I'm trying to solve a LP problem using Pyomo where my decision variable (Pst = Pst+ + Pst-) can be either positive or negative (Pst+ for discharging and Pst- for charging my battery) and it's the same for Pg. I would like to know if it is obligatory to specify the domain as binary so that the solver CPLEX can solve the problem or else how do I specify the domain of my decision variable.
import pyomo.environ as pyo
model = pyo.ConcreteModel()
#Défine parameters
model.Ppv = 5
model.Pch = 20
#Power max décharge battery (parameters)
model.Pbattdechargemax = 10
model.rendementdecharge = 0.85
#Power max charge battery (parameters)
model.Pbattchargemax = -10
model.rendementcharge = 0.85
#Défine decision variables
model.Pg_decharge = pyo.Var(domain=pyo.NonNegativeReals)
model.Pg_charge = pyo.Var(domain=pyo.NegativeReals)
model.Pst_decharge = pyo.Var(domain=pyo.NonNegativeReals)
model.Pst_charge = pyo.Var(domain=pyo.NegativeReals)
#Objective function
model.OBJ = pyo.Objective(sense = pyo.minimize, expr =
2*model.Pg_decharge + 2*model.Pg_charge +
3*model.Pst_decharge + 3*model.Pst_charge)
#Contraints
model.Constraint1 = pyo.Constraint(expr = model.Pg_decharge +
model.Ppv + model.Pst_decharge == model.Pch)
model.Constraint2 = pyo.Constraint(expr = model.Pst_decharge
<= model.Pbattdechargemax * model.rendementdecharge)
model.Constraint3 = pyo.Constraint(expr = model.Pst_charge >=
model.Pbattchargemax / model.rendementcharge)
ERROR: evaluating object as numeric value: Pg_decharge
(object: <class 'pyomo.core.base.var.ScalarVar'>)
No value for uninitialized NumericValue object Pg_decharge
ERROR: evaluating object as numeric value: OBJ
(object: <class 'pyomo.core.base.objective.ScalarObjective'>)
No value for uninitialized NumericValue object Pg_decharge
Related
I'm trying to practice using pymc3 on the kinds of data I come across in my research, but I'm having trouble thinking through how to fit the model when each person gives me multiple data points, and each person comes from a different group (so trying a hierarchical model).
Here's the practice scenario I'm using: Suppose we have 2 groups of people, N = 30 in each group. All 60 people go through a 10 question survey, where each person can response ("1") or not respond ("0") to each question. So, for each person, I have an array of length 10 with 1's and 0's.
To model these data, I assume each person has some latent trait "theta", and each item has a "discrimination" a and a "difficulty" b (this is just a basic item response model), and the probability of responding ("1") is given by: (1 + exp(-a(theta - b)))^(-1). (Logistic applied to a(theta - b) .)
Here is how I tried to fit it using pymc3:
traces = {}
for grp in range(2):
group = prac_data["Group ID"] == grp
data = prac_data[group]["Response"]
with pm.Model() as irt:
# Priors
a_tmp = pm.Normal('a_tmp',mu=0, sd = 1, shape = 10)
a = pm.Deterministic('a', np.exp(a_tmp))
# We do this transformation since we must have a >= 0
b = pm.Normal('b', mu = 0, sd = 1, shape = 10)
# Now for the hyperpriors on the groups:
theta_mu = pm.Normal('theta_mu', mu = 0, sd = 1)
theta_sigma = pm.Uniform('theta_sigma', upper = 2, lower = 0)
theta = pm.Normal('theta', mu = theta_mu,
sd = theta_sigma, shape = N)
p = getProbs(Disc, Diff, theta, N)
y = pm.Bernoulli('y', p = p, observed = data)
traces[grp] = pm.sample(1000)
The function "getProbs" is supposed to give me an array of probabilities for the Bernoulli random variable, as the probability of responding 1 changes across trials/survey questions for each person. But this method gives me an error because it says to "specify one of p or logit_p", but I thought I did with the function?
Here's the code for "getProbs" in case it's helpful:
def getProbs(Disc, Diff, THETA, Nprt):
# Get a large array of probabilities for the bernoulli random variable
n = len(Disc)
m = Nprt
probs = np.array([])
for th in range(m):
for t in range(n):
p = item(Disc[t], Diff[t], THETA[th])
probs = np.append(probs, p)
return probs
I added the Nprt parameter because if I tried to get the length of THETA, it would give me an error since it is a FreeRV object. I know I can try and vectorize the "item" function, which is just the logistic function I put above, instead of doing it this way, but that also got me an error when I tried to run it.
I think I can do something with pm.Data to fix this, but the documentation isn't exactly clear to me.
Basically, I'm used to building models in JAGS, where you loop through each data point, but pymc3 doesn't seem to work like that. I'm confused about how to build/index my random variables in the model to make sure that the probabilities change how I'd like them to from trial-to-trial, and to make sure that the parameters I'm estimating correspond to the right person in the right group.
Thanks in advance for any help. I'm pretty new to pymc3 and trying to get the hang of it, and wanted to try something different from JAGS.
EDIT: I was able to solve this by first building the array I needed by looping through the trials, then transforming the array using:
p = theano.tensor.stack(p, axis = 0)
I then put this new variable in the "p" argument of the Bernoulli instance and it worked! Here's the updated full model: (below, I imported theano.tensor as T)
group = group.astype('int')
data = prac_data["Response"]
with pm.Model() as irt:
# Priors
# Item parameters:
a = pm.Gamma('a', alpha = 1, beta = 1, shape = 10) # Discrimination
b = pm.Normal('b', mu = 0, sd = 1, shape = 10) # Difficulty
# Now for the hyperpriors on the groups: shape = 2 as there are 2 groups
theta_mu = pm.Normal('theta_mu', mu = 0, sd = 1, shape = 2)
theta_sigma = pm.Uniform('theta_sigma', upper = 2, lower = 0, shape = 2)
# Individual-level person parameters:
# group is a 2*N array that lets the model know which
# theta_mu to use for each theta to estimate
theta = pm.Normal('theta', mu = theta_mu[group],
sd = theta_sigma[group], shape = 2*N)
# Here, we're building an array of the probabilities we need for
# each trial:
p = np.array([])
for n in range(2*N):
for t in range(10):
x = -a[t]*(theta[n] - b[t])
p = np.append(p, x)
# Here, we turn p into a tensor object to put as an argument to the
# Bernoulli random variable
p = T.stack(p, axis = 0)
y = pm.Bernoulli('y', logit_p = p, observed = data)
# On my computer, this took about 5 minutes to run.
traces = pm.sample(1000, cores = 1)
print(az.summary(traces)) # Summary of parameter distributions
I have this 7 quasi-lorentzian curves which are fitted to my data.
and I would like to join them, to make one connected curved line. Do You have any ideas how to do this? I've read about ComposingModel at lmfit documentation, but it's not clear how to do this.
Here is a sample of my code of two fitted curves.
for dataset in [Bxfft]:
dataset = np.asarray(dataset)
freqs, psd = signal.welch(dataset, fs=266336/300, window='hamming', nperseg=16192, scaling='spectrum')
plt.semilogy(freqs[0:-7000], psd[0:-7000]/dataset.size**0, color='r', label='Bx')
x = freqs[100:-7900]
y = psd[100:-7900]
# 8 Hz
model = Model(lorentzian)
params = model.make_params(amp=6, cen=5, sig=1, e=0)
result = model.fit(y, params, x=x)
final_fit = result.best_fit
print "8 Hz mode"
print(result.fit_report(min_correl=0.25))
plt.plot(x, final_fit, 'k-', linewidth=2)
# 14 Hz
x2 = freqs[220:-7780]
y2 = psd[220:-7780]
model2 = Model(lorentzian)
pars2 = model2.make_params(amp=6, cen=10, sig=3, e=0)
pars2['amp'].value = 6
result2 = model2.fit(y2, pars2, x=x2)
final_fit2 = result2.best_fit
print "14 Hz mode"
print(result2.fit_report(min_correl=0.25))
plt.plot(x2, final_fit2, 'k-', linewidth=2)
UPDATE!!!
I've used some hints from user #MNewville, who posted an answer and using his code I got this:
So my code is similar to his, but extended with each peak. What I'm struggling now is replacing ready LorentzModel with my own.
The problem is when I do this, the code gives me an error like this.
C:\Python27\lib\site-packages\lmfit\printfuncs.py:153: RuntimeWarning:
invalid value encountered in double_scalars [[Model]] spercent =
'({0:.2%})'.format(abs(par.stderr/par.value))
About my own model:
def lorentzian(x, amp, cen, sig, e):
return (amp*(1-e)) / ((pow((1.0 * x - cen), 2)) + (pow(sig, 2)))
peak1 = Model(lorentzian, prefix='p1_')
peak2 = Model(lorentzian, prefix='p2_')
peak3 = Model(lorentzian, prefix='p3_')
# make composite by adding (or multiplying, etc) components
model = peak1 + peak2 + peak3
# make parameters for the full model, setting initial values
# using the prefixes
params = model.make_params(p1_amp=6, p1_cen=8, p1_sig=1, p1_e=0,
p2_ampe=16, p2_cen=14, p2_sig=3, p2_e=0,
p3_amp=16, p3_cen=21, p3_sig=3, p3_e=0,)
rest of the code is similar like at #MNewville
[![enter image description here][3]][3]
A composite model for 3 Lorentzians would look like this:
from lmfit import Model, LorentzianModel
peak1 = LorentzianModel(prefix='p1_')
peak2 = LorentzianModel(prefix='p2_')
peak3 = LorentzianModel(prefix='p3_')
# make composite by adding (or multiplying, etc) components
model = peak1 + peaks2 + peak3
# make parameters for the full model, setting initial values
# using the prefixes
params = model.make_params(p1_amplitude=10, p1_center=8, p1_sigma=3,
p2_amplitude=10, p2_center=15, p2_sigma=3,
p3_amplitude=10, p3_center=20, p3_sigma=3)
# perhaps set bounds to prevent peaks from swapping or crazy values
params['p1_amplitude'].min = 0
params['p2_amplitude'].min = 0
params['p3_amplitude'].min = 0
params['p1_sigma'].min = 0
params['p2_sigma'].min = 0
params['p3_sigma'].min = 0
params['p1_center'].min = 2
params['p1_center'].max = 11
params['p2_center'].min = 10
params['p2_center'].max = 18
params['p3_center'].min = 17
params['p3_center'].max = 25
# then do a fit over the full data range
result = model.fit(y, params, x=x)
I think the key parts you were missing were: a) just add models together, and b) use prefix to avoid name collisions of parameters.
I hope that is enough to get you started...
So what I'm trying to accomplish is the newsvendor problem, where the program is supposed to run and give me the demand that gives me the best chances of turning a profit as per this link.
But the issue is that when I run the below code it gives me the right demand along with the demand for it.
q = {5:0.2 ,6:0.25 ,7:0.3 ,8:.25}
w = 55
p = 80
s = 40
cul5 = 0.2
cul6 = 0.25
cul7 = 0.3
cul8 = .25
overage = w - s
underage = p - w
crit = overage/float((underage)+(overage)) # its better to use floats within the parenthesis in python 2
cumul_q = {}
cumulativevalue = 0
for key, value in sorted(q.iteritems()):
cumulativevalue = cumulativevalue + value
# print key , value
cumul_q[key] = cumulativevalue
# print cumul_q
previous_key = None
for key, value, in sorted(cumul_q.iteritems(),reverse = True):
cumulprob = 1 - value
cumulprob1 = float(cumulprob)
if crit < cumulprob1:
continue
elif crit > cumulprob1:
print key
previous_key = key
I'm trying to solve some convolution integrals but without any luck.
import matplotlib.pyplot as plt
import numpy as np
import sympy as sp
t1 = 0.01
t1_sym = sp.Symbol('t_1')
F0 = 30e3
F0_sym = sp.Symbol('F_0')
m = 4000
m_sym = sp.Symbol('m')
k = 5000e3
k_sym = sp.Symbol('k')
omega0_sym = sp.sqrt(k_sym/m_sym)
t = sp.Symbol('t')
tau = sp.Symbol(r'\tau')
F1_sym= 2*F0_sym*tau/t1_sym
x1_sym = sp.integrate((F1_sym*sp.sin(omega0_sym*(t-tau))/(omega0_sym*m_sym)) ,(tau,0,t))
I would be really grateful for any insights.
In general, it is a good practise to provide sympy any information available about the parameters and variables involved in an integral, instead of simply declaring them as symbols. From the numerical values you are using, it is clear that all variables and parameters are real and positive. You can provide sympy this information when you define them. With this information, sympy provides an expression for the integral (I will leave it to you to check if it is correct or not)
t1_sym = sp.Symbol('t_1', positive = True)
F0_sym = sp.Symbol('F_0', positive = True)
m_sym = sp.Symbol('m', positive = True)
k_sym = sp.Symbol('k', positive = True)
omega0_sym = sp.sqrt(k_sym/m_sym)
t = sp.Symbol('t', positive = True)
tau = sp.Symbol(r'\tau', positive = True)
F1_sym= 2*F0_sym*tau/t1_sym
x1_sym = sp.integrate((F1_sym*sp.sin(omega0_sym*(t-tau))/(omega0_sym*m_sym)) ,(tau,0,t))
print(x1_sym)
2*F_0*t/(k*t_1) - 2*F_0*sqrt(m)*sin(sqrt(k)*t/sqrt(m))/(k**(3/2)*t_1)
I need a function in python to return N random numbers from a skew normal distribution. The skew needs to be taken as a parameter.
e.g. my current use is
x = numpy.random.randn(1000)
and the ideal function would be e.g.
x = randn_skew(1000, skew=0.7)
Solution needs to conform with: python version 2.7, numpy v.1.9
A similar answer is here: skew normal distribution in scipy However this generates a PDF not the random numbers.
I start by generating the PDF curves for reference:
NUM_SAMPLES = 100000
SKEW_PARAMS = [-3, 0]
def skew_norm_pdf(x,e=0,w=1,a=0):
# adapated from:
# http://stackoverflow.com/questions/5884768/skew-normal-distribution-in-scipy
t = (x-e) / w
return 2.0 * w * stats.norm.pdf(t) * stats.norm.cdf(a*t)
# generate the skew normal PDF for reference:
location = 0.0
scale = 1.0
x = np.linspace(-5,5,100)
plt.subplots(figsize=(12,4))
for alpha_skew in SKEW_PARAMS:
p = skew_norm_pdf(x,location,scale,alpha_skew)
# n.b. note that alpha is a parameter that controls skew, but the 'skewness'
# as measured will be different. see the wikipedia page:
# https://en.wikipedia.org/wiki/Skew_normal_distribution
plt.plot(x,p)
Next I found a VB implementation of sampling random numbers from the skew normal distribution and converted it to python:
# literal adaption from:
# http://stackoverflow.com/questions/4643285/how-to-generate-random-numbers-that-follow-skew-normal-distribution-in-matlab
# original at:
# http://www.ozgrid.com/forum/showthread.php?t=108175
def rand_skew_norm(fAlpha, fLocation, fScale):
sigma = fAlpha / np.sqrt(1.0 + fAlpha**2)
afRN = np.random.randn(2)
u0 = afRN[0]
v = afRN[1]
u1 = sigma*u0 + np.sqrt(1.0 -sigma**2) * v
if u0 >= 0:
return u1*fScale + fLocation
return (-u1)*fScale + fLocation
def randn_skew(N, skew=0.0):
return [rand_skew_norm(skew, 0, 1) for x in range(N)]
# lets check they at least visually match the PDF:
plt.subplots(figsize=(12,4))
for alpha_skew in SKEW_PARAMS:
p = randn_skew(NUM_SAMPLES, alpha_skew)
sns.distplot(p)
And then wrote a quick version which (without extensive testing) appears to be correct:
def randn_skew_fast(N, alpha=0.0, loc=0.0, scale=1.0):
sigma = alpha / np.sqrt(1.0 + alpha**2)
u0 = np.random.randn(N)
v = np.random.randn(N)
u1 = (sigma*u0 + np.sqrt(1.0 - sigma**2)*v) * scale
u1[u0 < 0] *= -1
u1 = u1 + loc
return u1
# lets check again
plt.subplots(figsize=(12,4))
for alpha_skew in SKEW_PARAMS:
p = randn_skew_fast(NUM_SAMPLES, alpha_skew)
sns.distplot(p)
from scipy.stats import skewnorm
a=10
data= skewnorm.rvs(a, size=1000)
Here, a is a parameter which you can refer to:
https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.skewnorm.html
Adapted from rsnorm function from fGarch R package
def random_snorm(n, mean = 0, sd = 1, xi = 1.5):
def random_snorm_aux(n, xi):
weight = xi/(xi + 1/xi)
z = numpy.random.uniform(-weight,1-weight,n)
xi_ = xi**numpy.sign(z)
random = -numpy.absolute(numpy.random.normal(0,1,n))/xi_ * numpy.sign(z)
m1 = 2/numpy.sqrt(2 * numpy.pi)
mu = m1 * (xi - 1/xi)
sigma = numpy.sqrt((1 - m1**2) * (xi**2 + 1/xi**2) + 2 * m1**2 - 1)
return (random - mu)/sigma
return random_snorm_aux(n, xi) * sd + mean