I've discretised a diffusion equation with FEniCS as follows:
def DiscretiseEquation(h):
mesh = UnitSquareMesh(h, h)
V = FunctionSpace(mesh, 'Lagrange', 1)
def on_boundary(x, on_boundary):
return on_boundary
bc_value = Constant(0.0)
boundary_condition = DirichletBC(V, bc_value, on_boundary)
class RandomDiffusionField(Expression):
def __init__(self, m, n, element):
self._rand_field = np.exp(-np.random.randn(m, n))
self._m = m
self._n = n
self._ufl_element = element
def eval(self, value, x):
x_index = np.int(np.floor(self._m * x[0]))
y_index = np.int(np.floor(self._n * x[1]))
i = min(x_index, self._m - 1)
j = min(y_index, self._n - 1)
value[0] = self._rand_field[i, j]
def value_shape(self):
return(1, )
class RandomRhs(Expression):
def __init__(self, m, n, element):
self._rand_field = np.random.randn(m, n)
self._m = m
self._n = n
self._ufl_element = element
def eval(self, value, x):
x_index = np.int(np.floor(self._m * x[0]))
y_index = np.int(np.floor(self._n * x[1]))
i = min(x_index, self._m - 1)
j = min(y_index, self._n - 1)
value[0] = self._rand_field[i, j]
def value_shape(self):
return (1, )
u = TrialFunction(V)
v = TestFunction(V)
random_field = RandomDiffusionField(100, 100, element=V.ufl_element())
zero = Expression("0", element=V.ufl_element())
one = Expression("1", element=V.ufl_element())
diffusion = as_matrix(((random_field, zero), (zero, one)))
a = inner(diffusion * grad(u), grad(v)) * dx
L = RandomRhs(h, h, element=V.ufl_element()) * v * dx
A = assemble(a)
b = assemble(L)
boundary_condition.apply(A, b)
A = as_backend_type(A).mat()
(indptr, indices, data) = A.getValuesCSR()
mat = csr_matrix((data, indices, indptr), shape=A.size)
rhs = b.array()
#Solving
x = spsolve(mat, rhs)
#Conversion to a FEniCS function
u = Function(V)
u.vector()[:] = x
I am running the GMRES solver as normal. The callback argument is a separate iteration counter I've defined.
DiscretiseEquation(100)
A = mat
b = rhs
x, info = gmres(A, b, callback = IterCount())
The routine returns a NameError, stating that 'mat' is not defined:
NameError Traceback (most recent call last)
<ipython-input-18-e096b2eea097> in <module>()
1 DiscretiseEquation(200)
----> 2 A = mat
3 b = rhs
4 x_200, info_200 = gmres(A, b, callback = IterCount())
5 gmres_res = closure_variables["residuals"]
NameError: name 'mat' is not defined
As far as I'm aware, it should be defined when I call the DiscretiseEquation function?
Related
I am trying custom traning of the network using Gradient.Tape method.
This traning is unsupervised.
The details of network and cost function is as following,
My Network is,
def CreateNetwork(inplayer, hidlayer, outlayer,seed):
model = keras.Sequential()
model.add(Dense(hidlayer, input_dim=inplayer, kernel_initializer=initializers.RandomNormal(mean=0.0,stddev=1/np.sqrt(inplayer),seed=seed), bias_initializer=initializers.RandomNormal(mean=0.0,stddev=1/np.sqrt(inplayer),seed=seed), activation='tanh'))
model.add(Dense(outlayer, kernel_initializer=initializers.RandomNormal(mean=0.0,stddev=1/np.sqrt(hidlayer),seed=seed), bias_initializer=initializers.Zeros(), activation='linear'))
return model
and my custom cost function is defined as,
def H_tilda(J,U,nsamples,nsites,configs,out_matrix):
EigenValue = 0.0
for k in range(nsamples):
config = configs[k,:]
out_n = out_matrix[k,:]
exp = 0.0
for i in range(nsamples):
n = configs[i,:]
out_nprime = out_matrix[i,:]
#------------------------------------------------------------------------------------------------
# Calculation of Hopping Term
#------------------------------------------------------------------------------------------------
hop = 0.0
for j in range(nsites):
if j == 0:
k = [nsites-1,j+1]
elif j == (nsites - 1):
k = [j-1,0]
else:
k = [j-1,j+1]
if n[k[0]] != 0:
annihiliate1 = np.sqrt(n[k[0]])
n1 = np.copy(n)
n1[k[0]] = n1[k[0]] - 1
n1[j] = n1[j] +1
if (config == n1).all():
delta1 = 1
else:
delta1 = 0
else:
annihiliate1 = 0
n1 = np.zeros(nsites)
delta1 = 0
if n[k[1]] != 0:
annihiliate2 = np.sqrt(n[k[1]])
n2 = np.copy(n)
n2[k[1]] = n2[k[1]] -1
n2[j] = n2[j] + 1
if (config == n2).all():
delta2 = 1
else:
delta2 = 0
else:
annihiliate2 = 0
n2 = np.zeros(nsites)
delta2 = 0
create = np.sqrt(n[j] + 1)
hop = hop + create*(annihiliate1*delta1 + annihiliate2*delta2)
#------------------------------------------------------------------------------------------------
#------------------------------------------------------------------------------------------------
# Calculation of Onsite Term
#------------------------------------------------------------------------------------------------
if (config == n).all():
ons = np.sum(np.dot(np.square(n),n - 1))
else:
ons = 0.0
#------------------------------------------------------------------------------------------------
phi_value = phi(out_nprime.numpy())
exp = exp + ((hop + ons) * phi_value)
Phi_value = phi(out_n.numpy())
EigenValue = EigenValue + exp/Phi_value
return np.real(EigenValue/nsamples)
I want to do custom traning using GradientTape method, for which I used following lines ,
optimizer = optimizers.SGD(learning_rate=1e-3)
with tf.GradientTape(watch_accessed_variables=False) as tape:
tape.watch(tf.convert_to_tensor(configs))
out_matrix = model(configs)
print(out_matrix)
eival = H_tilda(J,U,nsamples,nsites,configs,out_matrix)
print(eival)
gradients = tape.gradient(tf.convert_to_tensor(eival), model.trainable_weights)
print(gradients)
But the gradient I am getting is NONE,
output: [None, None, None, None]
I'm super confusing with lambda layer during functional API
please tell me which is correct
For example
def magic(x):
x2=x+x
return x2
D = Input((n,))
E = Lambda(magic)(D)
d = Model(~~~~)
or
D = Input((n,))
E = magic(D)
d = Model(~~~~)
and another case
def magic():
x2=np.randn(3,1)
return x2
D = Input((n,))
E = Lambda(magic)
d = Model(~~~~)
or
D = Input((n,))
E = magic()
d = Model(~~~~)
please save me !!
You can use the first way if you want to wrap your custom operation as a Layer like follows
def magic(x):
x2 = x + x
return x2
inp = tf.keras.Input(shape=(2,))
x = tf.keras.layers.Dense(4, name="id_1")(inp)
# Ways to Define Custom Layer
x = tf.keras.layers.Lambda(lambda val: magic(val))(x)
# or
# x = tf.keras.layers.Lambda(magic)(x)
# or
# x = tf.keras.layers.Lambda(lambda val: (val + val))(x)
x = tf.keras.layers.Dense(4, name="id_2")(x)
model = tf.keras.Model(inputs=inp, outputs=x, name="inner_model")
print(model.summary())
I have 3D (64,64,64) shape (chair) when I reshape it using tf operation to (8,32,32,32) then do my operation Deep learning operation and then return it back using tf reshape to (64,64,64) the shape looks very bad, actually there is no shape only strange looks unknown shape (100% not looks like chair)
but if I use function that I build to slice 32 by 32 and I stack them as (8,32,32,32) I use it as input to my DL Model. the output (8,32,32,32) I use also combine function which I build to recombine by reversing the slice function I got good looking shape
the issue both function slice and combine numpy not tf. I have to train model end-to-end so I need equivalent function that slice or combine in tensorflow please
def slice(self,size, obj):
#print('inside')
oldi = 0
newi = 0
oldj = 0
newj = 0
oldk = 0
newk = 0
lst = []
s = obj.shape[0]
s += 1
for i in range(size, s, size):
if (newi == s - 1):
oldi = 0
else:
oldi = newi
for j in range(size, s, size):
if (newj == s - 1):
oldj = 0
else:
oldj = newj
for k in range(size, s, size):
newi = i
newj = j
newk = k
slc = obj[oldi:newi, oldj:newj, oldk:newk]
#print(oldi,':',newi,',',oldj,':',newj,',',oldk,':',newk)
#print(slc.shape)
lst.append(slc)
if (newk == s - 1):
oldk = 0
else:
oldk = newk
# print(slc.shape)
return lst
def combine(self,lst, shape, size):
oldi = 0
newi = 0
oldj = 0
newj = 0
oldk = 0
newk = 0
obj = np.zeros((shape, shape, shape))
s = shape
s += 1
counter = 0
for i in range(size, s, size):
if (newi == s - 1):
oldi = 0
else:
oldi = newi
for j in range(size, s, size):
if (newj == s - 1):
oldj = 0
else:
oldj = newj
for k in range(size, s, size):
newi = i
newj = j
newk = k
obj[oldi:newi, oldj:newj, oldk:newk] = lst[counter]
counter += 1
#print(oldi,':',newi,',',oldj,':',newj,',',oldk,':',newk)
# print(slc.shape)
if (newk == s - 1):
oldk = 0
else:
oldk = newk
return obj
in other words I want tensorflow operation mimic
the following function
def combine(self,lst, shape, size):
oldi = 0
newi = 0
oldj = 0
newj = 0
oldk = 0
newk = 0
obj = np.zeros((shape, shape, shape))
s = shape
s += 1
counter = 0
for i in range(size, s, size):
if (newi == s - 1):
oldi = 0
else:
oldi = newi
for j in range(size, s, size):
if (newj == s - 1):
oldj = 0
else:
oldj = newj
for k in range(size, s, size):
newi = i
newj = j
newk = k
obj[oldi:newi, oldj:newj, oldk:newk] = lst[counter]
counter += 1
#print(oldi,':',newi,',',oldj,':',newj,',',oldk,':',newk)
# print(slc.shape)
if (newk == s - 1):
oldk = 0
else:
oldk = newk
return obj
From deeplearning course on Coursera I've implemented logistic regression :
import numpy as np
from sklearn.datasets import load_iris
import matplotlib.pyplot as plt
def sigmoid(z):
s = 1 / (1 + np.exp(-z))
return s
def initialize_with_zeros(dim):
w = np.zeros(shape=(dim, 1))
b = 0
return w, b
def propagate(w, b, X, Y):
m = X.shape[1]
A = sigmoid(np.dot(w.T, X) + b) # compute activation
cost = (- 1 / m) * np.sum(Y * np.log(A) + (1 - Y) * (np.log(1 - A))) # compute cost
dw = (1 / m) * np.dot(X, (A - Y).T)
db = (1 / m) * np.sum(A - Y)
cost = np.squeeze(cost)
grads = {"dw": dw,
"db": db}
return grads, cost
def optimize(w, b, X, Y, num_iterations, learning_rate, print_cost = False):
costs = []
for i in range(num_iterations):
grads, cost = propagate(w, b, X, Y)
dw = grads["dw"]
db = grads["db"]
w = w - learning_rate * dw # need to broadcast
b = b - learning_rate * db
if i % 100 == 0:
costs.append(cost)
# Print the cost every 100 training examples
if print_cost and i % 100 == 0:
print ("Cost after iteration %i: %f" % (i, cost))
params = {"w": w,
"b": b}
grads = {"dw": dw,
"db": db}
return params, grads, costs
def predict(w, b, X):
m = X.shape[1]
Y_prediction = np.zeros((1, m))
w = w.reshape(X.shape[0], 1)
A = sigmoid(np.dot(w.T, X) + b)
for i in range(A.shape[1]):
# Convert probabilities a[0,i] to actual predictions p[0,i]
### START CODE HERE ### (≈ 4 lines of code)
print(A)
Y_prediction[0, i] = 1 if A[0, i] > 0.5 else 0
### END CODE HERE ###
assert(Y_prediction.shape == (1, m))
return Y_prediction
print ("sigmoid(0) = " + str(sigmoid(0)))
print ("sigmoid(9.2) = " + str(sigmoid(9.2)))
dim = 2
w, b = initialize_with_zeros(dim)
print ("w = " + str(w))
print ("b = " + str(b))
w, b, X, Y = np.array([[1], [2]]), 2, np.array([[-1,-2], [3,4]]), np.array([[1, 0]])
grads, cost = propagate(w, b, X, Y)
print ("dw = " + str(grads["dw"]))
print ("db = " + str(grads["db"]))
print ("cost = " + str(cost))
params, grads, costs = optimize(w, b, X, Y, num_iterations= 10000, learning_rate = 0.01, print_cost = False)
print ("w = " + str(params["w"]))
print ("b = " + str(params["b"]))
print ("dw = " + str(grads["dw"]))
print ("db = " + str(grads["db"]))
print("predictions = " + str(predict(w, b, X)))
def model(X_train, Y_train, X_test, Y_test, num_iterations=2000, learning_rate=0.5, print_cost=False):
w, b = initialize_with_zeros(X_train.shape[0])
parameters, grads, costs = optimize(w, b, X_train, Y_train, num_iterations, learning_rate, print_cost)
w = parameters["w"]
b = parameters["b"]
Y_prediction_test = predict(w, b, X_test)
Y_prediction_train = predict(w, b, X_train)
print("train accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_train - Y_train)) * 100))
print("test accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_test - Y_test)) * 100))
d = {"costs": costs,
"Y_prediction_test": Y_prediction_test,
"Y_prediction_train" : Y_prediction_train,
"w" : w,
"b" : b,
"learning_rate" : learning_rate,
"num_iterations": num_iterations}
return d
I'm attempting to use a generic dataset which contains 5 samples where each sample contain 4 elements :
train_set_x = np.array([[1,2,3,4],[4,3,2,1],[1,2,3,4],[4,3,2,1],[1,2,3,4]])
train_set_y = np.array([1,0,1,0,1])
test_set_x = np.array([[1,2,3,4],[4,3,2,1],[1,2,3,4],[4,3,2,1],[1,2,3,4]])
test_set_y = np.array([1,0,1,0,1])
train_set_x , train_set_y , test_set_x , test_set_y
d = model(train_set_x, train_set_y, test_set_x, test_set_y, num_iterations = 2000, learning_rate = 0.005, print_cost = True)
But the following error is thrown :
<ipython-input-409-bd4e233a8f4e> in propagate(w, b, X, Y)
18
19 A = sigmoid(np.dot(w.T, X) + b) # compute activation
---> 20 cost = (- 1 / m) * np.sum(Y * np.log(A) + (1 - Y) * (np.log(1 - A))) # compute cost
21
22 dw = (1 / m) * np.dot(X, (A - Y).T)
ValueError: operands could not be broadcast together with shapes (5,) (1,4)
Do I need to change the weight dimensions in order to compute the cost value ?
Update :
Using modification :
A = sigmoid(np.dot(X , w) + b) # compute activation
causes error :
<ipython-input-546-7a7980550834> in propagate(w, b, X, Y)
20 m = X.shape[1]
21
---> 22 A = sigmoid(np.dot(X , w) + b) # compute activation
23 print('w.T' , w.T , 'w' , w, 'X' , X , 'Y' , Y , 'A' , A)
24 cost = (- 1 / m) * np.sum(Y * np.log(A) + (1 - Y) * (np.log(1 - A))) # compute cost
ValueError: shapes (5,4) and (5,1) not aligned: 4 (dim 1) != 5 (dim 0)
I have two dataframes (X & Y). I would like to link them together and to predict the probability that each potential match is correct.
X = pd.DataFrame({'A': ["One", "Two", "Three"]})
Y = pd.DataFrame({'A': ["One", "To", "Free"]})
Method A
I have not yet fully understood the theory but there is an approach presented in:
Sayers, A., Ben-Shlomo, Y., Blom, A.W. and Steele, F., 2015. Probabilistic record linkage. International journal of epidemiology, 45(3), pp.954-964.
Here is my attempt to implementat it in Pandas:
# Probability that Matches are True Matches
m = 0.95
# Probability that non-Matches are True non-Matches
u = min(len(X), len(Y)) / (len(X) * len(Y))
# Priors
M_Pr = u
U_Pr = 1 - M_Pr
O_Pr = M_Pr / U_Pr # Prior odds of a match
# Combine the dataframes
X['key'] = 1
Y['key'] = 1
Z = pd.merge(X, Y, on='key')
Z = Z.drop('key',axis=1)
X = X.drop('key',axis=1)
Y = Y.drop('key',axis=1)
# Levenshtein distance
def Levenshtein_distance(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
distances = range(len(s1) + 1)
for i2, c2 in enumerate(s2):
distances_ = [i2+1]
for i1, c1 in enumerate(s1):
if c1 == c2:
distances_.append(distances[i1])
else:
distances_.append(1 + min((distances[i1], distances[i1 + 1], distances_[-1])))
distances = distances_
return distances[-1]
L_D = np.vectorize(Levenshtein_distance, otypes=[float])
Z["D"] = L_D(Z['A_x'], Z['A_y'])
# Max string length
def Max_string_length(X, Y):
return max(len(X), len(Y))
M_L = np.vectorize(Max_string_length, otypes=[float])
Z["L"] = M_L(Z['A_x'], Z['A_y'])
# Agreement weight
def Agreement_weight(D, L):
return 1 - ( D / L )
A_W = np.vectorize(Agreement_weight, otypes=[float])
Z["C"] = A_W(Z['D'], Z['L'])
# Likelihood ratio
def Likelihood_ratio(C):
return (m/u) - ((m/u) - ((1-m) / (1-u))) * (1-C)
L_R = np.vectorize(Likelihood_ratio, otypes=[float])
Z["G"] = L_R(Z['C'])
# Match weight
def Match_weight(G):
return math.log(G) * math.log(2)
M_W = np.vectorize(Match_weight, otypes=[float])
Z["R"] = M_W(Z['G'])
# Posterior odds
def Posterior_odds(R):
return math.exp( R / math.log(2)) * O_Pr
P_O = np.vectorize(Posterior_odds, otypes=[float])
Z["O"] = P_O(Z['R'])
# Probability
def Probability(O):
return O / (1 + O)
Pro = np.vectorize(Probability, otypes=[float])
Z["P"] = Pro(Z['O'])
I have verified that this gives the same results as in the paper. Here is a sensitivity check on m, showing that it doesn't make a lot of difference:
Method B
These assumptions won't apply to all applications but in some cases each row of X should match a row of Y. In that case:
The probabilities should sum to 1
If there are many credible candidates to match to then that should reduce the probability of getting the right one
then:
X["I"] = X.index
# Combine the dataframes
X['key'] = 1
Y['key'] = 1
Z = pd.merge(X, Y, on='key')
Z = Z.drop('key',axis=1)
X = X.drop('key',axis=1)
Y = Y.drop('key',axis=1)
# Levenshtein distance
def Levenshtein_distance(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
distances = range(len(s1) + 1)
for i2, c2 in enumerate(s2):
distances_ = [i2+1]
for i1, c1 in enumerate(s1):
if c1 == c2:
distances_.append(distances[i1])
else:
distances_.append(1 + min((distances[i1], distances[i1 + 1], distances_[-1])))
distances = distances_
return distances[-1]
L_D = np.vectorize(Levenshtein_distance, otypes=[float])
Z["D"] = L_D(Z['A_x'], Z['A_y'])
# Max string length
def Max_string_length(X, Y):
return max(len(X), len(Y))
M_L = np.vectorize(Max_string_length, otypes=[float])
Z["L"] = M_L(Z['A_x'], Z['A_y'])
# Agreement weight
def Agreement_weight(D, L):
return 1 - ( D / L )
A_W = np.vectorize(Agreement_weight, otypes=[float])
Z["C"] = A_W(Z['D'], Z['L'])
# Normalised Agreement Weight
T = Z .groupby('I') .agg({'C' : sum})
D = pd.DataFrame(T)
D.columns = ['T']
J = Z.set_index('I').join(D)
J['P1'] = J['C'] / J['T']
Comparing it against Method A:
Method C
This combines method A with method B:
# Normalised Probability
U = Z .groupby('I') .agg({'P' : sum})
E = pd.DataFrame(U)
E.columns = ['U']
K = Z.set_index('I').join(E)
K['P1'] = J['P1']
K['P2'] = K['P'] / K['U']
We can see that method B (P1) doesn't take account of uncertainty whereas method C (P2) does.