I have dataset of images which are half black in a upper triangular fashion, i.e. all pixels below the main diagonal are black.
Is there a way in Tensorflow to give such an image to a conv2d layer and mask or limit the convolution to only the relevant pixels?
If the black translates to 0 then you don't need to do anything. The convolution will multiply the 0 by whatever weight it has so it's not going to contribute to the result. If it's not you can multiply the data with a binary mask to make them 0.
For all black pixels you will still get any bias term if you have any.
You could multiply the result with a binary mask to 0 out the areas you don't want populated. This way you can also decide to drop results that have too many black cells, like around the diagonal.
You can also write your own custom operation that does what you want. I would recommend against it because you only get a speedup of at most 2 (the other operations will lower it). You probably get more performance by running on a GPU.
Related
I have a multi-dimensional, hyper-spectral image (channels, width, height = 15, 2500, 2500). I want to compress its 15 channel dimensions into 5 channels.So, the output would be (channels, width, height = 5, 2500, 2500). One simple way to do is to apply PCA. However, performance is not so good. Thus, I want to use Variational AutoEncoder(VAE).
When I saw the available solution in Tensorflow or keras library, it shows an example of clustering the whole images using Convolutional Variational AutoEncoder(CVAE).
https://www.tensorflow.org/tutorials/generative/cvae
https://keras.io/examples/generative/vae/
However, I have a single image. What is the best practice to implement CVAE? Is it by generating sample images by moving window approach?
One way of doing it would be to have a CVAE that takes as input (and output) values of all the spectral features for each of the spatial coordinates (the stacks circled in red in the picture). So, in the case of your image, you would have 2500*2500 = 6250000 input data samples, which are all vectors of length 15. And then the dimension of the middle layer would be a vector of length 5. And, instead of 2D convolutions that are normally used along the spatial domain of images, in this case it would make sense to use 1D convolution over the spectral domain (since the values of neighbouring wavelengths are also correlated). But I think using only fully-connected layers would also make sense.
As a disclaimer, I haven’t seen CVAEs used in this way before, but like this, you would also get many data samples, which is needed in order for the learning generalise well.
Another option would be indeed what you suggested -- to just generate the samples (patches) using a moving window (maybe with a stride that is the half size of the patch). Even though you wouldn't necessarily get enough data samples for the CVAE to generalise really well on all HSI images, I guess it doesn't matter (if it overfits), since you want to use it on that same image.
I have the task to simulate a camera with a full well capacity of 10.000 Photons per sensor element
in numpy. My first Idea was to do it like that:
camera = np.random.normal(0.0,1/10000,np.shape(img))
Imgwithnoise= img+camera
but it hardly shows an effect.
Has someone an idea how to do it?
From what I interpret from your question, if each physical pixel of the sensor has a 10,000 photon limit, this points to the brightest a digital pixel can be on your image. Similarly, 0 incident photons make the darkest pixels of the image.
You have to create a map from the physical sensor to the digital image. For the sake of simplicity, let's say we work with a grayscale image.
Your first task is to fix the colour bit-depth of the image. That is to say, is your image an 8-bit colour image? (Which usually is the case) If so, the brightest pixel has a brightness value = 255 (= 28 - 1, for 8 bits.) The darkest pixel is always chosen to have a value 0.
So you'd have to map from the range 0 --> 10,000 (sensor) to 0 --> 255 (image). The most natural idea would be to do a linear map (i.e. every pixel of the image is obtained by the same multiplicative factor from every pixel of the sensor), but to correctly interpret (according to the human eye) the brightness produced by n incident photons, often different transfer functions are used.
A transfer function in a simplified version is just a mathematical function doing this map - logarithmic TFs are quite common.
Also, since it seems like you're generating noise, it is unwise and conceptually wrong to add camera itself to the image img. What you should do, is fix a noise threshold first - this can correspond to the maximum number of photons that can affect a pixel reading as the maximum noise value. Then you generate random numbers (according to some distribution, if so required) in the range 0 --> noise_threshold. Finally, you use the map created earlier to add this noise to the image array.
Hope this helps and is in tune with what you wish to do. Cheers!
I have 209 cat/noncat images and I am looking to augment my dataset. In order to do so, this is the following code I am using to convert each NumPy array of RGB values to have a grey filter. The problem is I need their dimensions to be the same for my Neural Network to work, but they happen to have different dimensions.The code:
def rgb2gray(rgb):
return np.dot(rgb[...,:3], [0.2989, 0.5870, 0.1140])
Normal Image Dimension: (64, 64, 3)
After Applying the Filter:(64,64)
I know that the missing 3 is probably the RGB Value or something,but I cannot find a way to have a "dummy" third dimension that would not affect the actual image. Can someone provide an alternative to the rgb2gray function that maintains the dimension?
The whole point of applying that greyscale filter is to reduce the number of channels from 3 (i.e. R,G and B) down to 1 (i.e. grey).
If you really, really want to get a 3-channel image that looks just the same but takes 3x as much memory, just make all 3 channels equal:
grey = np.dstack((grey, grey, grey))
def rgb2gray(rgb):
return np.dot(rgb[...,:3], [[0.2989, 0.5870, 0.1140],[0.2989, 0.5870, 0.1140],[0.2989, 0.5870, 0.1140]])
Can anybody please explain this basic thing to me that how does a 192x28x28 input image gets reduced to a 16x28x28 feature maps using a 1x1 conv mapping. My question is about the understanding of what exactly happens when 192 goes to 16 ??
i know about ((I-2P-F)/S)+1, but what happens in the process of reducing depth.
The 1x1 Convolution compresses the whole 192*28*28 input image (which could be read as 192 feature maps of 28px * 28px pixels images) into a single 1*28*28 image. So far it reduces depth in the "feature map axis" to 1 while preserving the height and width of the original image.
But then... why do you get the 16? In a convolutional layer you can have different kernels. Basically each kernel is an indepentent filter with the same size. In your case it looks like your 1x1 Conv layer has 16 kernels by default, hence you get 16 28*28 images (one per kernel).
Is there a way to chose the x/y output axes range from np.fft2 ?
I have a piece of code computing the diffraction pattern of an aperture. The aperture is defined in a 2k x 2k pixel array. The diffraction pattern is basically the inner part of the 2D FT of the aperture. The np.fft2 gives me an output array same size of the input but with some preset range of the x/y axes. Of course I can zoom in by using the image viewer, but I have already lost detail. What is the solution?
Thanks,
Gert
import numpy as np
import matplotlib.pyplot as plt
r= 500
s= 1000
y,x = np.ogrid[-s:s+1, -s:s+1]
mask = x*x + y*y <= r*r
aperture = np.ones((2*s+1, 2*s+1))
aperture[mask] = 0
plt.imshow(aperture)
plt.show()
ffta= np.fft.fft2(aperture)
plt.imshow(np.log(np.abs(np.fft.fftshift(ffta))**2))
plt.show()
Unfortunately, much of the speed and accuracy of the FFT come from the outputs being the same size as the input.
The conventional way to increase the apparent resolution in the output Fourier domain is by zero-padding the input: np.fft.fft2(aperture, [4 * (2*s+1), 4 * (2*s+1)]) tells the FFT to pad your input to be 4 * (2*s+1) pixels tall and wide, i.e., make the input four times larger (sixteen times the number of pixels).
Begin aside I say "apparent" resolution because the actual amount of data you have hasn't increased, but the Fourier transform will appear smoother because zero-padding in the input domain causes the Fourier transform to interpolate the output. In the example above, any feature that could be seen with one pixel will be shown with four pixels. Just to make this fully concrete, this example shows that every fourth pixel of the zero-padded FFT is numerically the same as every pixel of the original unpadded FFT:
# Generate your `ffta` as above, then
N = 2 * s + 1
Up = 4
fftup = np.fft.fft2(aperture, [Up * N, Up * N])
relerr = lambda dirt, gold: np.abs((dirt - gold) / gold)
print(np.max(relerr(fftup[::Up, ::Up] , ffta))) # ~6e-12.
(That relerr is just a simple relative error, which you want to be close to machine precision, around 2e-16. The largest error between every 4th sample of the zero-padded FFT and the unpadded FFT is 6e-12 which is quite close to machine precision, meaning these two arrays are nearly numerically equivalent.) End aside
Zero-padding is the most straightforward way around your problem. But it does cost you a lot of memory. And it is frustrating because you might only care about a tiny, tiny part of the transform. There's an algorithm called the chirp z-transform (CZT, or colloquially the "zoom FFT") which can do this. If your input is N (for you 2*s+1) and you want just M samples of the FFT's output evaluated anywhere, it will compute three Fourier transforms of size N + M - 1 to obtain the desired M samples of the output. This would solve your problem too, since you can ask for M samples in the region of interest, and it wouldn't require prohibitively-much memory, though it would need at least 3x more CPU time. The downside is that a solid implementation of CZT isn't in Numpy/Scipy yet: see the scipy issue and the code it references. Matlab's CZT seems reliable, if that's an option; Octave-forge has one too and the Octave people usually try hard to match/exceed Matlab.
But if you have the memory, zero-padding the input is the way to go.