I have been trying for quite some time to implement my code to run on GPU, however with little success. I would really appreciate someone helping with the implementation.
Let me say a few words about the problem. I have a graph G with N nodes and a distribution mx on each node x. I would like to compute the distance between the distributions for every pair of nodes for all edges. For a given pair, (x,y), I use the code ot.sinkhorn(mx, my, dNxNy) from the python POT package to compute the distance. Again, mx, my are vectors of size Nx and Ny on nodes x and y and dNxNy is a Nx x Ny distance matrix.
Now, I discovered that there is a GPU implementation of this code ot.gpu.sinkhorn(mx, my, dNxNy). However, this is not good enough because I mx, my and dNxNy would need to be uploaded to the GPU at every iteration, which is a massive overhead. So, the idea is to parallelise this for all edges on GPU.
The essence of the code is as follows. mx_all is all the distributions
for i,e in enumerate(G.edges):
W[i] = W_comp(mx_all,dist,e)
def W_comp(mx_all, dist, e):
i = e[0]
j = e[1]
Nx = np.array(mx_all[i][1]).flatten()
Ny = np.array(mx_all[j][1]).flatten()
mx = np.array(mx_all[i][0]).flatten()
my = np.array(mx_all[j][0]).flatten()
dNxNy = dist[Nx,:][:,Ny].copy(order='C')
W = ot.sinkhorn2(mx, my, dNxNy, 1)
Below is a minimal working example. Please ignore everything except the part between dashed === signs.
import ot
import numpy as np
import scipy as sc
def main():
import networkx as nx
#some example graph
G = nx.planted_partition_graph(4, 20, 0.6, 0.3, seed=2)
L = nx.normalized_laplacian_matrix(G)
#this just computes all distributions (IGNORE)
mx_all = []
for i in G.nodes:
mx_all.append(mx_comp(L,1,1,i))
#some random distance matrix (IGNORE)
dist = np.random.randint(5,size=(nx.number_of_nodes(G),nx.number_of_nodes(G)))
# =============================================================================
#this is what needs to be parallelised on GPU
W = np.zeros(nx.Graph.size(G))
for i,e in enumerate(G.edges):
print(i)
W[i] = W_comp(mx_all,dist,e)
return W
def W_comp(mx_all, dist, e):
i = e[0]
j = e[1]
Nx = np.array(mx_all[i][1]).flatten()
Ny = np.array(mx_all[j][1]).flatten()
mx = np.array(mx_all[i][0]).flatten()
my = np.array(mx_all[j][0]).flatten()
dNxNy = dist[Nx,:][:,Ny].copy(order='C')
return ot.sinkhorn2(mx, my, dNxNy,1)
# =============================================================================
#some other functions (IGNORE)
def delta(i, n):
p0 = np.zeros(n)
p0[i] = 1.
return p0
# all neighbourhood densities
def mx_comp(L, t, cutoff, i):
N = np.shape(L)[0]
mx_all = sc.sparse.linalg.expm_multiply(-t*L, delta(i, N))
Nx_all = np.argwhere(mx_all > (1-cutoff)*np.max(mx_all))
return mx_all, Nx_all
if __name__ == "__main__":
main()
Thank you!!
There are some packages, which allow you to run code on your GPU.
You can use one of the following packages:
pyCuda
numba(Pro)
Theano
When you want to use numba, the Python Anaconda distribution is recommended for doing this. Also, Anaconda Accelerate is needed. You can install it using conda install accelerate. In this example, you can see how the usage of the GPU is achieved https://gist.githubusercontent.com/aweeraman/ae6e40f54a924f1f5832081be9521d92/raw/d6775c421aa4fa4c0d582e6c58873499d28b913a/gpu.py .
It's done by adding target='cuda' to the #vectorize decorator. Note the import from numba import vectorize. The vectorize decorator takes the signature of the function that is to be accelerated as input.
Good luck!
Sources:
https://weeraman.com/put-that-gpu-to-good-use-with-python-e5a437168c01
https://www.researchgate.net/post/How_do_I_run_a_python_code_in_the_GPU
Related
I have some discrete data points representing a path and I want to minimize the distance between a trajectory of an object to these path points along with some other constraints. I'm trying out gekko as a tool to solve this problem and for that I made a simple problem by making data points from a parabola and a constraint to the path. My attempt to solve it is
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
import time
#path data points
x_ref = np.linspace(0, 4, num=21)
y_ref = - np.square(x_ref) + 16
#constraint for visualization purposes
x_bound = np.linspace(0, 4, num=10)
y_bound = 1.5*x_bound + 4
def distfunc(x,y,xref,yref,p):
'''
Shortest distance from (x,y) to (xref, yref)
'''
dtemp = []
for i in range(len(xref)):
d = (x-xref[i])**2+(y-yref[i])**2
dtemp.append(dtemp)
min_id = dtemp.index(min(dtemp))
if min_id == 0:
next_id = min_id+1
elif min_id == len(x_ref):
next_id = min_id-1
else:
d2 = (x-xref[min_id-1])**2+(y-yref[min_id-1])**2
d1 = (x-xref[min_id+1])**2+(y-yref[mid_id+1])**2
d_next = [d2, d1]
next_id = min_id + 2*d_next.index(min(d_next)) - 1
n1 = xref[next_id] - xref[min_id]
n2 = yref[next_id] - yref[min_id]
nnorm = p.sqrt(n1**2+n2**2)
n1 = n1 / nnorm
n2 = n2 / nnorm
difx = x-xref[min_id]
dify = y-yref[min_id]
dot = difx*n1 + dify*n2
deltax = difx - dot*n1
deltay = dify - dot*n2
return deltax**2+deltay**2
v_ref = 3
now = time.time()
p = GEKKO(remote=False)
p.time = np.linspace(0,10,21)
x = p.Var(value=0)
y = p.Var(value=16)
vx = p.Var(value=1)
vy = p.Var(value=0)
ax = p.Var(value=0)
ay = p.Var(value=0)
p.options.IMODE = 6
p.options.SOLVER = 3
p.options.WEB = 0
x_refg = p.Param(value=x_ref)
y_refg = p.Param(value=y_ref)
x_refg = p.Param(value=x_ref)
y_refg = p.Param(value=y_ref)
v_ref = p.Const(value=v_ref)
p.Obj(distfunc(x,y,x_refg,y_refg,p))
p.Obj( (p.sqrt(vx**2+vy**2) - v_ref)**2 + ax**2 + ay**2)
p.Equation(x.dt()==vx)
p.Equation(y.dt()==vy)
p.Equation(vx.dt()==ax)
p.Equation(vy.dt()==ay)
p.Equation(y>=1.5*x+4)
p.solve(disp=False, debug=True)
print(f'run time: {time.time()-now}')
plt.plot(x_ref, y_ref)
plt.plot(x_bound, y_bound)
plt.plot(x1.value,x2.value)
plt.show()
This is the result that I get. As you can see, its not exactly the solution that one should expect. For reference to a solution that you may expect, here is what I get using the cost function below
p.Obj((x-x_refg)**2 + (y-y_refg)**2 + ax**2 + ay**2)
However since what I actually wanted is the shortest distance to a path described by these points I expect the distfunc to be closer to what I want since the shortest distance is most likely to some interpolated point. So my question is twofold:
Is this the correct gekko expression/formulation for the objective function?
My other goal is solution speed so is there a more efficient way of expressing this problem for gekko?
You can't define an objective function that changes based on conditions unless you insert logical conditions that are continuously differentiable such as with the if2 or if3 function. Gekko evaluates the symbolic model once and then passes that off to an executable for solution. It only calls the Python model build once because it is compiling the model to efficient byte-code for execution. You can see the model that you created with p.open_folder(). The model file ends in the apm extension: gk_model0.apm.
Model
Constants
i0 = 3
End Constants
Parameters
p1
p2
p3
p4
End Parameters
Variables
v1 = 0
v2 = 16
v3 = 1
v4 = 0
v5 = 0
v6 = 0
End Variables
Equations
v3=$v1
v4=$v2
v5=$v3
v6=$v4
v2>=(((1.5)*(v1))+4)
minimize (((((v1-0.0)-((((((v1-0.0))*((0.2/sqrt(0.04159999999999994))))+(((v2-16.0))&
*((-0.03999999999999915/sqrt(0.04159999999999994))))))*&
((0.2/sqrt(0.04159999999999994))))))^(2))+((((v2-16.0)&
-((((((v1-0.0))*((0.2/sqrt(0.04159999999999994))))+(((v2-16.0))&
*((-0.03999999999999915/sqrt(0.04159999999999994))))))&
*((-0.03999999999999915/sqrt(0.04159999999999994))))))^(2)))
minimize (((((sqrt((((v3)^(2))+((v4)^(2))))-i0))^(2))+((v5)^(2)))+((v6)^(2)))
End Equations
End Model
One strategy is to split your problem into multiple optimization problems that are all minimal time problems where you navigate to the first way-point and then re-initialize the problem to navigate to the second way-point, and so on. If you want to preserve momentum and anticipate the turning then you'll need to use more advanced methods such as shown in the Pigeon / Eagle tracking problem (see source files) or similar to a trajectory optimization with UAVs or HALE UAVs (see references below).
Martin, R.A., Gates, N., Ning, A., Hedengren, J.D., Dynamic Optimization of High-Altitude Solar Aircraft Trajectories Under Station-Keeping Constraints, Journal of Guidance, Control, and Dynamics, 2018, doi: 10.2514/1.G003737.
Gates, N.S., Moore, K.R., Ning, A., Hedengren, J.D., Combined Trajectory, Propulsion and Battery Mass Optimization for Solar-Regenerative High-Altitude Long Endurance Unmanned Aircraft, AIAA Science and Technology Forum (SciTech), 2019.
Since, random.choices() is not available in python 3.5, is there any alternative method for this function?
Copied the code from python 3.6 to as suggested by Mark Dickinson
from itertools import accumulate as _accumulate, repeat as _repeat
from bisect import bisect as _bisect
import random
def choices(population, weights=None, *, cum_weights=None, k=1):
"""Return a k sized list of population elements chosen with replacement.
If the relative weights or cumulative weights are not specified,
the selections are made with equal probability.
"""
n = len(population)
if cum_weights is None:
if weights is None:
_int = int
n += 0.0 # convert to float for a small speed improvement
return [population[_int(random.random() * n)] for i in _repeat(None, k)]
cum_weights = list(_accumulate(weights))
elif weights is not None:
raise TypeError('Cannot specify both weights and cumulative weights')
if len(cum_weights) != n:
raise ValueError('The number of weights does not match the population')
bisect = _bisect
total = cum_weights[-1] + 0.0 # convert to float
hi = n - 1
return [population[bisect(cum_weights, random.random() * total, 0, hi)]
for i in _repeat(None, k)]
now you can use choices function peacefully!
I wasn't falling head over heels in love with Venkatesh Mondi's answer, but it's totally functional.
import random
def random_choices(population, weights=None, k=1):
"""
This function achieves the same result.
If weights is not provided, it returns a list of k random choices made using random.
choice with no weighting. If weights is provided, it first normalizes the weights by
dividing each weight by the sum of all the weights, and then returns a list of k random
choices made using random.choice with the normalized weights.
"""
if weights is None:
return [random.choice(population) for _ in range(k)]
total = sum(weights)
weights = [weight/total for weight in weights]
cumulative_weights = [sum(weights[:i+1]) for i in range(len(weights))]
return [population[cumulative_weights.index(next(x for x in cumulative_weights if x >= random.random()))] for _ in range(k)]
#note you may also do something like:
return population[weights.index(random.choice(weights))]
print(random_choices(population, weights=weights, k=1)[0])
here's an alternative i have been using for a while now, I'm not sure if someone wrote it first or if i wrote it though.
I have a matrix (really a loaded image) in which every element is a L2 distance from some unknown center point.
Here is a trivial example
A = [1.4142 1.0000 1.4142 2.2361]
[1.0000 0.0000 1.0000 2.0000]
[1.4142 1.0000 1.4142 2.2361]
In this case, the center is obviously at coordinate (1,1) (index A[1,1] in a 0-indexed matrix or 2D array).
However, in the case where my centers are not constrained to be integer indices, it's no longer as obvious. For example, given this matrix B, where is my center coordinate?
B = [3.0292 1.9612 2.8932 5.8252]
[1.2292 0.1612 1.0932 4.0252]
[1.4292 0.3612 1.2932 4.2252]
How would you find that the answer in this case is at row 1.034 and column 1.4?
I am aware of the trilateration solution (having provided MATLAB code to visualize that in 3D previously), but is there a more efficient way (e.g. one without a matrix inversion)?
This question is sort of language agnostic, as I am looking more for algorithmic help. If you could stick to MATLAB, Python, or C++ though in a solution, that would be great ;-).
While having no experience with similar tasks, i read some stuff and also tried something.
When unfamiliar with this topic it's hard to grasp it seems and all those resources i found are a bit chaotic.
Still unclear in regards to theory for me:
is the problem as stated above a convex-optimization problem (local-minimum = global-minimum; would mean access to powerful solvers!)
there are much more resources about more generic problems (Sensor Network
Localization), which are non-convex and where extremely complex methods have been developed
is your trilateration-approach able to exploit > 3 points (trilateration vs. multilateration; at least this code does not seem like it can which means: bad performance with noise!)
Here some example code with two approaches:
A: Convex-optimization: SOCP-Relaxation
Follows SECOND-ORDER CONE PROGRAMMING RELAXATION OF SENSOR NETWORK LOCALIZATION
Not impressive performance, but should be powerful as approximation for big-data
Guaranteed global-optimum for this relaxation!
Implemented with cvxpy
B: Nonlinear-programming optimization
Implemented using scipy.optimize
Pretty much perfect in my synthetic experiments; even good results in noisy case; despite the fact we are using numerical-differentiation (automatic-diff hard to use here)
Some additional remark:
Your example B surely has some (pretty bad) noise or some other problem in my opinion, as my approaches are completely off; while especially approach B shines for my synthetic-data (at least that's my impression)
Code:
import numpy as np
import cvxpy as cvx
from scipy.spatial.distance import cdist
from scipy.optimize import minimize
np.random.seed(1)
""" Create noise-free (not anymore!) fake-problem """
real_x = np.random.random(size=2) * 3
M, N = 5, 10
NOISE_DISTS = 0.1
pos = np.array([(i,j) for i in range(M) for j in range(N)]) # ugly -> tile/repeat/stack
real_x_stacked = np.vstack([real_x for i in range(pos.shape[0])])
Y = cdist(pos, real_x[np.newaxis])
Y += np.random.normal(size=Y.shape)*NOISE_DISTS # Let's add some noise!
print('-----')
print('PROBLEM')
print('-------')
print('real x: ', real_x)
print('dist mat: ', np.round(Y,3).T)
""" Helper """
def cost(x, Y, pos):
res = np.linalg.norm(pos - x, ord=2, axis=1) - Y.ravel()
return np.linalg.norm(res, 2)
print('cost with real_x (check vs. noisy): ', cost(real_x, Y, pos))
""" SOLVER SOCP """
def solve_socp_relax(pos, Y):
x = cvx.Variable(2)
y = cvx.Variable(pos.shape[0])
fake_stack = [x for i in range(pos.shape[0])] # hacky
objective = cvx.sum_entries(cvx.norm(y - Y))
x_stacked = cvx.reshape(cvx.vstack(*fake_stack), pos.shape[0], 2) # hacky
constraints = [cvx.norm(pos - x_stacked, 2, axis=1) <= y]
problem = cvx.Problem(cvx.Minimize(objective), constraints)
problem.solve(solver=cvx.ECOS, verbose=False)
return x.value.T
""" SOLVER NLP """
def solve_nlp(pos, Y):
sol = minimize(cost, np.zeros(pos.shape[1]), args=(Y, pos), method='BFGS')
# print(sol)
return sol.x
""" TEST """
print('-----')
print('SOLVE')
print('-----')
socp_relax_sol = solve_socp_relax(pos, Y)
print('SOCP RELAX SOL: ', socp_relax_sol)
nlp_sol = solve_nlp(pos, Y)
print('NLP SOL: ', nlp_sol)
Output:
-----
PROBLEM
-------
real x: [ 1.25106601 2.16097348]
dist mat: [[ 2.444 1.599 1.348 1.276 2.399 3.026 4.07 4.973 6.118 6.746
2.143 1.149 0.412 0.766 1.839 2.762 3.851 4.904 5.734 6.958
2.377 1.432 0.856 1.056 1.973 2.843 3.885 4.95 5.818 6.84
2.711 2.015 1.689 1.939 2.426 3.358 4.385 5.22 6.076 6.97
3.422 3.153 2.759 2.81 3.326 4.162 4.734 5.627 6.484 7.336]]
cost with real_x (check vs. noisy): 0.665125233772
-----
SOLVE
-----
SOCP RELAX SOL: [[ 1.95749275 2.00607253]]
NLP SOL: [ 1.23560791 2.16756168]
Edit: Further speedup can be achieved (especially in large-scale) in using nonlinear-least-squares instead of the more general NLP-approach! My results are still the same (as expected if the problem would be convex). Timings between NLP/NLS can look like 9 vs. 0.5 seconds!
This is my recommended method!
def solve_nls(pos, Y):
def res(x, Y, pos):
return np.linalg.norm(pos - x, ord=2, axis=1) - Y.ravel()
sol = least_squares(res, np.zeros(pos.shape[1]), args=(Y, pos), method='lm')
# print(sol)
return sol.x
Especially the second-approach (NLP) will also run for much bigger instances (cvxpy's overhead hurts; that's not a downside of the SOCP-solver which should scale much much better!).
Here some output for M, N = 500, 1000 with some more noise:
-----
PROBLEM
-------
real x: [ 12.51066014 21.6097348 ]
dist mat: [[ 24.706 23.573 23.693 ..., 1090.29 1091.216
1090.817]]
cost with real_x (check vs. noisy): 353.354267797
-----
SOLVE
-----
NLP SOL: [ 12.51082419 21.60911561]
used: 5.9552763315495625 # SECONDS
So in my experiments it works, but i won't give any global-convergence guarantees or reconstruction-guarantees (still missing some theory).
At first i though about using the global optimum of the relaxed-SOCP-problem as initial-point in the NLP-solver, but i did not find any example where this is needed!
Some just-for-fun visuals using:
M, N = 20, 30
NOISE_DISTS = 0.2
...
import matplotlib.pyplot as plt
plt.imshow(Y.reshape(M, N), cmap='viridis', interpolation='none')
plt.colorbar()
plt.scatter(nlp_sol[1], nlp_sol[0], color='red', s=20)
plt.xlim((0, N))
plt.ylim((0, M))
plt.show()
And some super noisy case (nice performance!):
M, N = 50, 100
NOISE_DISTS = 5
-----
PROBLEM
-------
real x: [ 12.51066014 21.6097348 ]
dist mat: [[ 22.329 18.745 27.588 ..., 94.967 80.034 91.206]]
cost with real_x (check vs. noisy): 354.527196716
-----
SOLVE
-----
NLP SOL: [ 12.44158986 21.50164637]
used: 0.01050068340320306
If I understand correctly, you have a matrix A, where A[i,j] holds the distance from (i,j) to some unknown point (y,x). You could find (y,x) like this:
Square each element of A, to make a matrix B say.
We then want to find (y,x) so
(y-i)*(y-i) + (x-j)*(x-j) = B[i,j]
Subtracting each equation from the 0,0 equation and rearranging:
2*i*y + 2*j*x = B[0,0] + i*i + j*j - B[i,j]
This can be solved by linear least squares. Note that since there are 2 unknowns, the matix inversion (better, factorisation) involved will be on a 2x2 matrix and so not time consuming. You could indeed, given just the dimensions of A, work out the required matrix and its inverse analytically.
I have a very large absorbing Markov chain. I want to obtain the fundamental matrix of this chain to calculate the expected number of steps before absortion. From this question I know that this can be calculated by the equation
(I - Q)t=1
which can be obtained by using the following python code:
def expected_steps_fast(Q):
I = numpy.identity(Q.shape[0])
o = numpy.ones(Q.shape[0])
numpy.linalg.solve(I-Q, o)
However, I would like to calculate it using some kind of iterative method similar to the power iteration method used for calculate the PageRank. This method would allow me to calculate an approximation to the expected number of steps before absortion in a mapreduce-like system.
¿Does something similar exist?
If you have a sparse matrix, check if scipy.spare.linalg.spsolve works. No guarantees about numerical robustness, but at least for trivial examples it's significantly faster than solving with dense matrices.
import networkx as nx
import numpy as np
import scipy.sparse as sp
import scipy.sparse.linalg as spla
def example(n):
"""Generate a very simple transition matrix from a directed graph
"""
g = nx.DiGraph()
for i in xrange(n-1):
g.add_edge(i+1, i)
g.add_edge(i, i+1)
g.add_edge(n-1, n)
g.add_edge(n, n)
m = nx.to_numpy_matrix(g)
# normalize rows to ensure m is a valid right stochastic matrix
m = m / np.sum(m, axis=1)
return m
A = sp.csr_matrix(example(2000)[:-1,:-1])
Ad = np.array(A.todense())
def sp_solve(Q):
I = sp.identity(Q.shape[0], format='csr')
o = np.ones(Q.shape[0])
return spla.spsolve(I-Q, o)
def dense_solve(Q):
I = numpy.identity(Q.shape[0])
o = numpy.ones(Q.shape[0])
return numpy.linalg.solve(I-Q, o)
Timings for sparse solution:
%timeit sparse_solve(A)
1000 loops, best of 3: 1.08 ms per loop
Timings for dense solution:
%timeit dense_solve(Ad)
1 loops, best of 3: 216 ms per loop
Like Tobias mentions in the comments, I would have expected other solvers to outperform the generic one, and they may for very large systems. For this toy example, the generic solve seems to work well enough.
I arraived to this answer thanks to #tobias-ribizel's suggestion of using the Neumann series. If we part from the following equation:
Using the Neumann series:
If we multiply each term of the series by the vector 1 we could operate separately over each row of the matrix Q and approximate successively with:
This is the python code I use to calculate this:
def expected_steps_iterative(Q, n=10):
N = Q.shape[0]
acc = np.ones(N)
r_k_1 = np.ones(N)
for k in range(1, n):
r_k = np.zeros(N)
for i in range(N):
for j in range(N):
r_k[i] += r_k_1[j] * Q[i, j]
if np.allclose(acc, acc+r_k, rtol=1e-8):
acc += r_k
break
acc += r_k
r_k_1 = r_k
return acc
And this is the code using Spark. This code expects that Q is a RDD where each row is a tuple (row_id, dict of weights for that row of the matrix).
def expected_steps_spark(sc, Q, n=10):
def dict2np(d, sz):
vec = np.zeros(sz)
for k, v in d.iteritems():
vec[k] = v
return vec
sz = Q.count()
acc = np.ones(sz)
x = {i:1.0 for i in range(sz)}
for k in range(1, n):
bc_x = sc.broadcast(x)
x_old = x
x = Q.map(lambda (u, ol): (u, reduce(lambda s, j: s + bc_x.value[j]*ol[j], ol, 0.0)))
x = x.collectAsMap()
v_old = dict2np(x_old, sz)
v = dict2np(x, sz)
acc += v
if np.allclose(v, v_old, rtol=1e-8):
break
return acc
I would like to implement to implement the Dirichlet process example referenced in
Implementing Dirichlet processes for Bayesian semi-parametric models (source: here) in PyMC 3.
In the example the stick-breaking probabilities are computed using the pymc.deterministic
decorator:
v = pymc.Beta('v', alpha=1, beta=alpha, size=N_dp)
#pymc.deterministic
def p(v=v):
""" Calculate Dirichlet probabilities """
# Probabilities from betas
value = [u*np.prod(1-v[:i]) for i,u in enumerate(v)]
# Enforce sum to unity constraint
value[-1] = 1-sum(value[:-1])
return value
z = pymc.Categorical('z', p, size=len(set(counties)))
How would you implement this in PyMC 3 which is using Theano for the gradient computation?
edit:
I tried the following solution using the theano.scan method:
with pm.Model() as mod:
conc = Uniform('concentration', lower=0.5, upper=10)
v = Beta('v', alpha=1, beta=conc, shape=n_dp)
p, updates = theano.scan(fn=lambda stick, idx: stick * t.prod(1 - v[:idx]),
outputs_info=None,
sequences=[v, t.arange(n_dp)])
t.set_subtensor(p[-1], 1 - t.sum(p[:-1]))
category = Categorical('category', p, shape=n_algs)
sd = Uniform('precs', lower=0, upper=20, shape=n_dp)
means = Normal('means', mu=0, sd=100, shape=n_dp)
points = Normal('obs',
means[category],
sd=sd[category],
observed=data)
step1 = pm.Slice([conc, v, sd, means])
step3 = pm.ElemwiseCategoricalStep(var=category, values=range(n_dp))
trace = pm.sample(2000, step=[step1, step3], progressbar=True)
Which sadly is really slow and does not obtain the original parameters of the synthetic data.
Is there a better solution and is this even correct?
Not sure I have a good answer but perhaps this could be sped up by instead using a theano blackbox op which allows you to write a distribution (or deterministic) in python code. E.g.: https://github.com/pymc-devs/pymc3/blob/master/pymc3/examples/disaster_model_arbitrary_deterministic.py