Is there a neat way to compute a color histogram of an image? Maybe by abusing the internal code of tf.histogram_summary? From what I've seen, this code is not very modular and calls directly some C++ code.
Thanks in advance.
I would use tf.unsorted_segment_sum, where the "segment IDs" are computed from the color values and the thing you sum is a tf.ones vector. Note that tf.unsorted_segment_sum is probably better thought of as "bucket sum". It implements dest[segment] += thing_to_sum -- exactly the operation you need for a histogram.
In slightly pseudocode (meaning I haven't run this):
binned_values = tf.reshape(tf.floor(img_r * (NUM_BINS-1)), [-1])
binned_values = tf.cast(binned_values, tf.int32)
ones = tf.ones_like(binned_values, dtype=tf.int32)
counts = tf.unsorted_segment_sum(ones, binned_values, NUM_BINS)
You could accomplish this in one pass instead of separating out the r, g, and b values with a split if you wanted to cleverly construct your "ones" to look like "100100..." for red, "010010" for green, etc., but I suspect it would be slower overall, and harder to read. I'd just do the split that you proposed above.
This is what I'm using right now:
# Assumption: img is a tensor of the size [img_width, img_height, 3], normalized to the range [-1, 1].
with tf.variable_scope('color_hist_producer') as scope:
bin_size = 0.2
hist_entries = []
# Split image into single channels
img_r, img_g, img_b = tf.split(2, 3, img)
for img_chan in [img_r, img_g, img_b]:
for idx, i in enumerate(np.arange(-1, 1, bin_size)):
gt = tf.greater(img_chan, i)
leq = tf.less_equal(img_chan, i + bin_size)
# Put together with logical_and, cast to float and sum up entries -> gives count for current bin.
hist_entries.append(tf.reduce_sum(tf.cast(tf.logical_and(gt, leq), tf.float32)))
# Pack scalars together to a tensor, then normalize histogram.
hist = tf.nn.l2_normalize(tf.pack(hist_entries), 0)
tf.histogram_fixed_width
might be what you are looking for...
Full documentation on
https://www.tensorflow.org/api_docs/python/tf/histogram_fixed_width
Related
I have a batch of images img_batch, size [8,3,32,32], and I want to manipulate each image by setting randomly selected pixels to zero. I can do this using a for loop over each image but I'm not sure how to vectorize it so I'm not processing only one image at a time. This is my code using loops.
batch_size = 8
prct0 = 0.1
noise = torch.tensor([9, 14, 5, 7, 6, 14, 1, 3])
comb_img = []
for ind in range(batch_size):
img = img_batch[ind]
c, h, w = img.shape
prct = 1 - (1 - prct0)**noise[ind].item()
idx = random.sample(range(h*w), int(prct*h*w) )
img_noised = img.clone()
img_noised.view(c,1,-1)[:,0,idx] = 0
comb_img.append(img_noised)
comb_img = torch.stack(comb_img) # output is comb_img [8,3,32,32]
I'm new to pytorch and if you see any other improvements, please share.
First note: Do you need to use noise? It will be a lot easier if you treat all images the same and don't have a different set number of pixels to set to 0.
However, you can do it this way, but you still need a small for loop (in the list comprehension).
#don't want RGB masking, want the whole pixel
rng = torch.rand(*img_batch[:,0:1].shape)
#create binary mask
mask = torch.stack([rng[i] <= 1-(1-prct0)**noise[i] for i in range(batch_size)])
img_batch_masked = img_batch.clone()
#broadcast mask to 3 RGB channels
img_batch_masked[mask.tile([1,3,1,1])] = 0
You can check that the mask is set correctly by summing mask across the last 3 dims, and seeing if it matches your target percentage:
In [5]: print(mask.sum([1,2,3])/(mask.shape[2] * mask.shape[3]))
tensor([0.6058, 0.7716, 0.4195, 0.5162, 0.4739, 0.7702, 0.1012, 0.2684])
In [6]: print(1-(1-prct0)**noise)
tensor([0.6126, 0.7712, 0.4095, 0.5217, 0.4686, 0.7712, 0.1000, 0.2710])
You can easily do this without a loop in a fully vectorized manner:
Create noise tensor
Select a threshold and round the noise tensor to 0 or 1 based on above or below that threshold (prct0)
Element-wise multiply image tensor by noise tensor
I think calling the vector of power mutlipliers noise is a bit confusing, so I've renamed that vector power_vec in this example:
power_vec = noise
# create random noise - note one channel rather than 3 color channels
rand_noise = torch.rand(8,1,32,32)
noise = torch.pow(rand_noise,power_vec) # these tensors are broadcastable
# "round" noise based on threshold
z = torch.zeros(noise.shape)
o = torch.ones(noise.shape)
noise_rounded = torch.where(noise>prct0,o,z)
# apply noise mask to each color channel
output = img_batch * noise_rounded.expand(8,3,32,32)
For simplicity this solution uses your original batch size and image size but could be trivially extended to work on inputs of any image and batch size.
I want to understand how convolution works.
Here is some code:
import numpy
from scipy import misc
data = misc.imread("path_to_a_512x512_grayscale_image.png")
data = data/255.0
masque = numpy.array([[-1,0,1],
[-2,0,0],
[-1,0,1]],numpy.double)
def my_convolution(image, masque):
hauteur,largeur = image.shape
resultat = numpy.empty((hauteur,largeur))
for y in range(1,hauteur-1):
for x in range(1,largeur-1):
pixel = 0.0
for ym in range(3):
for xm in range(3):
pixel += masque[ym,xm]*image[y-1+ym,x-1+ym]
resultat[y,x]=pixel/9.0
return resultat
my_result = my_convolution(data,masque)
plt.imshow(my_result, cmap='gray')
Result is not exactly the same with this basic method bellow.
My previous method gives a a picture that seems to be darker
from scipy import signal
result2 = signal.convolve2d(data, masque)
result2 = result2[1:-1,1:-1]
plt.imshow(result2, cmap='gray')
Anyone call explain me with those 2 codes does not give the same result ?
I do not want to know which method is fastest, i know the first method is very ugly, i just want to understand.
Thanks
The convolution requires going backwards over one of the convolved functions, which means subtracting the inner indices, not adding them. Also, the second index in the image access expression has mismatched terms. So,
pixel += masque[ym,xm]*image[y-1+ym,x-1+ym]
should instead be more like
pixel += masque[ym,xm]*image[y-1-ym,x-1-xm]
For confirmation look at the code that runs when you call signal.convolve2d (specifically here and here). The inner indices match their respective outer indices, and they are subtracted during a convolution, not added.
I am wondering if it possible to apply a mask before performing theano.tensor.nnet.softmax?
This is the behavior I am looking for:
>>>a = np.array([[1,2,3,4]])
>>>m = np.array([[1,0,1,0]]) # ignore index 1 and 3
>>>theano.tensor.nnet.softmax(a,m)
array([[ 0.11920292, 0. , 0.88079708, 0. ]])
Note that a and m are matrices, so I would like the softmax with work on an entire matrix and perform row-wise masked softmax.
Also the output should be the same shape as a, so the solution can not do advanced indexing e.g. theano.tensor.softmax(a[0,[0,2]])
def masked_softmax(a, m, axis):
e_a = T.exp(a)
masked_e = e_a * m
sum_masked_e = T.sum(masked_e, axis, keepdims=True)
return masked_e / sum_masked_e
theano.tensor.switch is one way to do this.
In the computational graph you can do the following:
a_mask = theano.tensor.switch(m, a, np.NINF)
sm = theano.tensor.softmax(a_mask)
hope it helps others.
For each location in the result matrix, instead of storing the dot product of the corresponding row and column in the argument matrices, I would like like to store the element wise product, which will be a vector extending into a third dimension.
One idea would be to convert the argument matrices to vectors with vector entries, and then take their outer product, but I'm not sure how to do this either.
EDIT:
I figured it out before I saw there was a reply. Here is my solution:
def newdot(A, B):
A = A.reshape((1,) + A.shape)
B = B.reshape((1,) + B.shape)
A = A.transpose(2, 1, 0)
B = B.transpose(1, 0, 2)
return A * B
What I am doing is taking apart each row and column pair that will have their outer product taken, and forming two lists of them, which then get their contents matrix multiplied together in parallel.
It's a little convoluted (and difficult to explain) but this function should get you what you're looking for:
def f(m1, m2):
return (m2.A.T * m1.A.reshape(m1.shape[0],1,m1.shape[1]))
m3 = m1 * m2
m3_el = f(m1, m2)
m3[i,j] == sum(m3_el[i,j,:])
m3 == m3_el.sum(2)
The basic idea is to turn the matrices into arrays and do element-by-element multiplication. One of the arrays gets reshaped to have a size of one in its middle dimension, and array broadcasting rules expand this dimension out to match the height of the other array.
I am looking for algorithm to solve the following problem :
I have two sets of vectors, and I want to find the matrix that best approximate the transformation from the input vectors to the output vectors.
vectors are 3x1, so matrix is 3x3.
This is the general problem. My particular problem is I have a set of RGB colors, and another set that contains the desired color. I am trying to find an RGB to RGB transformation that would give me colors closer to the desired ones.
There is correspondence between the input and output vectors, so computing an error function that should be minimized is the easy part. But how can I minimize this function ?
This is a classic linear algebra problem, the key phrase to search on is "multiple linear regression".
I've had to code some variation of this many times over the years. For example, code to calibrate a digitizer tablet or stylus touch-screen uses the same math.
Here's the math:
Let p be an input vector and q the corresponding output vector.
The transformation you want is a 3x3 matrix; call it A.
For a single input and output vector p and q, there is an error vector e
e = q - A x p
The square of the magnitude of the error is a scalar value:
eT x e = (q - A x p)T x (q - A x p)
(where the T operator is transpose).
What you really want to minimize is the sum of e values over the sets:
E = sum (e)
This minimum satisfies the matrix equation D = 0 where
D(i,j) = the partial derivative of E with respect to A(i,j)
Say you have N input and output vectors.
Your set of input 3-vectors is a 3xN matrix; call this matrix P.
The ith column of P is the ith input vector.
So is the set of output 3-vectors; call this matrix Q.
When you grind thru all of the algebra, the solution is
A = Q x PT x (P x PT) ^-1
(where ^-1 is the inverse operator -- sorry about no superscripts or subscripts)
Here's the algorithm:
Create the 3xN matrix P from the set of input vectors.
Create the 3xN matrix Q from the set of output vectors.
Matrix Multiply R = P x transpose (P)
Compute the inverseof R
Matrix Multiply A = Q x transpose(P) x inverse (R)
using the matrix multiplication and matrix inversion routines of your linear algebra library of choice.
However, a 3x3 affine transform matrix is capable of scaling and rotating the input vectors, but not doing any translation! This might not be general enough for your problem. It's usually a good idea to append a "1" on the end of each of the 3-vectors to make then a 4-vector, and look for the best 3x4 transform matrix that minimizes the error. This can't hurt; it can only lead to a better fit of the data.
You don't specify a language, but here's how I would approach the problem in Matlab.
v1 is a 3xn matrix, containing your input colors in vertical vectors
v2 is also a 3xn matrix containing your output colors
You want to solve the system
M*v1 = v2
M = v2*inv(v1)
However, v1 is not directly invertible, since it's not a square matrix. Matlab will solve this automatically with the mrdivide operation (M = v2/v1), where M is the best fit solution.
eg:
>> v1 = rand(3,10);
>> M = rand(3,3);
>> v2 = M * v1;
>> v2/v1 - M
ans =
1.0e-15 *
0.4510 0.4441 -0.5551
0.2220 0.1388 -0.3331
0.4441 0.2220 -0.4441
>> (v2 + randn(size(v2))*0.1)/v1 - M
ans =
0.0598 -0.1961 0.0931
-0.1684 0.0509 0.1465
-0.0931 -0.0009 0.0213
This gives a more language-agnostic solution on how to solve the problem.
Some linear algebra should be enough :
Write the average squared difference between inputs and outputs ( the sum of the squares of each difference between each input and output value ). I assume this as definition of "best approximate"
This is a quadratic function of your 9 unknown matrix coefficients.
To minimize it, derive it with respect to each of them.
You will get a linear system of 9 equations you have to solve to get the solution ( unique or a space variety depending on the input set )
When the difference function is not quadratic, you can do the same but you have to use an iterative method to solve the equation system.
This answer is better for beginners in my opinion:
Have the following scenario:
We don't know the matrix M, but we know the vector In and a corresponding output vector On. n can range from 3 and up.
If we had 3 input vectors and 3 output vectors (for 3x3 matrix), we could precisely compute the coefficients αr;c. This way we would have a fully specified system.
But we have more than 3 vectors and thus we have an overdetermined system of equations.
Let's write down these equations. Say that we have these vectors:
We know, that to get the vector On, we must perform matrix multiplication with vector In.In other words: M · I̅n = O̅n
If we expand this operation, we get (normal equations):
We do not know the alphas, but we know all the rest. In fact, there are 9 unknowns, but 12 equations. This is why the system is overdetermined. There are more equations than unknowns. We will approximate the unknowns using all the equations, and we will use the sum of squares to aggregate more equations into less unknowns.
So we will combine the above equations into a matrix form:
And with some least squares algebra magic (regression), we can solve for b̅:
This is what is happening behind that formula:
Transposing a matrix and multiplying it with its non-transposed part creates a square matrix, reduced to lower dimension ([12x9] · [9x12] = [9x9]).
Inverse of this result allows us to solve for b̅.
Multiplying vector y̅ with transposed x reduces the y̅ vector into lower [1x9] dimension. Then, by multiplying [9x9] inverse with [1x9] vector we solved the system for b̅.
Now, we take the [1x9] result vector and create a matrix from it. This is our approximated transformation matrix.
A python code:
import numpy as np
import numpy.linalg
INPUTS = [[5,6,2],[1,7,3],[2,6,5],[1,7,5]]
OUTPUTS = [[3,7,1],[3,7,1],[3,7,2],[3,7,2]]
def get_mat(inputs, outputs, entry_len):
n_of_vectors = inputs.__len__()
noe = n_of_vectors*entry_len# Number of equations
#We need to construct the input matrix.
#We need to linearize the matrix. SO we will flatten the matrix array such as [a11, a12, a21, a22]
#So for each row we combine the row's variables with each input vector.
X_mat = []
for in_n in range(0, n_of_vectors): #For each input vector
#populate all matrix flattened variables. for 2x2 matrix - 4 variables, for 3x3 - 9 variables and so on.
base = 0
for col_n in range(0, entry_len): #Each original unknown matrix's row must be matched to all entries in the input vector
row = [0 for i in range(0, entry_len ** 2)]
for entry in inputs[in_n]:
row[base] = entry
base+=1
X_mat.append(row)
Y_mat = [item for sublist in outputs for item in sublist]
X_np = np.array(X_mat)
Y_np = np.array([Y_mat]).T
solution = np.dot(np.dot(numpy.linalg.inv(np.dot(X_np.T,X_np)),X_np.T),Y_np)
var_mat = solution.reshape(entry_len, entry_len) #create square matrix
return var_mat
transf_mat = get_mat(INPUTS, OUTPUTS, 3) #3 means 3x3 matrix, and in/out vector size 3
print(transf_mat)
for i in range(0,INPUTS.__len__()):
o = np.dot(transf_mat, np.array([INPUTS[i]]).T)
print(f"{INPUTS[i]} x [M] = {o.T} ({OUTPUTS[i]})")
The output is as such:
[[ 0.13654096 0.35890767 0.09530002]
[ 0.31859558 0.83745124 0.22236671]
[ 0.08322497 -0.0526658 0.4417611 ]]
[5, 6, 2] x [M] = [[3.02675088 7.06241873 0.98365224]] ([3, 7, 1])
[1, 7, 3] x [M] = [[2.93479472 6.84785436 1.03984767]] ([3, 7, 1])
[2, 6, 5] x [M] = [[2.90302805 6.77373212 2.05926064]] ([3, 7, 2])
[1, 7, 5] x [M] = [[3.12539476 7.29258778 1.92336987]] ([3, 7, 2])
You can see, that it took all the specified inputs, got the transformed outputs and matched the outputs to the reference vectors. The results are not precise, since we have an approximation from the overspecified system. If we used INPUT and OUTPUT with only 3 vectors, the result would be exact.