage("xxxx")
image = actual.image[,2070:4000]
image1 = thresh(image)
image1 = fillHull(image1)
As there are several objects in the image, I used the following to label
image1 = bwlabel(image1)
I generated features using the built in function
features = data.frame(computeFeatures(image1,image))
Now, computeFeatures gives max radius and min radius. I need all the radii of all the objects it has computed for my analysis. At least if I get the coordinates of boundaries of all objects, I can compute the radii through some other code.
I know images are stored as matrices and can come up with a convoluted way to find the boundaries and then compute radii. But, was wondering if there a more elegant method?
You could try extracting each object + some padding, and plotting the x and y axis intensity profiles for each object. The intensity profiles is simply the sum of rows / columns which can be computed using rowSums and colSums in R
Then you could find where it dropps by splitting each intensity profiles in half and computing the nearest minimum value.
Maybe an example would help clear things up:
Hopefully this makes sense
I'm trying to find similar or equivalent function of Matlabs "Bwareaopen" function in OpenCV?
In MatLab Bwareaopen(image,P) removes from a binary image all connected components (objects) that have fewer than P pixels.
In my 1 channel image I want to simply remove small regions that are not part of bigger ones? Is there any trivial way to solve this?
Take a look at the cvBlobsLib, it has functions to do what you want. In fact, the code example on the front page of that link does exactly what you want, I think.
Essentially, you can use CBlobResult to perform connected-component labeling on your binary image, and then call Filter to exclude blobs according to your criteria.
There is not such a function, but you can
1) find contours
2) Find contours area
3) filter all external contours with area less then threshhold
4) Create new black image
5) Draw left contours on it
6) Mask it with a original image
I had the same problem and came up with a function that uses connectedComponentsWithStats():
def bwareaopen(img, min_size, connectivity=8):
"""Remove small objects from binary image (approximation of
bwareaopen in Matlab for 2D images).
Args:
img: a binary image (dtype=uint8) to remove small objects from
min_size: minimum size (in pixels) for an object to remain in the image
connectivity: Pixel connectivity; either 4 (connected via edges) or 8 (connected via edges and corners).
Returns:
the binary image with small objects removed
"""
# Find all connected components (called here "labels")
num_labels, labels, stats, centroids = cv2.connectedComponentsWithStats(
img, connectivity=connectivity)
# check size of all connected components (area in pixels)
for i in range(num_labels):
label_size = stats[i, cv2.CC_STAT_AREA]
# remove connected components smaller than min_size
if label_size < min_size:
img[labels == i] = 0
return img
For clarification regarding connectedComponentsWithStats(), see:
How to remove small connected objects using OpenCV
https://www.programcreek.com/python/example/89340/cv2.connectedComponentsWithStats
https://python.hotexamples.com/de/examples/cv2/-/connectedComponentsWithStats/python-connectedcomponentswithstats-function-examples.html
The closest OpenCV solution to your question is the morphological closing or opening.
Say you have white regions in your image that you need to remove. You can use morphological opening. Opening is erosion + dilation, in that order. Erosion is when the white regions in your image are shrunk. Dilation is (the opposite) where white regions in your image are enlarged. When you perform an opening operation, your small white region is eroded until it vanishes. Larger white features will not vanish but will be eroded from the boundary. The subsequent dilation step restores their original size. However, since the small element(s) vanished during the erosion step, they will not appear in the final image after dilation.
For example consider this image where we want to remove the small white regions but retain the 3 large white ellipses. Running the following code removes the white regions and displays the clean image
import cv2
im = cv2.imread('sample.png')
clean = cv2.morphologyEx(im, cv2.MORPH_OPEN, np.ones((10, 10)))
cv2.imshwo("Clean image", clean)
The clean image output would be like this.
The command above uses a square block of size 10 as the kernel. You can modify this to suit your requirement. You can even generate a more advanced kernel using the function getStructuringElement().
Note that if your image is inverted, i.e., with black noise on a white background, you simply need to use the morphological closing operation (cv2.MORPH_CLOSE method) instead of opening. This reverses the order of operation - first the image is eroded and then dilated.
See also: Why is my image rotation algorithm not working?
This question isn't language specific, and is a math problem. I will however use some C++ code to explain what I need as I'm not experienced with the mathematic equations needed to express the problem (but if you know about this, I’d be interested to learn).
Here's how the image is composed:
ImageMatrix image;
image[0][0][0] = 1;
image[0][1][0] = 2;
image[0][2][0] = 1;
image[1][0][0] = 0;
image[1][1][0] = 0;
image[1][2][0] = 0;
image[2][0][0] = -1;
image[2][1][0] = -2;
image[2][2][0] = -1;
Here's the prototype for the function I'm trying to create:
ImageMatrix rotateImage(ImageMatrix image, double angle);
I'd like to rotate only the first two indices (rows and columns) but not the channel.
The usual way to solve this is by doing it backwards. Instead of calculating where each pixel in the input image ends up in the output image, you calculate where each pixel in the output image is located in the input image (by rotationg the same amount in the other direction. This way you can be sure that all pixels in the output image will have a value.
output = new Image(input.size())
for each pixel in input:
{
p2 = rotate(pixel, -angle);
value = interpolate(input, p2)
output(pixel) = value
}
There are different ways to do interpolation. For the formula of rotation I think you should check https://en.wikipedia.org/wiki/Rotation_matrix#In_two_dimensions
But just to be nice, here it is (rotation of point (x,y) angle degrees/radians):
newX = cos(angle)*x - sin(angle)*y
newY = sin(angle)*x + cos(angle)*y
To rotate an image, you create 3 points:
A----B
|
|
C
and rotate that around A. To get the new rotated image you do this:
rotate ABC around A in 2D, so this is a single euler rotation
traverse in the rotated state from A to B. For every pixel you traverse also from left to right over the horizontal line in the original image. So if the image is an image of width 100, height 50, you'll traverse from A to B in 100 steps and from A to C in 50 steps, drawing 50 lines of 100 pixels in the area formed by ABC in their rotated state.
This might sound complicated but it's not. Please see this C# code I wrote some time ago:
rotoZoomer by me
When drawing, I alter the source pointers a bit to get a rubber-like effect, but if you disable that, you'll see the code rotates the image without problems. Of course, on some angles you'll get an image which looks slightly distorted. The sourcecode contains comments what's going on so you should be able to grab the math/logic behind it easily.
If you like Java better, I also have made a java version once, 14 or so years ago ;) ->
http://www.xs4all.nl/~perseus/zoom/zoom.java
Note there's another solution apart from rotation matrices, that doesn't loose image information through aliasing.
You can separate 2D image rotation into skews and scalings, which preserve the image quality.
Here's a simpler explanation
It seems like the example you've provided is some edge detection kernel. So if what you want to is detect edges of different angles you'd better choose some continuous function (which in your case might be a parametrized gaussian of x1 multiplied by x2) and then rotate it according to formulae provided by kigurai. As a result you would be able to produce a diskrete kernel more efficiently and without aliasing.