I would like to solve the above formulation in Scipy and solve it using milp(). For a given graph (V, E), f_ij and x_ij are the decision variables. f_ij is the flow from i to j (it can be continuous). x_ij is the number of vehicles from i to j. p is the price. X is the available number vehicles in a region. c is the capacity.
I have difficulty in translating the formulation to Scipy milp code. I would appreciate it if anyone could give me some pointers.
What I have done:
The code for equation (1):
f_obj = [p[i] for i in Edge]
x_obj = [0]*len(Edge)
obj = f_obj + v_obj
Integrality:
f_cont = [0 for i in Edge] # continous
x_int = [1]*len(Edge) # integer
integrality = f_cont + x_int
Equation (2):
def constraints(self):
b = []
A = []
const = [0]*len(Edge) # for f_ij
for i in v: # for x_ij
for e in Edge:
if e[0] == i:
const.append(1)
else:
const.append(0)
A.append(const)
b.append(self.accInit[i])
const = [0]*len(Edge) # for f_ij
return A, b
Equation (4):
[(0, demand[e]) for e in Edge]
I'm going to do some wild guessing, given how much you've left open to interpretation. Let's assume that
this is a maximisation problem, since the minimisation problem is trivial
Expression (1) is actually the maximisation objective function, though you failed to write it as such
p and d are floating-point vectors
X is an integer vector
c is a floating-point scalar
the graph edges, since you haven't described them at all, do not matter for problem setup
The variable names are not well-chosen and hide what they actually contain. I demonstrate potential replacements.
import numpy as np
from numpy.random._generator import Generator
from scipy.optimize import milp, Bounds, LinearConstraint
import scipy.sparse
from numpy.random import default_rng
rand: Generator = default_rng(seed=0)
N = 20
price = rand.uniform(low=0, high=10, size=N) # p
demand = rand.uniform(low=0, high=10, size=N) # d
availability = rand.integers(low=0, high=10, size=N) # X aka. accInit
capacity = rand.uniform(low=0, high=10) # c
c = np.zeros(2*N) # f and x
c[:N] = -price # (1) f maximized with coefficients of 'p'
# x not optimized
CONTINUOUS = 0
INTEGER = 1
integrality = np.empty_like(c, dtype=int)
integrality[:N] = CONTINUOUS # f
integrality[N:] = INTEGER # x
upper = np.empty_like(c)
upper[:N] = demand # (4) f
upper[N:] = availability # (2) x
eye_N = scipy.sparse.eye(N)
A = scipy.sparse.hstack((-eye_N, capacity*eye_N)) # (3) 0 <= -f + cx
result = milp(
c=c, integrality=integrality,
bounds=Bounds(lb=np.zeros_like(c), ub=upper),
constraints=LinearConstraint(lb=np.zeros(N), A=A),
)
print(result.message)
flow = result.x[:N]
vehicles = result.x[N:].astype(int)
I have coded the laplacien function for a non-regular mesh (created with the scipy.spatial.Delaunay function).
I have not errors but the results are not correct : the eigenvectors are correct but the eigenvalues are too high (in absolute value).
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import scipy.spatial
def rect_drum(L,H,U):
vals = []
val = 0
k = 1
l = 1
while val >= -U:
while val >= -U:
val = -np.pi**2*((k/L)**2+(l/H)**2)
if val >= -U:
vals.append(val)
l += 1
l = 1
k += 1
val = -np.pi**2*((k/L)**2+(l/H)**2)
return np.array(vals)
def count_vp(tab,U):
#count the n eigenvalues greater than equal to -U in the array tab
return tab[tab>=-U]
def in_curve(f,fargs,shape,a):
points = [] # the points inside the curve
for j in range(shape[0]):
for i in range(shape[1]):
if f(i*a,j*a,*fargs) < 0:
points.append([i*a,j*a])
return np.array(points)
def triang(points,a,f,fargs,bord):
tri_points = points.copy()
tri_points[:,1] *= np.sqrt(3)
tri_points2 = np.vstack((points,bord))
tri_points2[:,1] *= np.sqrt(3)
tri_points2[:,0] += a/2
tri_points2[:,1] += np.sqrt(3)/2*a
fin = np.vstack((tri_points,tri_points2))
i = 0
eps = 0.01
while i < len(fin):
if f(fin[i,0]+eps,fin[i,1]+eps,*fargs) > 0:
fin = np.delete(fin,i,0)
i -= 1
i += 1
return np.vstack((fin,bord)),len(fin),len(bord)
def tri_ang(points,ind,p0):
# sort the points in trigonometric order
vec=np.arctan2((points-p0)[:,1],(points-p0)[:,0])
values = []
dtype = [('val',float),('n',int)]
for i in range(len(vec)):
values.append((vec[i],i))
values = np.sort(np.array(values,dtype),order='val')
new_points = []
new_ind = []
for tup in values:
new_points.append(points[tup[1]])
new_ind.append(ind[tup[1]])
return np.array(new_points),np.array(new_ind)
def M(points,tri,Nint):
indptr,ind = tri.vertex_neighbor_vertices
W = np.zeros((Nint,Nint)) # cotangents matrix
A = np.zeros((Nint,1)) # surfaces vertex array for each point i (A[i])
for i in range(Nint):
tot = 0
nhb_ind = ind[indptr[i]:indptr[i+1]] # indices of the points close to the point of index k
nhb = points[nhb_ind] # their coordinates
nhb,nhb_ind = tri_ang(nhb,nhb_ind,points[i]) #the coordinates (nhb) and (nhb_ind) of each neighbor of i
for j in range(len(nhb_ind)):
vec = nhb[j]-points[i] # a vector connecting the point to his neighbor of index 0
vec_av = nhb[j-1]-points[i] # another vector but with the Vosin from before
if j+1 >= len(nhb_ind):
k = 0
else:
k = j+1
vec_ap = nhb[k]-points[i] # another vector but with the next neighbor
# another vector but with the next neighbor
A[i] += 0.5/3*np.linalg.norm(np.cross(vec,vec_av))
if nhb_ind[j] < Nint:
# we use the vector and scalar product to calculate the cotangents: A.B/||AxB||
cotan_alpha = np.dot(vec_av,vec_av-vec)/np.linalg.norm(np.cross(vec_av,vec_av-vec))
cotan_beta = np.dot(vec_ap,vec_ap-vec)/np.linalg.norm(np.cross(vec_ap,vec_ap-vec))
# Wij value :
W[i,nhb_ind[j]] = -0.5*(cotan_alpha+cotan_beta)
tot += cotan_alpha+cotan_beta
W[i,i] = -0.5*tot # diagonal values
return (1/A)*W
def rect(x,y,L,H,x0=0,y0=0):
if 0<x-x0<L and 0<y-y0<H:
return -1
else:
return 1
def rect_rim(L,H,a,x0=0,y0=0):
tab1 = np.arange(x0,L+x0,a)[:,np.newaxis]
h = np.hstack((tab1,H*np.ones((len(tab1),1))+y0))
b = np.hstack((tab1,np.zeros((len(tab1),1))+y0))
tab2 = np.arange(y0+a,H+y0,a)[:,np.newaxis]
g = np.hstack((np.zeros((len(tab2),1))+x0,tab2))
d = np.hstack((L*np.ones((len(tab2),1))+x0,tab2))
hp = np.array([[L+x0,H+y0]])
bp = np.array([[L+x0,0]])
return np.vstack((h,b,g,d,hp,bp))
# sample with a square 1*1
L = 1
H = 1
dl = 0.05
sol = in_curve(rect,[L,H],(100,100),dl)
sol_tri,Nint,Nbord = triang(sol,dl,rect,[L,H],rect_rim(L,H,dl))
# plt.plot(sol_tri[:,0],sol_tri[:,1],linestyle="",marker="+",label="tri")
# plt.plot(sol[:,0],sol[:,1],linestyle="",marker="x")
# plt.legend()
# plt.show()
# triangulation
tri = scipy.spatial.Delaunay(sol_tri)
# plt.triplot(sol_tri[:,0],sol_tri[:,1],tri.simplices)
# plt.show()
M = M(sol_tri,tri,Nint)
valp,vecp = np.linalg.eig(M) # eigenvalues and eigenvectors
vecp = np.real(vecp)
# comparison with the exact solution:
T = 1000
U = np.arange(0,T,1)
NUsim = np.array([len(count_vp(valp,u)) for u in U])
NU = np.array([len(rect_drum(L,H,u)) for u in U])
plt.plot(U,NUsim,label='simulation')
plt.plot(U,NU,label='exacts')
plt.legend()
plt.show()
# 3D plot of an eigenvector
vecp_tot = np.vstack((vecp,np.zeros((Nbord,Nint))))
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot_trisurf(sol_tri[:,0],sol_tri[:,1],vecp_tot[:,0],triangles=tri.simplices)
plt.show()
The laplacian is the function named "M".
The "in_curve function" return the points inside a curve defined by f(x,y,*fargs) < 0 (a square in the sample).
The "triang" function return points with added points (triangle meshs). The fonction uses an another function for the rim of the curve (for most precision), in the sample it is the "rect_rim" function.
I used the formula given at https://en.wikipedia.org/wiki/Discrete_Laplace_operator ("mesh laplacians").
I have solve my problem : it's a sign and a rim problems.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import scipy.spatial
def rect_drum(L,H,U):
vals = []
val = 0
k = 1
l = 1
while val >= -U:
while val >= -U:
val = -np.pi**2*((k/L)**2+(l/H)**2)
if val >= -U:
vals.append(val)
l += 1
l = 1
k += 1
val = -np.pi**2*((k/L)**2+(l/H)**2)
return np.array(vals)
def count_vp(tab,U):
#count the n eigenvalues greater than equal to -U in the array tab
return tab[tab>=-U]
def in_curve(f,fargs,shape,a):
points = [] # the points inside the curve
for j in range(shape[0]):
for i in range(shape[1]):
if f(i*a,j*a,*fargs) < 0:
points.append([i*a,j*a])
return np.array(points)
def triang(points,a,f,fargs,bord):
tri_points = points.copy()
tri_points[:,1] *= np.sqrt(3)
tri_points2 = np.vstack((points,bord))
tri_points2[:,1] *= np.sqrt(3)
tri_points2[:,0] += a/2
tri_points2[:,1] += np.sqrt(3)/2*a
fin = np.vstack((tri_points,tri_points2))
i = 0
eps = 0.01
while i < len(fin):
if f(fin[i,0]+eps,fin[i,1]+eps,*fargs) > 0:
fin = np.delete(fin,i,0)
i -= 1
i += 1
return np.vstack((fin,bord)),len(fin),len(bord)
def tri_ang(points,ind,p0):
# sort the points in trigonometric order
vec=np.arctan2((points-p0)[:,1],(points-p0)[:,0])
values = []
dtype = [('val',float),('n',int)]
for i in range(len(vec)):
values.append((vec[i],i))
values = np.sort(np.array(values,dtype),order='val')
new_points = []
new_ind = []
for tup in values:
new_points.append(points[tup[1]])
new_ind.append(ind[tup[1]])
return np.array(new_points),np.array(new_ind)
def Laplacian(points,tri,Nint):
indptr,ind = tri.vertex_neighbor_vertices
W = np.zeros((Nint,Nint)) # cotangents matrix
A = np.zeros((Nint,1)) # surfacesvertex aray of point i (A[i])
for i in range(Nint):
tot = 0
nhb_ind = ind[indptr[i]:indptr[i+1]] # indices of the points close to the point of index k
nhb = points[nhb_ind] # their coordinates
nhb,nhb_ind = tri_ang(nhb,nhb_ind,points[i]) #the coordinates (nhb) and (nhb_ind) of each neighbor of i
for j in range(len(nhb_ind)):
vec = nhb[j]-points[i] # a vector connecting the point to his neighbor of index 0
vec_av = nhb[j-1]-points[i] # another vector but with the Vosin from before
if j+1 >= len(nhb_ind):
k = 0
else:
k = j+1
vec_ap = nhb[k]-points[i] # another vector but with the next neighbor
# we use the cross product to calculate the areas of the triangles: ||AxB||/2:
A[i] += 0.5/3*np.linalg.norm(np.cross(vec,vec_av))
# we use the cross product and scalar product to calculate the cotangents: A.B/||AxB||
cotan_alpha = np.dot(vec_av,vec_av-vec)/np.linalg.norm(np.cross(vec_av,vec_av-vec))
cotan_beta = np.dot(vec_ap,vec_ap-vec)/np.linalg.norm(np.cross(vec_ap,vec_ap-vec))
tot += cotan_alpha+cotan_beta
if nhb_ind[j] < Nint:
W[i,nhb_ind[j]] = 0.5*(cotan_alpha+cotan_beta)
W[i,i] = -0.5*tot # diagonal values
return (1/A)*W
def rect(x,y,L,H,x0=0,y0=0):
if 0<x-x0<L and 0<y-y0<H:
return -1
else:
return 1
def rect_rim(L,H,a,x0=0,y0=0):
tab1 = np.arange(x0,L+x0,a)[:,np.newaxis]
h = np.hstack((tab1,H*np.ones((len(tab1),1))+y0))
b = np.hstack((tab1,np.zeros((len(tab1),1))+y0))
tab2 = np.arange(y0+a,H+y0,a)[:,np.newaxis]
g = np.hstack((np.zeros((len(tab2),1))+x0,tab2))
d = np.hstack((L*np.ones((len(tab2),1))+x0,tab2))
hp = np.array([[L+x0,H+y0]])
bp = np.array([[L+x0,0]])
return np.vstack((h,b,g,d,hp,bp))
# sample with a square 1*1
L = 1
H = 1
dl = 0.04
sol = in_curve(rect,[L,H],(100,100),dl)
sol_tri,Nint,Nbord = triang(sol,dl,rect,[L,H],rect_rim(L,H,dl))
# triangulation
tri = scipy.spatial.Delaunay(sol_tri)
M = Laplacian(sol_tri,tri,Nint)
valp,vecp = np.linalg.eig(M) # eigenvalues and eigenvectors
vecp = np.real(vecp)
# comparison with the exact solution:
T = 1000
U = np.arange(0,T,1)
NUsim = np.array([len(count_vp(valp,u)) for u in U])
NU = np.array([len(rect_drum(L,H,u)) for u in U])
plt.plot(U,NUsim,label='simulation')
plt.plot(U,NU,label='exacts')
plt.legend()
plt.show()
# 3D plot of an eigenvector
mode = 0 # change this for an another mode
vecp_tot = np.vstack((vecp,np.zeros((Nbord,Nint))))
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot_trisurf(sol_tri[:,0],sol_tri[:,1],vecp_tot[:,mode],triangles=tri.simplices)
plt.show()
Notes :
1- The hight eigenvalues are false : it's an effect of discretisation.
2- If dl is too small, we have false eigenvectors and eigenvalues (at the top of valp and firsts vectors of vecp), it's probably due to the quality of the meshing.
Python code:
I have used the Python code as below. Here, machine is trained by using Logistic Regression algorithm and wine dataset. Here, problem is that weights are not getting updated. I don't understand where is the problem.
from sklearn import datasets
import numpy as np
import pandas as pd
from sklearn.model_selection import train_test_split
dataset = datasets.load_wine()
x = dataset.data
y = dataset.target
y = y.reshape(178,1)
x_train,x_test,y_train,y_test = train_test_split(x,y,test_size=0.15,shuffle=True)
print(x_train.shape)
class log_reg():
def __init__(self):
pass
def sigmoid(self,x):
return 1 / (1 + np.exp(-x))
def train(self,x,y,w1,w2,alpha,iterations):
cost_history = [0] * iterations
Y_train = np.zeros([y.shape[0],3])
for i in range(Y_train.shape[0]):
for j in range(Y_train.shape[1]):
if(y[i] == j):
Y_train[i,j] = 1
for iteration in range(iterations):
z1 = x.dot(w1)
a1 = self.sigmoid(z1)
z2 = a1.dot(w2)
a2 = self.sigmoid(z2)
sig_sum = np.sum(np.exp(a2),axis=1)
sig_sum = sig_sum.reshape(a2.shape[0],1)
op = np.exp(a2) / sig_sum
loss = (Y_train * np.log(op))
dl = (op-Y_train)
dz1 = ((dl*(self.sigmoid(z2))*(1-self.sigmoid(z2))).dot(w2.T))*(self.sigmoid(z1))*(1-self.sigmoid(z1))
dz2 = (dl * (self.sigmoid(z2))*(1-self.sigmoid(z2)))
dw1 = x.T.dot(dz1)
dw2 = a1.T.dot(dz2)
w1 += alpha * dw1
w2 += alpha * dw2
cost_history[iteration] = (np.sum(loss)/len(loss))
return w1,w2,cost_history
def predict(self,x,y,w1,w2):
z1 = x.dot(w1)
a1 = self.sigmoid(z1)
z2 = a1.dot(w2)
a2 = self.sigmoid(z2)
sig_sum = np.sum(np.exp(a2),axis=1)
sig_sum = sig_sum.reshape(a2.shape[0],1)
op = np.exp(a2) / sig_sum
y_preds = np.argmax(op,axis=1)
acc = self.accuracy(y_preds,y)
return y_preds,acc
def accuracy(self,y_preds,y):
y_preds = y_preds.reshape(len(y_preds),1)
correct = (y_preds == y)
accuracy = (np.sum(correct) / len(y)) * 100
return (accuracy)
if __name__ == "__main__":
network = log_reg()
w1 = np.random.randn(14,4) * 0.01
w2 = np.random.randn(4,3) * 0.01
X_train = np.ones([x_train.shape[0],x_train.shape[1]+1])
X_train[:,:-1] = x_train
X_test = np.ones([x_test.shape[0],x_test.shape[1]+1])
X_test[:,:-1] = x_test
new_w1,new_w2,cost = network.train(X_train,y_train,w1,w2,0.0045,10000)
y_preds,accuracy = network.predict(X_test,y_test,new_w1,new_w2)
print(y_preds,accuracy)
In the above code, parameters are mentioned as below
x--training set,
y--target(output),
w1--weights for first layer,
w2--weights for second layer,
I used logistic regression with 2 hidden layers.
I am trying to train dataset wine from sklearn.I don't know where the problem is, but weights are not updating. Any help would be appreciated.
Your weights are updating , but I think you cant see them changing because you are printing them after execution. Python has a object reference method for numpy arrays so when you passed w1 , its values change values too so new_w1 and w1 become the same .
Take this example
import numpy as np
x=np.array([1,2,3,4])
def change(x):
x+=3
return x
print(x)
change(x)
print(x)
if you see the output it comes out as
[1 2 3 4]
[4 5 6 7]
I recommend that you add a bias and fix your accuracy function as I get my accuracy as 1000.
My execution when i run the code
the w1 and w2 values are indeed changing .
the only thing i changed was the main code and enabled the original data set , please do the same and tell if your weights are still not updating
if __name__ == "__main__":
network = log_reg()
w1 = np.random.randn(13,4) * 0.01
w2 = np.random.randn(4,3) * 0.01
print(w1)
print(" ")
print(w2)
print(" ")
new_w1,new_w2,cost = network.train(x_train,y_train,w1,w2,0.0045,10000)
print(w1)
print(" ")
print(w2)
print(" ")
y_preds,accuracy = network.predict(x_test,y_test,new_w1,new_w2)
print(y_preds,accuracy)
I've never used incremental PCA which exists in sklearn and I'm a bit confused about it's parameters and not able to find a good explanation of them.
I see that there is batch_size in the constructor, but also, when using partial_fit method you can again pass only a part of your data, I've found the following way:
n = df.shape[0]
chunk_size = 100000
iterations = n//chunk_size
ipca = IncrementalPCA(n_components=40, batch_size=1000)
for i in range(0, iterations):
ipca.partial_fit(df[i*chunk_size : (i+1)*chunk_size].values)
ipca.partial_fit(df[iterations*chunk_size : n].values)
Now, what I don't understand is the following - when using partial fit, does the batch_size play any role at all, or not? And how are they related?
Moreover, if both are considered, how should I change their values properly, when wanting to increase the precision while increasing memory footprint (and the other way around, decrease the memory consumption for the price of decreased accuracy)?
The docs say:
batch_size : int or None, (default=None)
The number of samples to use for each batch. Only used when calling fit...
This param is not used within partial_fit, where the batch-size is controlled by the user.
Bigger batches will increase memory-consumption, smaller ones will decrease it.
This is also written in the docs:
This algorithm has constant memory complexity, on the order of batch_size, enabling use of np.memmap files without loading the entire file into memory.
Despite some checks and parameter-heuristics, the whole fit-function looks like this:
for batch in gen_batches(n_samples, self.batch_size_):
self.partial_fit(X[batch], check_input=False)
Here is some an incremental PCA code based on https://github.com/kevinhughes27/pyIPCA which is an implementation of CCIPCA method.
import scipy.sparse as sp
import numpy as np
from scipy import linalg as la
import scipy.sparse as sps
from sklearn import datasets
class CCIPCA:
def __init__(self, n_components, n_features, amnesic=2.0, copy=True):
self.n_components = n_components
self.n_features = n_features
self.copy = copy
self.amnesic = amnesic
self.iteration = 0
self.mean_ = None
self.components_ = None
self.mean_ = np.zeros([self.n_features], np.float)
self.components_ = np.ones((self.n_components,self.n_features)) / \
(self.n_features*self.n_components)
def partial_fit(self, u):
n = float(self.iteration)
V = self.components_
# amnesic learning params
if n <= int(self.amnesic):
w1 = float(n+2-1)/float(n+2)
w2 = float(1)/float(n+2)
else:
w1 = float(n+2-self.amnesic)/float(n+2)
w2 = float(1+self.amnesic)/float(n+2)
# update mean
self.mean_ = w1*self.mean_ + w2*u
# mean center u
u = u - self.mean_
# update components
for j in range(0,self.n_components):
if j > n: pass
elif j == n: V[j,:] = u
else:
# update the components
V[j,:] = w1*V[j,:] + w2*np.dot(u,V[j,:])*u / la.norm(V[j,:])
normedV = V[j,:] / la.norm(V[j,:])
normedV = normedV.reshape((self.n_features, 1))
u = u - np.dot(np.dot(u,normedV),normedV.T)
self.iteration += 1
self.components_ = V / la.norm(V)
return
def post_process(self):
self.explained_variance_ratio_ = np.sqrt(np.sum(self.components_**2,axis=1))
idx = np.argsort(-self.explained_variance_ratio_)
self.explained_variance_ratio_ = self.explained_variance_ratio_[idx]
self.components_ = self.components_[idx,:]
self.explained_variance_ratio_ = (self.explained_variance_ratio_ / \
self.explained_variance_ratio_.sum())
for r in range(0,self.components_.shape[0]):
d = np.sqrt(np.dot(self.components_[r,:],self.components_[r,:]))
self.components_[r,:] /= d
You can test it with
import pandas as pd, ccipca
df = pd.read_csv('iris.csv')
df = np.array(df)[:,:4].astype(float)
pca = ccipca.CCIPCA(n_components=2,n_features=4)
S = 10
print df[0, :]
for i in range(150): pca.partial_fit(df[i, :])
pca.post_process()
The resulting eigenvectors / values will not exaactly be the same as the batch PCA. Results are approximate, but they are useful.