I would like to apply a periodic boundary condition in FEniCS for Julia language but all examples I have found are in either C++ or Python. How to create a periodic boundary condition using Julia? It seems difficult because Julia does not have classes.
here is a minimal example:
using FEniCS
using PyCall
length=2.2
height=0.41
channel = Rectangle(Point([0.0, 0.0]), Point([length, height]))
domain = channel
mesh = generate_mesh(domain, 64)
# insert function here for PeriodicBoundarycondition
Q = FunctionSpace(mesh, "P", 1,constrained_domain=#the function that i am looking for)
I looked at the julia code that makes up FEniCS, the example of periodic boundary condition on the fenics page, and some of my old python codes for fenics and it inspired me to write this:
using FEniCS
using PyCall
#pyimport fenics
py"""
from dolfin import *
from mshr import *
length=2.2
height=0.41
channel = Rectangle(Point([0.0, 0.0]), Point([length, height]))
domain = channel
mesh = generate_mesh(domain, 64)
subdomains = MeshFunction("size_t", mesh, 1)
subdomains.set_all(0)
class Wall(SubDomain):
def inside(self,x,on_boundary):
return (near(x[1],height) or near(x[1],height)) and on_boundary
wall=Wall()
class PeriodicBoundary(SubDomain):
# Left boundary is "target domain" G
def inside(self, x, on_boundary):
return bool(x[0] < DOLFIN_EPS and x[0] > -DOLFIN_EPS and on_boundary)
# Map right boundary (H) to left boundary (G)
def map(self, x, y):
y[0] = x[0] - length
y[1] = x[1]
pbc=PeriodicBoundary()
"""
Q=FunctionSpace(fenics.VectorFunctionSpace(py"mesh", "P", 1,constrained_domain=py"pbc"))
the solution is not optimal as it just does the entire thing in python but I think I will have to live with it.
Related
I am trying to calculate the gradient of a functional of a stochastic differential equation (SDE) solution given a specific realization of the noise. I can successfully calculate these gradients if I leave the noise unspecified, as shown in DiffEqFlux.jl: Using Other Differential Equations. I can also successfully obtain the solution to my SDE for a specific noise realization, like shown in DifferentialEquations.jl: NoiseWrapper Example. When I try and put the two together, though, the code returns an error.
Here is a minimal working example adapted from the two separate examples referenced above:
using StochasticDiffEq, DiffEqBase, DiffEqNoiseProcess, DiffEqSensitivity, Zygote
function lotka_volterra(du,u,p,t)
x, y = u
α, β, δ, γ = p
du[1] = dx = α*x - β*x*y
du[2] = dy = -δ*y + γ*x*y
end
function lotka_volterra_noise(du,u,p,t)
du[1] = 0.1u[1]
du[2] = 0.1u[2]
end
dt = 1//2^(4)
u0 = [1.0,1.0]
p = [2.2, 1.0, 2.0, 0.4]
prob1 = SDEProblem(lotka_volterra,lotka_volterra_noise,u0,(0.0,10.0),p)
sol1 = solve(prob1,EM(),dt=dt,save_noise=true)
W2 = NoiseWrapper(sol1.W)
prob2 = SDEProblem(lotka_volterra,lotka_volterra_noise,u0,(0.0,10.0),p,noise=W2)
sol2 = solve(prob2,EM(),dt=dt)
function predict_sde1(p)
Array(concrete_solve(remake(prob1,p=p),EM(),dt=dt,sensealg=ForwardDiffSensitivity(),saveat=0.1))
end
loss_sde1(p) = sum(abs2,x-1 for x in predict_sde1(p))
loss_sde1(p)
# This gradient is successfully calculated
Zygote.gradient(loss_sde1,p)
function predict_sde2(p)
W2 = NoiseWrapper(sol1.W)
Array(concrete_solve(remake(prob2,p=p,noise=W2),EM(),dt=dt,sensealg=ForwardDiffSensitivity(),saveat=0.1))
end
loss_sde2(p) = sum(abs2,x-1 for x in predict_sde2(p))
# This loss is successfully calculated
loss_sde2(p)
# This gradient calculation raises and error
Zygote.gradient(loss_sde2,p)
The error I get at the end of running this code is
TypeError: in setfield!, expected Float64, got ForwardDiff.Dual{Nothing,Float64,4}
Stacktrace:
[1] setproperty! at ./Base.jl:21 [inlined]
...
followed by an interminable conclusion to the stacktrace (I can post it if you think it would be helpful, but since it's longer than the rest of this question I'd rather not clutter things up off the bat).
Is calculating gradients for SDE problems with specified noise realizations not currently supported, or am I just not making the appropriate function calls? I could easily believe the latter, since it was a bit of a struggle just to get to the point where the working parts of the above code worked, but I couldn't find any clue as to what I had incorrectly supplied after stepping through this code with the Juno debugger.
As a StackOverflow solution, you can use ForwardDiffSensitivity(convert_tspan=false) to work around this. Working code:
using StochasticDiffEq, DiffEqBase, DiffEqNoiseProcess, DiffEqSensitivity, Zygote
function lotka_volterra(du,u,p,t)
x, y = u
α, β, δ, γ = p
du[1] = dx = α*x - β*x*y
du[2] = dy = -δ*y + γ*x*y
end
function lotka_volterra_noise(du,u,p,t)
du[1] = 0.1u[1]
du[2] = 0.1u[2]
end
dt = 1//2^(4)
u0 = [1.0,1.0]
p = [2.2, 1.0, 2.0, 0.4]
prob1 = SDEProblem(lotka_volterra,lotka_volterra_noise,u0,(0.0,10.0),p)
sol1 = solve(prob1,EM(),dt=dt,save_noise=true)
W2 = NoiseWrapper(sol1.W)
prob2 = SDEProblem(lotka_volterra,lotka_volterra_noise,u0,(0.0,10.0),p,noise=W2)
sol2 = solve(prob2,EM(),dt=dt)
function predict_sde1(p)
Array(concrete_solve(remake(prob1,p=p),EM(),dt=dt,sensealg=ForwardDiffSensitivity(convert_tspan=false),saveat=0.1))
end
loss_sde1(p) = sum(abs2,x-1 for x in predict_sde1(p))
loss_sde1(p)
# This gradient is successfully calculated
Zygote.gradient(loss_sde1,p)
function predict_sde2(p)
Array(concrete_solve(prob2,EM(),prob2.u0,p,dt=dt,sensealg=ForwardDiffSensitivity(convert_tspan=false),saveat=0.1))
end
loss_sde2(p) = sum(abs2,x-1 for x in predict_sde2(p))
# This loss is successfully calculated
loss_sde2(p)
# This gradient calculation raises and error
Zygote.gradient(loss_sde2,p)
As a developer... this isn't a nice solution and our default should be better here. I'll work on this. You can track the development here https://github.com/JuliaDiffEq/DiffEqSensitivity.jl/issues/204. It'll probably get solved in an hour or so.
Edit: The fix is released and your original code works.
I'm implementing an RBF network by using some beginer examples from Pytorch Website. I have a problem when implementing the kernel bandwidth differentiation for the network. Also, Iwould like to know whether my attempt ti implement the idea is fine. This is a code sample to reproduce the issue. Thanks
# -*- coding: utf-8 -*-
import torch
from torch.autograd import Variable
def kernel_product(x,y, mode = "gaussian", s = 1.):
x_i = x.unsqueeze(1)
y_j = y.unsqueeze(0)
xmy = ((x_i-y_j)**2).sum(2)
if mode == "gaussian" : K = torch.exp( - xmy/s**2) )
elif mode == "laplace" : K = torch.exp( - torch.sqrt(xmy + (s**2)))
elif mode == "energy" : K = torch.pow( xmy + (s**2), -.25 )
return torch.t(K)
class MyReLU(torch.autograd.Function):
"""
We can implement our own custom autograd Functions by subclassing
torch.autograd.Function and implementing the forward and backward passes
which operate on Tensors.
"""
#staticmethod
def forward(ctx, input):
"""
In the forward pass we receive a Tensor containing the input and return
a Tensor containing the output. ctx is a context object that can be used
to stash information for backward computation. You can cache arbitrary
objects for use in the backward pass using the ctx.save_for_backward method.
"""
ctx.save_for_backward(input)
return input.clamp(min=0)
#staticmethod
def backward(ctx, grad_output):
"""
In the backward pass we receive a Tensor containing the gradient of the loss
with respect to the output, and we need to compute the gradient of the loss
with respect to the input.
"""
input, = ctx.saved_tensors
grad_input = grad_output.clone()
grad_input[input < 0] = 0
return grad_input
dtype = torch.cuda.FloatTensor
N, D_in, H, D_out = 64, 1000, 100, 10
# Create random Tensors to hold input and outputs, and wrap them in Variables.
x = Variable(torch.randn(N, D_in).type(dtype), requires_grad=False)
y = Variable(torch.randn(N, D_out).type(dtype), requires_grad=False)
# Create random Tensors for weights, and wrap them in Variables.
w1 = Variable(torch.randn(H, D_in).type(dtype), requires_grad=True)
w2 = Variable(torch.randn(H, D_out).type(dtype), requires_grad=True)
# I've created this scalar variable (the kernel bandwidth)
s = Variable(torch.randn(1).type(dtype), requires_grad=True)
learning_rate = 1e-6
for t in range(500):
# To apply our Function, we use Function.apply method. We alias this as 'relu'.
relu = MyReLU.apply
# Forward pass: compute predicted y using operations on Variables; we compute
# ReLU using our custom autograd operation.
# y_pred = relu(x.mm(w1)).mm(w2)
y_pred = relu(kernel_product(w1, x, s)).mm(w2)
# Compute and print loss
loss = (y_pred - y).pow(2).sum()
print(t, loss.data[0])
# Use autograd to compute the backward pass.
loss.backward()
# Update weights using gradient descent
w1.data -= learning_rate * w1.grad.data
w2.data -= learning_rate * w2.grad.data
# Manually zero the gradients after updating weights
w1.grad.data.zero_()
w2.grad.data.zero_()
However I get this error, which dissapears when I simply use a fixed scalar in the default input parameter of kernel_product():
RuntimeError: eq() received an invalid combination of arguments - got (str), but expected one of:
* (float other)
didn't match because some of the arguments have invalid types: (str)
* (Variable other)
didn't match because some of the arguments have invalid types: (str)
Well, you are calling kernel_product(w1, x, s) where w1, x and s are torch Variable while the definition of the function is: kernel_product(x,y, mode = "gaussian", s = 1.). Seems like s should be a string specifying the mode.
I'm trying to go through a .pdb file, calculate distance between alpha carbon atoms from different residues on chains A and B of a protein complex, then store the distance in a dictionary, together with the chain identifier and residue number.
For example, if the first alpha carbon ("CA") is found on residue 100 on chain A and the one it binds to is on residue 123 on chain B I would want my dictionary to look something like d={(A, 100):[B, 123, distance_between_atoms]}
from Bio.PDB.PDBParser import PDBParser
parser=PDBParser()
struct = parser.get_structure("1trk", "1trk.pdb")
def getAlphaCarbons(chain):
vec = []
for residue in chain:
for atom in residue:
if atom.get_name() == "CA":
vec = vec + [atom.get_vector()]
return vec
def dist(a,b):
return (a-b).norm()
chainA = struct[0]['A']
chainB = struct[0]['B']
vecA = getAlphaCarbons(chainA)
vecB = getAlphaCarbons(chainB)
t={}
model=struct[0]
for model in struct:
for chain in model:
for residue in chain:
for a in vecA:
for b in vecB:
if dist(a,b)<=8:
t={(chain,residue):[(a, b, dist(a, b))]}
break
print t
It's been running the programme for ages and I had to abort the run (have I made an infinite loop somewhere??)
I was also trying to do this:
t = {i:[((a, b, dist(a,b)) for a in vecA) for b in vecB if dist(a, b) <= 8] for i in chainA}
print t
But it's printing info about residues in the following format:
<Residue PHE het= resseq=591 icode= >: []
It's not printing anything related to the distance.
Thanks a lot, I hope everything is clear.
Would strongly suggest using C libraries while calculating distances. I use mdtraj for this sort of thing and it works much quicker than all the for loops in BioPython.
To get all pairs of alpha-Carbons:
import mdtraj as md
def get_CA_pairs(self,pdbfile):
traj = md.load_pdb(pdbfile)
topology = traj.topology
CA_index = ([atom.index for atom in topology.atoms if (atom.name == 'CA')])
pairs=list(itertools.combinations(CA_index,2))
return pairs
Then, for quick computation of distances:
def get_distances(self,pdbfile,pairs):
#returns list of resid1, resid2,distances between CA-CA
traj = md.load_pdb(pdbfile)
pairs=self.get_CA_pairs(pdbfile)
dist=md.compute_distances(traj,pairs)
#make dictionary you desire.
dict=dict(zip(CA, pairs))
return dict
This includes all alpha-Carbons. There should be a chain identifier too in mdtraj to select CA's from each chain.
I watch out this example: http://scikit-learn.org/stable/auto_examples/plot_digits_classification.html#example-plot-digits-classification-py
on handwritten digits in scikit-learn python library.
i would like to prepare a 3d array (N * a* b) where N is my images number (75) and a* b is the matrix of an image (like in the example a 8x8 shape).
My problem is: i have signs in a different shapes for every image: (202, 230), (250, 322).. and give me
this error: ValueError: array dimensions must agree except for d_0 in this code:
#here there is the error:
grigiume = np.dstack(listagrigie)
print(grigiume.shape)
grigiume=np.rollaxis(grigiume,-1)
print(grigiume.shape)
There is a manner to resize all images in a standard size (i.e. 200x200) or a manner to have a 3d array with matrix(a,b) where a != from b and do not give me an error in this code:
data = digits.images.reshape((n_samples, -1))
classifier.fit(data[:n_samples / 2], digits.target[:n_samples / 2])
My code:
import os
import glob
import numpy as np
from numpy import array
listagrigie = []
path = 'resize2/'
for infile in glob.glob( os.path.join(path, '*.jpg') ):
print("current file is: " + infile )
colorato = cv2.imread(infile)
grigiscala = cv2.cvtColor(colorato,cv2.COLOR_BGR2GRAY)
listagrigie.append(grigiscala)
print(len(listagrigie))
#here there is the error:
grigiume = np.dstack(listagrigie)
print(grigiume.shape)
grigiume=np.rollaxis(grigiume,-1)
print(grigiume.shape)
#last step
n_samples = len(digits.images)
data = digits.images.reshape((n_samples, -1))
# Create a classifier: a support vector classifier
classifier = svm.SVC(gamma=0.001)
# We learn the digits on the first half of the digits
classifier.fit(data[:n_samples / 2], digits.target[:n_samples / 2])
# Now predict the value of the digit on the second half:
expected = digits.target[n_samples / 2:]
predicted = classifier.predict(data[n_samples / 2:])
print "Classification report for classifier %s:\n%s\n" % (
classifier, metrics.classification_report(expected, predicted))
print "Confusion matrix:\n%s" % metrics.confusion_matrix(expected, predicted)
for index, (image, prediction) in enumerate(
zip(digits.images[n_samples / 2:], predicted)[:4]):
pl.subplot(2, 4, index + 5)
pl.axis('off')
pl.imshow(image, cmap=pl.cm.gray_r, interpolation='nearest')
pl.title('Prediction: %i' % prediction)
pl.show()
You have to resize all your images to a fixed size. For instance using the Image class of PIL or Pillow:
from PIL import Image
image = Image.open("/path/to/input_image.jpeg")
image.thumbnail((200, 200), Image.ANTIALIAS)
image.save("/path/to/output_image.jpeg")
Edit: the above won't work, try instead resize:
from PIL import Image
image = Image.open("/path/to/input_image.jpeg")
image = image.resize((200, 200), Image.ANTIALIAS)
image.save("/path/to/output_image.jpeg")
Edit 2: there might be a way to preserve the aspect ratio and pad the rest with black pixels but I don't know how to do in a few PIL calls. You could use PIL.Image.thumbnail and use numpy to do the padding though.
I'm building a demonstration any-dimensional Vector class to show some functional programming in Python.
class Vector():
def __init__(self, *coords):
self.coords = coords
def __add__(this, that):
return Point(*[(x+y) for x,y in zip(this.coords, that.coords)])
#...
While trying to come up with an example of a static #classmethod in this example, I decided it'd be nice to have a class method giving me an n-dimensional base of vectors for any n. That is:
>>> Vector.get_base(dimensions = 2)
[Vector(1,0), Vector(0,1)]
>>> Vector.get_base(3)
[Vector(1,0,0), Vector(0,1,0), Vector(0,0,1)]
>>> Vector.get_base(1)
[Vector(1)]
I'm however having a huge brain fart however and am stumbling on the problem of how to "properly" generate those lists.
What I can think up right now is a declarative solution:
def get_base(dimensions):
arrays = []
zeros = [0] * dimensions
for i in range(dimensions):
item = zeros
item[i] = 1
arrays.append(Vector(*array))
return arrays
There has to be a better way! How can I rewrite this function in a hopefully more concise or Pythonic functional style?
Well, you could do this:
def get_base(dimensions):
return [Vector(*coords) for coords in
[[(0,1)[i==j] for i in range(dimensions)] for j in range(dimensions)]]
but I would break it down a little:
def get_base(dimensions):
arrays = [[(0,1)[i==j] for i in range(dimensions)] for j in range(dimensions)]
return [Vector(*coords) for coords in arrays]
Which is a little better. Remember, not everything has to be a one-liner.
How about the next:
>>> def get_base(dimensions):
... for points in set(itertools.permutations([0] * (dimensions - 1) + [1], dimensions)):
... yield Vector(*points)