I made a Rational class in python and made the __add__, __mull__, etc. operators for that class. Here is my code for the __init__ and __add__ functions:
class Rational(object):
def __init__(self, p, q=None): # p/q
if q is None: q = 1
if q == 0: raise ZeroDivisionError("division by zero!")
self.p = p
self.q = q
self.simplify() # simplify the fraction
def __iadd__(self, other): # +=
self.p = self.p * other.q + self.q * other.p
self.q = self.q * other.q
self.simplify()
return self
def __add__(self, other):
r = Rational(self.p, self.q)
r += other
return r
Now I would want python to understand things like 1 + Rational(1, 3) (one plus a third). How could I implement this? Do I have to add an __radd__ function and add cases to when other is int? Or is there a simpler way?
I believe this might be what you're looking for:
class Rational(object):
def __init__(self, p, q=None): # p/q
if q is None: q = 1
if q == 0: raise ZeroDivisionError("division by zero!")
self.p = p
self.q = q
self.simplify() # simplify the fraction
def __iadd__(self, other): # +=
self.p = self.p * other.q + self.q * other.p
self.q = self.q * other.q
self.simplify()
return self
def __add__(self, other):
if type(other) == int :
r = Rational(self.p + other * self.q , self.q)
else:
r = Rational(self.p, self.q)
r += other
return r
This way you can define what happens when you add your Rational to another type, in this case you're adding an int so it relies on the following identity:
(p/q)+r=(p+qr)/q
I hope this helps!
EDIT: made a dumb mathematical error, fixed now
Related
I'm trying to implement Gaussian Mixture Model with Expectation–Maximization algorithm and I get this error.
This is the Gaussian Mixture model that I used:
class GaussianMixture:
"Model mixture of two univariate Gaussians and their EM estimation"
def __init__(self, data, mu_min=min(data), mu_max=max(data), sigma_min=.1, sigma_max=1, mix=.5):
self.data = data
#init with multiple gaussians
self.one = Gaussian(uniform(mu_min, mu_max),
uniform(sigma_min, sigma_max))
self.two = Gaussian(uniform(mu_min, mu_max),
uniform(sigma_min, sigma_max))
#as well as how much to mix them
self.mix = mix
def Estep(self):
"Perform an E(stimation)-step, freshening up self.loglike in the process"
# compute weights
self.loglike = 0. # = log(p = 1)
for datum in self.data:
# unnormalized weights
wp1 = self.one.pdf(datum) * self.mix
wp2 = self.two.pdf(datum) * (1. - self.mix)
# compute denominator
den = wp1 + wp2
# normalize
wp1 /= den
wp2 /= den
# add into loglike
self.loglike += log(wp1 + wp2)
# yield weight tuple
yield (wp1, wp2)
def Mstep(self, weights):
"Perform an M(aximization)-step"
# compute denominators
(left, rigt) = zip(*weights)
one_den = sum(left)
two_den = sum(rigt)
# compute new means
self.one.mu = sum(w * d / one_den for (w, d) in zip(left, data))
self.two.mu = sum(w * d / two_den for (w, d) in zip(rigt, data))
# compute new sigmas
self.one.sigma = sqrt(sum(w * ((d - self.one.mu) ** 2)
for (w, d) in zip(left, data)) / one_den)
self.two.sigma = sqrt(sum(w * ((d - self.two.mu) ** 2)
for (w, d) in zip(rigt, data)) / two_den)
# compute new mix
self.mix = one_den / len(data)
def iterate(self, N=1, verbose=False):
"Perform N iterations, then compute log-likelihood"
def pdf(self, x):
return (self.mix)*self.one.pdf(x) + (1-self.mix)*self.two.pdf(x)
def __repr__(self):
return 'GaussianMixture({0}, {1}, mix={2.03})'.format(self.one,
self.two,
self.mix)
def __str__(self):
return 'Mixture: {0}, {1}, mix={2:.03})'.format(self.one,
self.two,
self.mix)
And then , while training I get that error in the conditional statement.
# Check out the fitting process
n_iterations = 5
best_mix = None
best_loglike = float('-inf')
mix = GaussianMixture(data)
for _ in range(n_iterations):
try:
#train!
mix.iterate(verbose=True)
if mix.loglike > best_loglike:
best_loglike = mix.loglike
best_mix = mix
except (ZeroDivisionError, ValueError, RuntimeWarning): # Catch division errors from bad starts, and just throw them out...
pass
Any ideas why I have the following error?
AttributeError: GaussianMixture instance has no attribute 'loglike'
The loglike attribute is only created when you call the Estep method.
def Estep(self):
"Perform an E(stimation)-step, freshening up self.loglike in the process"
# compute weights
self.loglike = 0. # = log(p = 1)
You didn't call Estep between creating the GaussianMixture instance and mix.loglike:
mix = GaussianMixture(data)
for _ in range(n_iterations):
try:
#train!
mix.iterate(verbose=True)
if mix.loglike > best_loglike:
And the iterate method is empty (It looks like you forgot some code here).
def iterate(self, N=1, verbose=False):
"Perform N iterations, then compute log-likelihood"
So, there'll be no loglike attribute set on the mix instance by the time you do if mix.loglike. Hence, the AttributeError.
You need to do one of the following:
Call the Estep method (since you set self.loglike there)
Define a loglike attribute in __init__
I am trying migrating the ampl car problem that comes in the Ipopt source code tarball as example. I am having got problems with the end condition (reach a place with zero speed at final iteration) and with the cost function (minimize final time).
Can someone help me revise the following model?
# min tf
# dx/dt = 0
# dv/dt = a - R*v^2
# x(0) = 0; x(tf) = 100
# v(0) = 0; v(tf) = 0
# -3 <= a <= 1 (a is the control variable)
#!Python3.5
from pyomo.environ import *
from pyomo.dae import *
N = 20;
T = 10;
L = 100;
m = ConcreteModel()
# Parameters
m.R = Param(initialize=0.001)
# Variables
def x_init(m, i):
return i*L/N
m.t = ContinuousSet(bounds=(0,1000))
m.x = Var(m.t, bounds=(0,None), initialize=x_init)
m.v = Var(m.t, bounds=(0,None), initialize=L/T)
m.a = Var(m.t, bounds=(-3.0,1.0), initialize=0)
# Derivatives
m.dxdt = DerivativeVar(m.x, wrt=m.t)
m.dvdt = DerivativeVar(m.v, wrt=m.t)
# Objetives
m.obj = Objective(expr=m.t[N])
# DAE
def _ode1(m, i):
if i==0:
return Constraint.Skip
return m.dxdt[i] == m.v[i]
m.ode1 = Constraint(m.t, rule=_ode1)
def _ode2(m, i):
if i==0:
return Constraint.Skip
return m.dvdt[i] == m.a[i] - m.R*m.v[i]**2
m.ode2 = Constraint(m.t, rule=_ode2)
# Constraints
def _init(m):
yield m.x[0] == 0
yield m.v[0] == 0
yield ConstraintList.End
m.init = ConstraintList(rule=_init)
'''
def _end(m, i):
if i==N:
return m.x[i] == L amd m.v[i] == 0
return Constraint.Skip
m.end = ConstraintList(rule=_end)
'''
# Discretize
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, nfe=N, wrt=m.t, scheme='BACKWARD')
# Solve
solver = SolverFactory('ipopt', executable='C:\\EXTERNOS\\COIN-OR\\win32-msvc12\\bin\\ipopt')
results = solver.solve(m, tee=True)
Currently, a ContinuousSet in Pyomo has to be bounded. This means that in order to solve a minimum time optimal control problem using this tool, the problem must be reformulated to remove the time scaling from the ContinuousSet. In addition, you have to introduce an extra variable to represent the final time. I've added an example to the Pyomo github repository showing how this can be done for your problem.
After having seen the nice implementation of the "ampl car example" in Pyomo repository, I would like to keep extending the problem with new features and constraints, but I have found the next problems during development. Is someone able of fix them?
1) Added new constraint "electric car": Now the acceleration is limited by adherence until a determined speed and then constant power model is used. I am not able of implement this constraint as i would think. It is commented in the, but Pyomo complains about that a constraint is related to a variable. (now Umax depends of the car speed).
2) Added new comfort acceleration and jerk constraints. It seems they are working right, but should be nice if a Pyomo guru supervise them and tell me if they are really implemented in the correct way.
3) About last one, in order of reducing verbosity. Is there any way of combine accelerationL and accelerationU in a unique constraint? Same for jerkL and jerkU.
4) The last feature is a speed limit constraint divided in two steps. Again, I am not able of getting it works, so it is commented in code. Does anybody dare to fix it?
# Ampl Car Example (Extended)
#
# Shows how to convert a minimize final time optimal control problem
# to a format pyomo.dae can handle by removing the time scaling from
# the ContinuousSet.
#
# min tf
# dx/dt = v
# dv/dt = u - R*v^2
# x(0)=0; x(tf)=L
# v(0)=0; v(tf)=0
# -3 <= u <= 1 (engine constraint)
#
# {v <= 7m/s ===> u < 1
# u <= { (electric car constraint)
# {v > 7m/s ===> u < 1*7/v
#
# -1.5 <= dv/dt <= 0.8 (comfort constraint -> smooth driving)
# -0.5 <= d2v/dt2 <= 0.5 (comfort constraint -> jerk)
# v <= Vmax (40 kmh[0-500m] + 25 kmh(500-1000m])
from pyomo.environ import *
from pyomo.dae import *
m = ConcreteModel()
m.R = Param(initialize=0.001) # Friction factor
m.L = Param(initialize=1000.0) # Final position
m.T = Param(initialize=50.0) # Estimated time
m.aU = Param(initialize=0.8) # Acceleration upper bound
m.aL = Param(initialize=-1.5) # Acceleration lower bound
m.jU = Param(initialize=0.5) # Jerk upper bound
m.jL = Param(initialize=-0.5) # Jerk lower bound
m.NFE = Param(initialize=100) # Number of finite elements
'''
def _initX(m, i):
return m.x[i] == i*m.L/m.NFE
def _initV(m):
return m.v[i] == m.L/50
'''
m.tf = Var()
m.tau = ContinuousSet(bounds=(0,1)) # Unscaled time
m.t = Var(m.tau) # Scaled time
m.x = Var(m.tau, bounds=(0,m.L))
m.v = Var(m.tau, bounds=(0,None))
m.u = Var(m.tau, bounds=(-3,1), initialize=0)
m.dt = DerivativeVar(m.t)
m.dx = DerivativeVar(m.x)
m.dv = DerivativeVar(m.v)
m.da = DerivativeVar(m.v, wrt=(m.tau, m.tau))
m.obj = Objective(expr=m.tf)
def _ode1(m, i):
if i==0:
return Constraint.Skip
return m.dt[i] == m.tf
m.ode1 = Constraint(m.tau, rule=_ode1)
def _ode2(m, i):
if i==0:
return Constraint.Skip
return m.dx[i] == m.tf * m.v[i]
m.ode2 = Constraint(m.tau, rule=_ode2)
def _ode3(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] == m.tf*(m.u[i] - m.R*m.v[i]**2)
m.ode3 = Constraint(m.tau, rule=_ode3)
def _accelerationL(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] >= m.aL*m.tf
m.accelerationL = Constraint(m.tau, rule=_accelerationL)
def _accelerationU(m, i):
if i==0:
return Constraint.Skip
return m.dv[i] <= m.aU*m.tf
m.accelerationU = Constraint(m.tau, rule=_accelerationU)
def _jerkL(m, i):
if i==0:
return Constraint.Skip
return m.da[i] >= m.jL*m.tf**2
m.jerkL = Constraint(m.tau, rule=_jerkL)
def _jerkU(m, i):
if i==0:
return Constraint.Skip
return m.da[i] <= m.jU*m.tf**2
m.jerkU = Constraint(m.tau, rule=_jerkU)
'''
def _electric(m, i):
if i==0:
return Constraint.Skip
elif value(m.v[i])<=7:
return m.a[i] <= 1
else:
return m.v[i] <= 1*7/m.v[i]
m.electric = Constraint(m.tau, rule=_electric)
'''
'''
def _speed(m, i):
if i==0:
return Constraint.Skip
elif value(m.x[i])<=500:
return m.v[i] <= 40/3.6
else:
return m.v[i] <= 25/3.6
m.speed = Constraint(m.tau, rule=_speed)
'''
def _initial(m):
yield m.x[0] == 0
yield m.x[1] == m.L
yield m.v[0] == 0
yield m.v[1] == 0
yield m.t[0] == 0
m.initial = ConstraintList(rule=_initial)
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, nfe=value(m.NFE), wrt=m.tau, scheme='BACKWARD')
#discretizer = TransformationFactory('dae.collocation')
#discretizer.apply_to(m, nfe=value(m.NFE), ncp=4, wrt=m.tau, scheme='LAGRANGE-RADAU')
solver = SolverFactory('ipopt')
solver.solve(m,tee=True)
print("final time = %6.2f" %(value(m.tf)))
t = []
x = []
v = []
a = []
u = []
for i in m.tau:
t.append(value(m.t[i]))
x.append(value(m.x[i]))
v.append(3.6*value(m.v[i]))
a.append(10*value(m.u[i] - m.R*m.v[i]**2))
u.append(10*value(m.u[i]))
import matplotlib.pyplot as plt
plt.plot(x, v, label='v (km/h)')
plt.plot(x, a, label='a (dm/s2)')
plt.plot(x, u, label='u (dm/s2)')
plt.xlabel('distance')
plt.grid('on')
plt.legend()
plt.show()
Thanks a lot in advance,
Pablo
(1) You should not think of Pyomo constraint rules as callbacks that are used by the solver. You should think of them more as a function to generate a container of constraint objects that gets called once for each index when the model is constructed. Meaning it is invalid to use a variable in an if statement unless you are really only using its initial value to define the constraint expression. There are ways to express what I think you are trying to do, but they involve introducing binary variables into the problem, in which case you can no longer use Ipopt.
(2) Can't really provide any help. Syntax looks fine.
(3) Pyomo allows you to return double-sided inequality expressions (e.g., L <= f(x) <= U) from constraint rules, but they can not involve variable expressions in the L and U locations. It doesn't look like the constraints you are referring to can be combined into this form.
(4) See (1)
What would be the best approach to create a type that is a saturated integer in python ?
i.e.:
v = SaturatedInteger(0, 100)
# That should create an integer that will always be in between 0 and 100,
# and have all default operations
v += 68719
print v #Should print '100'.
I can think of inheriting int type, but where should the saturating logic be implemented then ?
If you need a new (quick and dirty) class for it, I would implement it as follows.
class SaturatedInteger:
def __init__(self, val, lo, hi):
self.real, self.lo, self.hi = val, lo, hi
def __add__(self, other):
return min(self.real + other.real, self.hi)
def __sub__(self, other):
return max(self.real - other.real, self.lo)
...
Add as many of the other operators in the docs as you feel you will need (and their 'r' variants).
By storing the value in the instance name real, you can do your arithmetic with regular integers, floats, etc. too:
a = SaturatedInteger(60, 0, 100)
print(a)
60
print(a+30)
90
print(a+40)
100
print(a+50.)
100
print(a-70.)
0
print(a+a)
100
Though, of course you only add the real part if you're adding a complex number to your SaturatedInteger, so watch out. (For a much more complete and robust version, #jonrsharpe's answer is the way to go).
In general, I would implement using a #property to protect an instance's value attribute, then emulate a numeric type, rather than inheriting from int:
class SaturatedInteger(object):
"""Emulates an integer, but with a built-in minimum and maximum."""
def __init__(self, min_, max_, value=None):
self.min = min_
self.max = max_
self.value = min_ if value is None else value
#property
def value(self):
return self._value
#value.setter
def value(self, new_val):
self._value = min(self.max, max(self.min, new_val))
#staticmethod
def _other_val(other):
"""Get the value from the other object."""
if hasattr(other, 'value'):
return other.value
return other
def __add__(self, other):
new_val = self.value + self._other_val(other)
return SaturatedInteger(self.min, self.max, new_val)
__radd__ = __add__
def __eq__(self, other):
return self.value == self._other_val(other)
if __name__ == '__main__':
v = SaturatedInteger(0, 100)
v += 68719
assert v == 100
assert 123 + v == 100
I've only implemented __add__, __radd__ and __eq__, but you can probably see how the rest could be built out as required. You might want to think about what happens when two SaturatedIntegers are used together - should the result have e.g. min(self.min, other.min) as its own min?
I wrote a sample class that has an add function:
class SatInt:
def __init__(self, low, up):
self.lower = low
self.upper = up
self.value = 0
def add(self, n):
if n+self.value > self.upper:
self.value = self.upper
else:
self.value = self.value + n
x = SatInt(0,100)
x.add(32718)
print(x.value)
100
i have a function that give me the result that im expecting in console mode, but if i try to use the function with Django, the page never load and just have a loop calculating and never end.
Any idea ?
*sorry with my english
Console function (WORK GREAT):
def sum_digitos(n):
sum = 0;
while n != 0:
sum += n % 10
n /= 10
if sum > 9:
x = str(sum)
y =list(x)
sum = int(y[0]) + int(y[1])
return sum
print sum_digitos(2461978)
Django views:
def Calcular(request):
if request.method == 'POST':
form = NumerologiaForm(request.POST)
if form.is_valid():
sum = 0;
ano = str(request.POST['fecha_year'])
mes = str(request.POST['fecha_month'])
dia = str(request.POST['fecha_day'])
data = dia + mes + ano
fecha = int(data)
while fecha != 0:
f = fecha
sum += f % 10
f /= 10
if sum > 9:
x = str(sum)
y =list(x)
sum = int(y[0]) + int(y[1])
resultado = get_object_or_404(Numero,numero = sum)
return HttpResponseRedirect(resultado.get_absolute_url())
else:
form = NumerologiaForm()
return render_to_response('numerologiaForm.html',{'form':form})
Try:
f = fecha
while f!= 0:
sum += f % 10
f /= 10
if sum > 9:
x = str(sum)
y =list(x)
sum = int(y[0]) + int(y[1])
It seems you were changing f, but checking fecha for the looping.
Sanjay's answer is the correct one, and I recommend it. I just wanted to ask why you didn't just do:
from numerology import sum_digitos
def Calcular(request):
# In your code, you return HttpResponseRedirect using a nonexistent
# "resultado" variable if the form is not valid. This will raise an
# exception. I think you meant to indent "return Http..." one step more.
if request.method == 'POST':
form = NumerologiaForm(request.POST)
else:
form = NumerologiaForm()
# "or..." part of next line not needed if form.is_valid() returns
# False for a blank form.
if not form.is_valid() or form == NumerologiaForm():
return render_to_response('numerologiaForm.html', {'form': form})
ano = str(request.POST['fecha_year'])
mes = str(request.POST['fecha_month'])
dia = str(request.POST['fecha_day'])
resultado = get_object_or_404(Numero,
numero=sum_digitos(int(dia + mes + ano)))
return HttpResponseRedirect(resultado.get_absolute_url())
You had a working function in Python already... why not just import it and use it?
There's no need to go to all that work to sum the digits in that number, because the sum of the digits is num % 9. If num % 9 is zero, then the actual sum of digits is 9.
By changing your method to
def sum_digitos(n):
sum_ = n % 9
return sum_ if sum_ != 0 else 9
You will completely avoid whatever issue was happening inside your original method.
You don't say what the rest of your environment is like, but you should be using f //= 10 to ensure that you're performing integer division.