I am trying to reconstruct the following matrix of shape (256 x 256 x 2) with SVD components as
U.shape = (256, 256, 256)
s.shape = (256, 2)
vh.shape = (256, 2, 2)
I have already tried methods from documentation of numpy and scipy to reconstruct the original matrix but failed multiple times, I think it maybe 3D matrix has a different way of reconstruction.
I am using numpy.linalg.svd for decompostion.
From np.linalg.svd's documentation:
"... If a has more than two dimensions, then broadcasting rules apply, as explained in :ref:routines.linalg-broadcasting. This means that SVD is
working in "stacked" mode: it iterates over all indices of the first
a.ndim - 2 dimensions and for each combination SVD is applied to the
last two indices."
This means that you only need to handle the s matrix (or tensor in general case) to obtain the right tensor. More precisely, what you need to do is pad s appropriately and then take only the first 2 columns (or generally, the number of rows of vh which should be equal to the number of columns of the returned s).
Here is a working code with example for your case:
import numpy as np
mat = np.random.randn(256, 256, 2) # Your matrix of dim 256 x 256 x2
u, s, vh = np.linalg.svd(mat) # Get the decomposition
# Pad the singular values' arrays, obtain diagonal matrix and take only first 2 columns:
s_rep = np.apply_along_axis(lambda _s: np.diag(np.pad(_s, (0, u.shape[1]-_s.shape[0])))[:, :_s.shape[0]], 1, s)
mat_reconstructed = u # s_rep # vh
mat_reconstructed equals to mat up to precision error.
Related
I have an idea for a tensor operation that would not be difficult to implement via iteration, with batch size one. However I would like to parallelize it as much as possible.
I have two tensors with shape (n, 5) called X and Y. X is actually supposed to represent 5 one-dimensional tensors with shape (n, 1): (x_1, ..., x_n). Ditto for Y.
I would like to compute a tensor with shape (n, 25) where each column represents the output of the tensor operation f(x_i, y_j), where f is fixed for all 1 <= i, j <= 5. The operation f has output shape (n, 1), just like x_i and y_i.
I feel it is important to clarify that f is essentially a fully-connected layer from the concatenated [...x_i, ...y_i] tensor with shape (1, 10), to an output layer with shape (1,5).
Again, it is easy to see how to do this manually with iteration and slicing. However this is probably very slow. Performing this operation in batches, where the tensors X, Y now have shape (n, 5, batch_size) is also desirable, particularly for mini-batch gradient descent.
It is difficult to really articulate here why I desire to create this network; I feel it is suited for my domain of 'itemized tabular data' and cuts down significantly on the number of weights per operation, compared to a fully connected network.
Is this possible using tensorflow? Certainly not using just keras.
Below is an example in numpy per AloneTogether's request
import numpy as np
features = 16
batch_size = 256
X_batch = np.random.random((features, 5, batch_size))
Y_batch = np.random.random((features, 5, batch_size))
# one tensor operation to reduce weights in this custom 'layer'
f = np.random.random((features, 2 * features))
for b in range(batch_size):
X = X_batch[:, :, b]
Y = Y_batch[:, :, b]
for i in range(5):
x_i = X[:, i:i+1]
for j in range(5):
y_j = Y[:, j:j+1]
x_i_y_j = np.concatenate([x_i, y_j], axis=0)
# f(x_i, y_j)
# implemented by a fully-connected layer
f_i_j = np.matmul(f, x_i_y_j)
All operations you need (concatenation and matrix multiplication) can be batched.
Difficult part here is, that you want to concatenate features of all items in X with features of all items in Y (all combinations).
My recommended solution is to expand the dimensions of X to [batch, features, 5, 1], expand dimensions of Y to [batch, features, 1, 5]
Than tf.repeat() both tensors so their shapes become [batch, features, 5, 5].
Now you can concatenate X and Y. You will have a tensor of shape [batch, 2*features, 5, 5]. Observe that this way all combinations are built.
Next step is matrix multiplication. tf.matmul() can also do batch matrix multiplication, but I use here tf.einsum() because I want more control over which dimensions are considered as batch.
Full code:
import tensorflow as tf
import numpy as np
batch_size=3
features=6
items=5
x = np.random.uniform(size=[batch_size,features,items])
y = np.random.uniform(size=[batch_size,features,items])
f = np.random.uniform(size=[2*features,features])
x_reps= tf.repeat(x[:,:,:,tf.newaxis], items, axis=3)
y_reps= tf.repeat(y[:,:,tf.newaxis,:], items, axis=2)
xy_conc = tf.concat([x_reps,y_reps], axis=1)
f_i_j = tf.einsum("bfij, fg->bgij", xy_conc,f)
f_i_j = tf.reshape(f_i_j , [batch_size,features,items*items])
I have a matrix X of size (1875, 77). For each column, I want to compute the covariance matrix, i.e., x_1 # x_1.T where x_1 has a shape (1875, 1). Ideally, I want to do this in one-go without a for loop. Is there an easy way to do this?
I was thinking about padding with zeros for each column up and down based on the column index(so x_1 will have 76 zeros columns, x_2 will have one (77, 1) zero column pad on top and 75 zero column pads), but this seems to complicate things more.
You probably want this:
import numpy as np
r, c = 1875, 77
X = np.random.rand(r, c)
covs = X.T[..., None] # X.T[:, None, :]
covs.shape
# (77, 1875, 1875)
This simply performs c number of matrix multiplications, 1 for each column of X. Here X.T[..., None] is of shape (c, r, 1) and X.T[:, None, :] is of shape (c, 1, r) and this makes the matrix multiple between them compatible.
I have a situation similar to the following:
import numpy as np
a = np.random.rand(55, 1, 3)
b = np.random.rand(55, 626, 3)
Here the shapes represent the number of observations, then the number of time slices per observation, then the number of dimensions of the observation at the given time slice. So b is a full representation of 3 dimensions for each of the 55 observations at one new time interval.
I'd like to stack a and b into an array with shape 55, 627, 3. How can one accomplish this in numpy? Any suggestions would be greatly appreciated!
To follow up on Divakar's answer above, the axis argument in numpy is the index of a given dimension within an array's shape. Here I want to stack a and b by virtue of their middle shape value, which is at index = 1:
import numpy as np
a = np.random.rand(5, 1, 3)
b = np.random.rand(5, 100, 3)
# create the desired result shape: 55, 627, 3
stacked = np.concatenate((b, a), axis=1)
# validate that a was appended to the end of b
print(stacked[:, -1, :], '\n\n\n', a.squeeze())
This returns:
[[0.72598529 0.99395887 0.21811998]
[0.9833895 0.465955 0.29518207]
[0.38914048 0.61633291 0.0132326 ]
[0.05986115 0.81354865 0.43589306]
[0.17706517 0.94801426 0.4567973 ]]
[[0.72598529 0.99395887 0.21811998]
[0.9833895 0.465955 0.29518207]
[0.38914048 0.61633291 0.0132326 ]
[0.05986115 0.81354865 0.43589306]
[0.17706517 0.94801426 0.4567973 ]]
A purist might use instead np.all(stacked[:, -1, :] == a.squeeze()) to validate this equivalence. All glory to #Divakar!
Strictly for the curious, the use case for this concatenation is a kind of wonky data preparation pipeline for a Long Short Term Memory Neural Network. In that kind of network, the training data shape should be number_of_observations, number_of_time_intervals, number_of_dimensions_per_observation. I am generating new predictions of each object at a new time interval, so those predictions have shape number_of_observations, 1, number_of_dimensions_per_observation. To visualize the sequence of observations' positions over time, I want to add the new positions to the array of previous positions, hence the question above.
In the case of a matrix mat n x n, i can do the following
sym = 0.5 * (mat + mat.T)
the operation gives the desired result sym[i,j] = sym[j,i]
Suppose we have a 3D array ndarr[i,j,k], where i,j,k 0,1,...n,
then ndarr is n x n x n. The idea is to obtain the following "symmetric" form
nsym[i,j,k] = nsym[j,i,k] using ndarr. I tried this:
import numpy as np
# Generate some random matrix, n = 5
ndarr = np.random.beta(0.1,1,(5,5,5))
# First attempt to symmetrize
sym1 = np.array([0.5*(ndarr[:,:,k]+ndarr[:,:,k].T) for k in range(5)])
The problem here is that sym1[i,j,k] != sym1[j,i,k] as it is required. In fact I obtain sym1[i,j,k] = sym1[i,k,j], symmetric under the exchange of the last two symbols!
# Second attempt
sym2 = 0.5*(ndarr+ndarr.T)
Same problem here and sym2 is symmetric with respect the second index sym2[i,j,k]=sym2[k,j,i].
To resume, the goal is to find a symmetric form for a 3D array with respect to the third index and to preserve the values in the diagonal for the original ndarr[i,i,i].
The problem here is that you're not using the correct transpose:
sym = 0.5 * (ndarr + np.transpose(ndarr, (1, 0, 2)))
By default, np.transpose and the .T property will reverse the order of the axes. In your case, we want to only flip the first two axes: (0,1,2) -> (1,0,2).
EDIT: The reason your first attempt failed is because you were concatenating each symmetrized matrix along the first axis, not the last. It's more clear if you make ndarr with shape (5, 5, 3):
In [16]: sym = np.array([0.5*(ndarr[:,:,k]+ndarr[:,:,k].T) for k in range(3)])
In [17]: sym.shape
Out[17]: (3L, 5L, 5L)
In any case, the version above with np.transpose is cleaner and more efficient.
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.