I have estimated nested logit in R using the mlogit package. However, I encountered some problems when trying to estimate the marginal effect. Below is the code I implemented.
library(mlogit)
# data
data2 = read.csv(file = "neat_num_energy.csv")
new_ener2 <- mlogit.data(
data2,
choice="alter4", shape="long",
alt.var="energy_altern",chid.var="id")
# estimate model
nest2 <- mlogit(
alter4 ~ expendmaint + expendnegy |
educ + sex + ppa_power_sp + hu_price_powersupply +
hu_2price +hu_3price + hu_9price + hu_10price +
hu_11price + hu_12price,
data = data2,
nests = list(
Trad = c('Biomas_Trad', 'Solar_Trad'),
modern = c('Biomas_Modern', 'Solar_Modern')
), unscaled=FALSE)
# create Z variable
z3 <- with(data2, data.frame(
expendnegy = tapply(expendnegy, idx(nest2,2), mean),
expendmaint= tapply(expendmaint, idx(nest2,2), mean),
educ= mean(educ),
sex = mean(sex),
hu_price_powersupply = mean(hu_price_powersupply),
ppa_power_sp = mean(ppa_power_sp),
hu_2price = mean(hu_2price),
hu_3price = mean(hu_3price),
hu_9price = mean(hu_9price),
hu_10price = mean(hu_10price),
hu_11price = mean(hu_11price),
ppa_power_sp = mean(ppa_power_sp),
hu_12price = mean(hu_12price)
))
effects(nest2, covariate = "sex", data = z3, type = "ar")
#> ** Error in Solve.default (H, g[!fixed]): Lapack routine dgesv: #> system is exactly singular:U[6,6] =0.**
My data is in long format with expendmaint and expendnegy being the only alternative specific while every other variable is case specific.
altern4 is a nominal variable representing each alternative
Related
I am trying to use julai as main language for my work. But I find that this plot is different than python (Which outputs the right plot)
Here is the python code and output
import numpy as np
import math
import matplotlib.pyplot as plt
u = 9.27*10**(-21)
k = 1.38*10**(-16)
j2 = 7/2
nrr = 780
h = 1000
na = 6*10**(23)
rho = 7.842
mgd = 157.25
a = mgd
d = na*rho*u/a
m_f = []
igd = 7.0
for t in range(1,401):
while True:
h1 = h+d*nrr*igd
x2 = (7*u*h1)/(k*t)
x4 = 2*j2
q2 = (x4+1)/x4
m = abs(7*(q2*math.tanh(q2*x2)**-1 - (1/x4)*math.tanh(x2/x4)**-1))
if abs(m - igd) < 10**(-12):
break
else:
igd = m
m_f.append(abs(m))
plt.plot(range(1,401), m_f)
plt.savefig("Py_plot.pdf")
and it gives the following right plot
The right plot as expected
But when I do the same calculations in julia it gives different output than python, here is my julia code
using Plots
u = 9.27*10^(-21)
k = 1.38*10^(-16)
j2 = 7/2
nrr = 780
h = 1000
na = 6*10^(23)
rho = 7.842
mgd = 157.25
a = mgd
d = na*rho*u/a
igd = 7.0
m = 0.0
m_f = Float64[]
for t in 1:400
while true
h1 = h+d*nrr*igd
x2 = (7*u*h1)/(k*t)
x4 = 2*j2
q2 = (x4+1)/x4
m = 7*(q2*coth(rad2deg(q2*x2))-(1/x4)*coth(rad2deg(x2/x4)))
if abs(abs(m)-igd) < 10^(-10)
break
else
igd = m
end
end
push!(m_f, abs(m))
end
plot(1:400, m_f)
and this is the unexpected julia output
unexpected wrong output from julia
I wish for help....
Code:
using Plots
const u = 9.27e-21
const k = 1.38e-16
const j2 = 7/2
const nrr = 780
const h = 1000
const na = 6.0e23
const rho = 7.842
const mgd = 157.25
const a = mgd
const d = na*rho*u/a
function plot_graph()
igd = 7.0
m = 0.0
trange = 1:400
m_f = Vector{Float64}(undef, length(trange))
for t in trange
while true
h1 = h+d*nrr*igd
x2 = (7*u*h1)/(k*t)
x4 = 2*j2
q2 = (x4+1)/x4
m = abs(7*(q2*coth(q2*x2)-(1/x4)*coth(x2/x4)))
if isapprox(m, igd, atol = 10^(-10))
break
else
igd = m
end
end
m_f[t] = m
end
plot(trange, m_f)
end
Plot:
Changes for correctness:
Changed na = 6.0*10^(23) to na = 6.0e23.
Since ^ has a higher precedence than *, 10^23 is evaluated first, and since the operands are Int values, the result is also an Int. However, Int (i.e. Int64) can only hold numbers up to approximately 9 * 10^18, so 10^23 overflows and gives a wrong result.
julia> 10^18
1000000000000000000
julia> 10^19 #overflow starts here
-8446744073709551616
julia> 10^23 #and gives a wrong value here too
200376420520689664
6.0e23 avoids this problem by directly using the scientific e-notation to create a literal Float64 value (Float64 can hold this value without overflowing).
Removed the rad2deg calls when calling coth. Julia trigonometric functions by default take radians, so there's no need to make this conversion.
Other changes
Marked all the constants as const, and moved the rest of the code into a function. See Performance tip: Avoid non-constant global variables
Changed the abs(m - igd) < 10^-10 to isapprox(m, igd, atol = 10^-10) which performs basically the same check, but is clearer and more flexible (for eg. if you wanted to change to a relative tolerance rtol later).
Stored the 1:400 as a named variable trange. This is just because it's used multiple times, so it's easier to manage as a variable.
Changed m_f = Float64[] to m_f = Vector{Float64}(undef, length(trange)) (and the push! at the end to an assignment). If the size of the array is known beforehand (as it is in this case), it's better for performance to pre-allocate it with undef values and then assign to it.
Changed u and k also to use the scientific e-notation, for consistency and clarity (thanks to #DNF for suggesting the use of this notation in the comments).
i´ve a problem with running the following function to perform a Wald Test on my Data. It always displays me the error message that the variable theta is not defined. I tried to define it in struct WaldTestFun{F, T, Z} but this does not work. The error message im getting is
Error Message = UndefVarError: theta not defined
Here is a part of the code:
using LinearAlgebra
using Optim
using PrettyTables
using Printf
using Statistics
using StatsBase
using StatsFuns
struct WaldTest
tbl::NamedTuple
rankmin::Int64
rankₘₐₓ::Int64
end
struct WaldTestFun{F, T, Z}
f::F
r::Int64
vecsigma::T
Vhat::Z
end
(wf::WaldTestFun)(theta) = wf.f(theta, wf.r, wf.vecsigma, wf.Vhat) #Here the error occurs
function waldobjfun(th, r, vecsigma, Vhat)
r,k = size(theta) #Here the error occurs
theta = reshape(theta, r+1, length(th)÷(r+1))
sigmamat = diagm(0=>theta[1,:].^2) .+ theta[2:r+1,:]'*theta[2:r+1,:]
tempsigma = sigmamat[findall(tril(ones(size(sigmamat))).==1)]
(vecsigma -tempsigma)' /Vhat *(vecsigma - tempsigma)
end
X = randn(100,10);
fm = Factotum.FactorModel(X, 3)
function waldtest(fm::FactorModel, minrank::Int = 0, maxrank::Int = 2)
X = copy(fm.X)
T, n = size(X)
## Normalize factor
Xs = X / diagm(0=>sqrt.(diag(cov(X))))
covX = cov(Xs)
meanX = mean(Xs, dims=1)
vecsigma = vech(covX)
bigN = length(vecsigma)
Vhat = Array{Float64}(undef, bigN, bigN)
varvecsig = zeros(n,n,n,n);
I have an expression like :
b = IndexedBase('b')
k = IndexedBase('k')
w = IndexedBase('w')
r = IndexedBase('r')
z = IndexedBase('z')
i = symbols("i", cls=Idx)
omega = symbols("omega", cls=Idx)
e_p = (-k[i, omega]*r[i]/w[i] + k[i, omega]*r[i]/(b[omega]*w[i]))**b[omega]*k[i, omega]*r[i]/(-b[omega]*k[i, omega]*k[i, omega]**b[omega]*r[i]*z[omega]/w[i] + k[i, omega]*k[i, omega]**b[omega]*r[i]*z[omega]/w[i])
e_p = simplify(e_p)
print(type(e_p))
print(e_p)
<class 'sympy.core.mul.Mul'>
-(-(b[omega] - 1)*k[i, omega]*r[i]/(b[omega]*w[i]))**b[omega]*k[i, omega]**(-b[omega])*w[i]/((b[omega] - 1)*z[omega])
So k[i, omega] should be canceled out when I use simplify() function but do nothing. How can I get rid of unnecessary variables and coefficients?
The following example is for anyone who is building a Cox Proportional Hazards models and trying to produce prediction error curves, but get an error stating:
Error in coxModelFrame.coxph(object) : invalid object
set x=TRUE in the call to coxph.
Here is the code to reproduce the error:
LIBRARIES
library(survival)
library(survminer)
library(pec)
library(Hmisc)
library(rms)
library(riskRegression)
#install.packages("doMC", repos="http://R-Forge.R-project.org")
library(doMC)
The Data
#Load and store the data
lcOrig <- read.csv("cancer.csv")
#Replace all the 1's with 0's (censored)
lcOrig$status <- gsub(pattern = "1", replacement = "0", x = lcOrig$status, fixed = TRUE)
#Replace all the 2's with 1's (death)
lcOrig$status <- gsub (pattern = "2", replacement = "1", x = lcOrig$status, fixed = TRUE)
#Do the same thing for sex (0 = Males, 1 = Females)
lcOrig$sex <- gsub(pattern = "1", replacement = "0", x = lcOrig$sex, fixed = TRUE)
lcOrig$sex <- gsub(pattern = "2", replacement = "1", x = lcOrig$sex, fixed = TRUE)
#Change the class of these variables to integer.
lcOrig$status <- as.integer(lcOrig$status)
lcOrig$sex <- as.integer(lcOrig$sex)
lcOrig$ph.ecog <- as.integer(lcOrig$ph.ecog)
#Remove missing values and column with over 20% missing data.
apply(lcOrig, 2, function(x) sum(is.na(x))/length(x))
lcOrig <- lcOrig[, c(1:9, 11)]
lc <- lcOrig[complete.cases(lcOrig), ]
Cox Proportional Hazards
fitform1 <- Surv(time, status) ~ inst + age + sex + ph.ecog + ph.karno + pat.karno + wt.loss
cox1 <- coxph(fitform1, data = lc)
PREDICTION ERROR CURVES
extends <- function(...) TRUE
library("doMC")
registerDoMC()
set.seed(0692)
fitpec1 <- pec(list("CPH" = cox1), data = lc, formula = fitform1, splitMethod = "cv10", B = 5, keep.index = TRUE, keep.matrix = TRUE)
The last line of code results in the following error:
Error in coxModelFrame.coxph(object) : invalid object
set x=TRUE in the call to coxph
SOLUTION
Change:
cox1 <- coxph(fitform1, data = lc)
To:
cox1 <- coxph(fitform1, data = lc, x = TRUE)
This did not use to be a requirement 2 years ago, but is now. I hope this helps save you some time!
I need some help with a quiete simple problem in FiPy. My goal is to simulate a fluid flowing through a concrete block while phase change.
But first of all I tried to do a simple 1D simulation assumed a fluid massflow and a constant wall temperature without any phase change.
from fipy import *
from fipy.meshes import CylindricalGrid2D, Grid1D
import matplotlib.pyplot as plt
import numpy as np
#%%
L = 1.5 #length transfer surface
bS = 0.75 #wide
AV = L * bS #transfer surface
tS0 = 350. #tWall
rhoWF = 880. #density fluid
mWF = 0.036 #mass flow
u = 5e-4 #Fluid speed
hWF = mWF / AV / rhoWF / u #height "fluid block"
nx = 50
VWF = hWF * L * bS/nx #fluid volumen
lambdaWF = 0.6 # thermal conductivity
alpha = 500. #heat transfer coefficient
tWF0 = 371.
mesh = Grid1D(dx=L/nx, nx=nx)
tWF = CellVariable(name="Fluid",
mesh=mesh,
value= tWF0,
hasOld=True)
tS = CellVariable(name="storage",
mesh=mesh,
value=tS0,
hasOld=True)
sourceWF=CellVariable(name="source Fluid", #Variable der Konvektion
mesh=mesh,
value=0.)
cvrho = CellVariable(name = 'cprho',#Fluid
mesh = mesh,
value = rhoWF * 4215.2,
hasOld = True)
tWF.constrain(tWF0, mesh.facesLeft()) #constant inlet temperature
t = 6*3600. #time
timeStepDuration = 1e2
#outflow boundary condition
outlet = mesh.facesRight
ConvCoeff = FaceVariable(mesh,value=u,rank=1)
exteriorCoeff = FaceVariable(mesh,value=0.,rank=1)
exteriorCoeff.setValue(value=ConvCoeff, where=outlet)
ConvCoeff.setValue(0., where=outlet)
residual1 = 1.
elapsedTime = 0.
tWFall = np.zeros(nx)[None,:]
while elapsedTime < t:
tWF.updateOld()
it = 0 #iterations
while residual1> 1e-2:
sourceWF.value = - AV / nx * alpha*(tWF - tS)/ cvrho / VWF #this will be a variable convection source
eq1 = HybridConvectionTerm(coeff=ConvCoeff) + TransientTerm(coeff=1.) == \
+ sourceWF\
- ImplicitSourceTerm(exteriorCoeff.divergence) \
#+ DiffusionTerm(coeff= lambdaWF / cvrho) #not necessary(?)
residual1 = eq1.sweep(dt = timeStepDuration, var = tWF)
print('res1: ' + str(residual1) )
it += 1
if it > 10:
raise ValueError (r'MaxIter reached')
elapsedTime += timeStepDuration ; print('t= ' + str(round(elapsedTime,2)))
residual1 = 1.
tWFall = np.r_[tWFall, tWF.value[None,:]] #value collection
#%% outlet fluid temperature and storage temperature
plt.plot(np.linspace(0,t/3600.,int(t/timeStepDuration)), tWFall[1:,-1], label=r'$\vartheta_{WF}$')
plt.legend()
I would expect a constant fluid outlet temperature because of the constant wall temperature and constant fluid inlet temperature. I have not defined the wall temperature as a boundary condition because some day I would like to analyse heat conduction and variable temperature gradients too. Running my mwe you can see that the fluid temperature at the outlet declines.
Could someone please help at this case?
Thanks in advance!
I changed the script around and it seem to give a constant temperature of 371.0. See this link.
The sourceWF term has been removed. I was unsure what this was for, but I think it would take time for the wall temperature to adjust to this.
The equation declaration has been moved outside the loop. This is the correct way to use FiPy, but shouldn't impact the results in this case.