I have 3-dimensional tensor ("tensor3" -- an array of matrices), and I'd like to compute the determinant (theano.sandbox.linalg.det) of each matrix. Is there a way to compute each determinant without using theano.scan? When I try calling det directly on the tensor I get the error
3-dimensional array given. Array must be two-dimensional.
But I read that scan is slow and doesn't parallelize well, and that one should use only tensor operations if possible. Is that so? Can I avoid using scan in this case?
I see 3 possibilities:
If you know before compiling the Theano function the number of matrix in the tensor3 variable, you could use the split() op or just call det() on all matrix in the tensor3.
If you don't know the shape, you can make your own op, that will loop over the input and call the numpy fct. See for an example on how to make an op.
Use scan. It is easy to use it for this case. See this example, just change the call from tensordot to det().
Related
I'm currently working to diagonalize a 5000x5000 Hermitian matrix, and I find that when I use Julia's eigen function in the LinearAlgebra module, which produces both the eigenvalues and eigenvectors, I get different results for the eigenvectors compared to when I solve the problem using numpy's np.linalg.eigh function. I believe both of them use BLAS, but I'm not sure what else they may be using that is different.
Has anyone else experienced this/knows what is going on?
numpy.linalg.eigh(a, UPLO='L') is a different algorithm. It assumes the matrix is symmetric and takes the lower triangular matrix (as a default) to more efficiently compute the decomposition.
The equivalent to Julia's LinearAlgebra.eigen() is numpy.linalg.eig. You should get the same result if you turn your matrix in Julia into a Symmetric(A, uplo=:L) matrix before feeding it into LinearAlgebra.eigen().
Check out numpy's docs on eig and eigh. Whilst Julia's standard LinearAlgebra capabilities are here. If you go down to the special matrices sections, it details what special methods it uses depending on the type of special matrix thanks to multiple dispatch.
I am trying to understand how to use the "fftw_plan_dft_r2c_2d". First, I generate a matrix with a non symmetric gaussian (sigma_x != sigma_y) and I make the fft using fftw_plan_dft_c2c_2D. But it requires a temporary fftw_complexe array where I just copy the gaussian matrix and set the imaginary part to 0.0.
In order to save some memory, I am trying to avoid the temporary fftw_complexe and use directly the "fftw_plan_dft_r2c_2D" expecting the same "half" result.
It looks like I missing something.. can anyone help please ?
thanks,
You need to shift the center frequency for the inverse transformation as described here for numpy:
https://numpy.org/doc/stable/reference/generated/numpy.fft.fftshift.html
I was wondering if there was a direct way of computing the iteration matrix for nth Linear Block Gauss Seidel iteration within OpenMDAO?
thank you
If I understand you correctly, you are referring to the matrix-form of the Gauss Seidel algorithm where you take Ax=b, and break A up into the Diagonal (D), Lower (L) and Upper (U) parts, then use those parts to compute the next iterate.
Specifically you compute [D-L]^-1. This, I believe is what you are referring to as the "iteration matrix" (I am not familiar with this terminology, but based on the algorithm I'm comfortable making an educated guess).
This formulation of the algorithm is useful to think about and a simple way to implement it, but OpenMDAO takes a different approach. The LBGS algorithm implemented in OpenMDAO is set up to work in a matrix-free manner. That means it only interacts with the linear operator methods solve_linear and apply_linear and never explicitly assembles the A matrix at all. Hence there isn't an opportunity to split A up into D, L, U.
Depending on the way you constructed the model, the A matrix you would need might or might not be there at all because OpenMDAO is capable of working in a completely matrix free context. However, if all of your components use the compute_partials or linearize methods to provide partial derivatives then the data you would need for the A matrix does exist in memory.
You'll have to dig for it a bit, and ironically the best place to see how to do that is in the direct solver which does actually require the matrix be formed to compute a factorization.
Also, in that code you'll see a function can iteratively call the linear operator to construct a dense matrix even if the underlying components don't provide their partials directly. Please note that this approach for assembling the matrix is extremely slow and is not recommended for normal operations.
I am getting a function (a learned dynamical system) through a neural network and want to pass it to JiTCODE to calculate trajectories, Lyapunov exponents, etc. As per the JiTCODE documentation, the function f has to be a symbolic function. Is there any way to change this since ultimately JiTCODE is going to lambdify the symbolic function?
Basically, this is what I'm doing right now:
# learns derviates from the Neural net model
# returns an array of numbers [\dot{x},\dot{y}] for input [x,y]
learned_fn = lambda t, y0: NN_model(t, y0)
ODE = jitcode_lyap(learned_fn, n_lyap=2)
ODE.set_integrator("vode")
First beware that JiTCODE does not take regular functions like your learned_fn as an input. It takes either iterables of symbolic expressions or generator functions returning symbolic expressions. This is why your example code will likely produce an error.
What you are asking for
You can “inject” any derivative with the right signature into JiTCODE by changing the f property and telling it that it failed compiling the actual derivative. Here is a minimal example doing this:
from jitcode import jitcode, y
ODE = jitcode([0])
ODE.f = lambda t,y: y[0]
ODE.compile_attempt = False
ODE.set_integrator("dopri5")
ODE.set_initial_value([1],0.0)
for time in range(30):
print(time,*ODE.integrate(time))
Why you probably do not want to do this
Ignoring Lyapunov exponents for a second, the entire point of JiTCODE is to hard-code your derivative for you and pass it to SciPy’s ode or solve_ivp who perform the actual integration. Thus the above example code is just an overly complicated way of passing a function to one SciPy’s standard integrators (here ode), with no advantage. If your NN_model is very efficiently implemented in the first place, you may not even gain a speed boost from JiTCODE’s auto-compilation.
The main reason to use JiTCODE’s Lyapunov-exponent capabilities is that it automatically obtains the Jacobian and the ODE for the tangent-vector evolution (needed for the Benettin method) from the symbolic representation of the derivative. Without a symbolic input, it cannot possibly do this. You could theoretically inject a tangent-vector ODE as well, but then again you would leave little for JiTCODE to do and you would probably better off using SciPy’s ode or solve_ivp directly.
What you probably need
If you want to use JiTCODE, you need to write a small piece of code that translates the output of your neural-network training to a symbolic representation of your ODE as needed by JiTCODE. This is probably much less scary than it sounds. You just need to obtain the trained coefficients and insert it in the equations of the general form of the neural network.
If you are lucky and your NN_model fully supports duck typing (and ), you may do something like this:
from jitcode import t,y
n = 10 # dimension of your ODE
NN_input = [y(i) for i in range(n)]
learned_fn = NN_model(t,NN_input)[1]
The idea is that you feed NN_model once with abstract symbolic input (t and NN_input). NN_model then once acts on this abstract input providing you an abstract result (here you need the duck-typing support). If I interpreted the output of your NN_model correctly, the second component of this result should be the abstract derivative as required by JiTCODE as an input.
Note that your NN_model appears to expect dimensions to be indices, but JiTCODE’s y expects dimensions to be function arguments. Thus you cannot just choose NN_input = y, but you have to transform it as above.
To quote directly from the linked documentation
JiTCODE takes an iterable (or generator function or dictionary) of symbolic expressions, which it translates to C code, compiles on the fly,
so there is no lambdification going on, the function is parsed, not just evaluated.
But in general that should be no problem, you just use the JITCODE provided symbolic vector y and symbol t instead of the function arguments t,y of the right side of the ODE.
I'm trying to manipulate individual weights of different neural nets to see how their performance degrades. As part of these experiments, I'm required to sample randomly from their weight tensors, which I've come to understand as sampling with replacement (in the statistical sense). However, since it's high-dimensional, I've been stumped by how to do this in a fair manner. Here are the approaches and research I've put into considering this problem:
This was previously implemented by selecting a random layer and then selecting a random weight in that layer (ignore the implementation of picking a random weight). Since layers are different sizes, we discovered that weights were being sampled unevenly.
I considered what would happen if we sampled according to the numpy.shape of the tensor; however, I realize now that this encounters the same problem as above.
Consider what happens to a rank 2 tensor like this:
[[*, *, *],
[*, *, *, *]]
Selecting a row randomly and then a value from that row results in an unfair selection. This method could work if you're able to assert that this scenario never occurs, but it's far from a general solution.
Note that this possible duplicate actually implements it in this fashion.
I found people suggesting flattening the tensor and use numpy.random.choice to select randomly from a 1D array. That's a simple solution, except I have no idea how to invert the flattened tensor back into its original shape. Further, flattening millions of weights would be a somewhat slow implementation.
I found someone discussing tf.random.multinomial here, but I don't understand enough of it to know whether it's applicable or not.
I ran into this paper about resevoir sampling, but again, it went over my head.
I found another paper which specifically discusses tensors and sampling techniques, but it went even further over my head.
A teammate found this other paper which talks about random sampling from a tensor, but it's only for rank 3 tensors.
Any help understanding how to do this? I'm working in Python with Keras, but I'll take an algorithm in any form that it exists. Thank you in advance.
Before I forget to document the solution we arrived at, I'll talk about the two different paths I see for implementing this:
Use a total ordering on scalar elements of the tensor. This is effectively enumerating your elements, i.e. flattening them. However, you can do this while maintaining the original shape. Consider this pseudocode (in Python-like syntax):
def sample_tensor(tensor, chosen_index: int) -> Tuple[int]:
"""Maps a chosen random number to its index in the given tensor.
Args:
tensor: A ragged-array n-tensor.
chosen_index: An integer in [0, num_scalar_elements_in_tensor).
Returns:
The index that accesses this element in the tensor.
NOTE: Entirely untested, expect it to be fundamentally flawed.
"""
remaining = chosen_index
for (i, sub_list) in enumerate(tensor):
if type(sub_list) is an iterable:
if |sub_list| > remaining:
remaining -= |sub_list|
else:
return i joined with sample_tensor(sub_list, remaining)
else:
if len(sub_list) <= remaining:
return tuple(remaining)
First of all, I'm aware this isn't a sound algorithm. The idea is to count down until you reach your element, with bookkeeping for indices.
We need to make crucial assumptions here. 1) All lists will eventually contain only scalars. 2) By direct consequence, if a list contains lists, assume that it also doesn't contain scalars at the same level. (Stop and convince yourself for (2).)
We also need to make a critical note here too: We are unable to measure the number of scalars in any given list, unless the list is homogeneously consisting of scalars. In order to avoid measuring this magnitude at every point, my algorithm above should be refactored to descend first, and subtract later.
This algorithm has some consequences:
It's the fastest in its entire style of approaching the problem. If you want to write a function f: [0, total_elems) -> Tuple[int], you must know the number of preceding scalar elements along the total ordering of the tensor. This is effectively bound at Theta(l) where l is the number of lists in the tensor (since we can call len on a list of scalars).
It's slow. It's too slow compared to sampling nicer tensors that have a defined shape to them.
It begs the question: can we do better? See the next solution.
Use a probability distribution in conjunction with numpy.random.choice. The idea here is that if we know ahead of time what the distribution of scalars is already like, we can sample fairly at each level of descending the tensor. The hard problem here is building this distribution.
I won't write pseudocode for this, but lay out some objectives:
This can be called only once to build the data structure.
The algorithm needs to combine iterative and recursive techniques to a) build distributions for sibling lists and b) build distributions for descendants, respectively.
The algorithm will need to map indices to a probability distribution respective to sibling lists (note the assumptions discussed above). This does require knowing the number of elements in an arbitrary sub-tensor.
At lower levels where lists contain only scalars, we can simplify by just storing the number of elements in said list (as opposed to storing probabilities of selecting scalars randomly from a 1D array).
You will likely need 2-3 functions: one that utilizes the probability distribution to return an index, a function that builds the distribution object, and possibly a function that just counts elements to help build the distribution.
This is also faster at O(n) where n is the rank of the tensor. I'm convinced this is the fastest possible algorithm, but I lack the time to try to prove it.
You might choose to store the distribution as an ordered dictionary that maps a probability to either another dictionary or the number of elements in a 1D array. I think this might be the most sensible structure.
Note that (2) is truly the same as (1), but we pre-compute knowledge about the densities of the tensor.
I hope this helps.