First thing's first; my skills in R are somewhat lacking, so there is a chance I may be using something incorrectly in the following. If I go wrong somewhere, please let me know.
I've been having a problem in Rstudio where I try to create 2 functions for formulae, then use nls() to create a model using those, with which I will make a plot. When I try to run the line for creating it, I get an error message saying an object is missing. It is always the last object in the function of the first "formula", in this case, 'p'.
I'll provide my code here then explain what I am trying to do for a little context;
DATA <- read.csv(file.choose(), as.is=T)
formula <- function(m, h, g, p){(2*m)/(m+(sqrt(m^2+1)))*p*g*(h^2/2)}
formula.2 <- function(P, V, g){P*V*g}
m = 0.85
p = 766.42
g = 9.81
P = 0.962
h = DATA$lithothick
V = DATA$Vol
fit.1 <- nls(formula (P, V, g) ~ formula(m, h, g, p), data = DATA)
If I run it how it is shown, I get the error;
Error in (2 * m)/(m + (sqrt(m^2 + 1))) * p : 'p' is missing
However it will show h if I rearrange the objects in the formula to (m,g,p,h)
Error in h^2 : 'h' is missing
Now, what I'm trying to do is this; I have a .csv file with 3 thicknesses (0.002, 0.004, 0.006 meters) and 3 volumes (10, 25, 50 milliliters). I am trying to see how the rates of strength and buoyancy increase (in relation to each other) as the thickness and volume for each object (respectively) increases. I was hoping to come out with a graph showing the upward trend for each property (strength and buoyancy), as I believe them to be unequal (one exponential the other linear). I hope that isn't more confusing than clarifying, but any pointers would be GREATLY appreciated.
You cannot overload functions this way in R, what you can do is provide optional arguments (which is a kind of overload) with syntax function(mandatory, optionnal="")
For what you are trying to do, you have to use formula.2 if you want to use the 3-arguments formula.
A workaround could be to use one function with one optionnal argument and check if this argument has been used. Something like :
formula = function(m, h, g, p="") {
if (is.numeric(p)) {
(2*m)/(m+(sqrt(m^2+1)))*p*g*(h^2/2)
} else {
m*h*g
}
}
This is ugly and a very bad way to do it (your variables do not really mean the same thing from one call to the other) but it works.
Related
I am trying to run a Bayesian clustering model where the number of clusters is random with binomial distribution.
This is my Jags model:
model{
for(i in 1:n){
y[ i ,1:M] ~ dmnorm( mu[z[i] , 1:M] , I[1:M, 1:M])
z[i] ~ dcat(omega[1:M])
}
for(j in 1:M){
mu[j,1:M] ~ dmnorm( mu_input[j,1:M] , I[1:M, 1:M] )
}
M ~ dbin(p, Mmax)
omega ~ ddirich(rep(1,Mmax))
}
to run it, we need to define the parameters anche the initial values for the variables, which is done in this R script
Mmax=10
y = matrix(0,100,Mmax)
I = diag(Mmax)
y[1:50,] = mvrnorm(50, rep(0,Mmax), I)
y[51:100,] = mvrnorm(50, rep(5,Mmax), I)
plot(y[,1:2])
z = 1*((1:100)>50) + 1
n = dim(y)[1]
M=2
mu=matrix(rnorm(Mmax^2),nrow=Mmax)
mu_input=matrix(2.5,Mmax,Mmax) ### prior mean
p=0.5
omega=rep(1,Mmax)/Mmax
data = list(y = y, I = I, n = n, mu_input=mu_input, Mmax = Mmax, p = p)
inits = function() {list(mu=mu,
M=M,
omega = omega) }
require(rjags)
modelRegress=jags.model("cluster_variabile.txt",data=data,inits=inits,n.adapt=1000,n.chains=1)
however, running the last command, one gets
Error in jags.model("cluster_variabile.txt", data = data, inits = inits,
: RUNTIME ERROR: Compilation error on line 6.
Unknown variable M Either supply values
for this variable with the data or define it on the left hand side of a relation.
which for me makes no sense, since the error is at line 6 even if M already appears at line 4 of the model! What is the actual problem in running this script?
So JAGS is not like R or other programming procedural languages in that it doesn't actually run line by line, it is a declarative language meaning the order of commands doesn't actually matter at least in terms of how the errors pop up. So just because it didn't throw an error on line 4 doesn't mean something isn't also wrong there. Im not positive, but I believe the error is occuring because JAGS tries to build the array first before inputting values, so M is not actually defined at this stage, but nothing you can do about that on your end.
With that aside, there should be a fairly easy work around for this, it is just less efficient. Instead of looping from 1:M make the loop iterate from 1:MMax that way the dimensions don't actually change, it is always an MMax x MMax. Then line 7 just assigns 1:M of those positions to a value. The downside of this is that it will require you to do some processing after the model is fit. So on each iteration, you will need to pull the sampled M and filter the matrix mu to be M x M, but that shouldn't be too tough. Let me know if you need more help.
So, I think the main problem is that you can't change the dimensionality of the stochastic node you're updating. This seems like a problem for reversible jump MCMC, though I don't think you can do this in JAGS.
I would like to fit 2-dim plot by straight line (a*x+b) using zfit like the following figure.
That is very easy work by a probfit package, but it has been deprecated by scikit-hep. https://nbviewer.jupyter.org/github/scikit-hep/probfit/blob/master/tutorial/tutorial.ipynb
How can I fit such 2dim plots by any function?
I've checked zfit examples, but it seems to be assumed some distribution (histogram) thus zfit requires dataset like 1d array and I couldn't reach how to pass 2d data to zfit.
There is no direct way in zfit currently to implement this out-of-the-box (with one line), since a corresponding loss is simply not added.
However, the SimpleLoss (zfit.loss.SimpleLoss) allows you to construct any loss that you can think of (have a look at the example as well in the docstring). In your case, this would look along this:
x = your_data
y = your_targets # y-value
obs = zfit.Space('x', (lower, upper))
param1 = zfit.Parameter(...)
param2 = zfit.Parameter(...)
...
model = Func(...) # a function is the way to go here
data = zfit.Data.from_numpy(array=x, obs=obs)
def mse():
prediction = model.func(data)
value = tf.reduce_mean((prediction - y) ** 2) # or whatever you want to have
return value
loss = zfit.loss.SimpleLoss(mse, [param1, param2])
# etc.
On another note, it would be a good idea to add such a loss. If you're interested to contribute I recommend to get in contact with the authors and they will gladly help you and guide you to it.
UPDATE
The loss function itself consists presumably of three to four things: x, y, a model and maybe an uncertainty on y. The chi2 loss looks like this:
def chi2():
y_pred = model.func(x)
return tf.reduce_sum((y_pred - y) / y_error) ** 2)
loss = zfit.loss.SimpleLoss(chi2, model.get_params())
That's all, 4 lines of code. x is a zfit.Data object, model is in this case a Func.
Does that work?
That's all.
I am using the function plkhci from library Bhat to construct Profile-likelihood based confidence intervals and I got this warning:
Warning message: In dqstep(list(label = x$label, est = btrf(xt, x$low,
x$upp), low = x$low, : oops: unable to find stepsize, use default
when i run
r <- dfp(x,f=nlogf)
Can I ignore this warning as I still can get the output?
Following is the complete coding:
library(Bhat)
beta0<--8
beta1<-0.03
gamma<-0.0105
alpha<-0.05
n<-100
u<-runif(n)
u
x<-rnorm(n)
x
c<-rexp(100,1/1515)
c
t1<-(1/gamma)*log(1-((gamma/(exp(beta0+beta1*x)))*(log(1-u))))
t1
t<-pmin(t1,c)
t
delta<-1*(t1>c)
delta
length(delta)
cp<-length(delta[delta==1])/n
cp
delta[delta==1]<-ifelse(rbinom(length(delta[delta==1]),1,0.5),1,2)
delta
deltae<-ifelse(delta==0, 1,0)
deltar<-ifelse(delta==1, 1,0)
deltai<-ifelse(delta==2, 1,0)
dat=data.frame(t,delta, deltae,deltar,deltai,x)
dat$interval[delta==2] <- as.character(cut(dat$t[delta==2], breaks=seq(0, 600, 100)))
labs <- cut(dat$t[delta==2], breaks=seq(0, 600, 100))
dat$lower[delta==2]<-as.numeric( sub("\\((.+),.*", "\\1", labs) )
dat$upper[delta==2]<-as.numeric( sub("[^,]*,([^]]*)\\]", "\\1", labs) )
data0<-dat[which(dat$delta==0),]#uncensored data
data1<-dat[which(dat$delta==1),]#right censored data
data2<-dat[which(dat$delta==2),]#interval censored data
nlogf<-function(para)
{
b0<-para[1]
b1<-para[2]
g<-para[3]
e<-sum((b0+b1*data0$x)+g*data0$t+(1/g)*exp(b0+b1*data0$x)*(1-exp(g*data0$t)))
r<-sum((1/g)*exp(b0+b1*data1$x)*(1-exp(g*data1$t)))
i<-sum(log(exp((1/g)*exp(b0+b1*data2$x)*(1-exp(g*data2$lower)))-exp((1/g)*exp(b0+b1*data2$x)*(1-exp(g*data2$upper)))))
l<-e+r+i
return(-l)
}
x <- list(label=c("beta0","beta1","gamma"),est=c(-8,0.03,0.0105),low=c(-10,0,0),upp=c(10,1,1))
r <- dfp(x,f=nlogf)
x$est <- r$est
plkhci(x,nlogf,"beta0")
plkhci(x,nlogf,"beta1")
plkhci(x,nlogf,"gamma")
I am giving you a super long answer, but it will help you see that you can chase down your own error messages (most of the time, sometimes this means of looking at functions will not work). It is good to see what is happening inside a method when it throws an warning because sometimes it is fine and sometimes you need to fix your data.
This function is REALLY involved! You can look at it by typing dfp into the R command line (NO TRAILING PARENTHESES) and it will print out the whole function.
17 lines from the end, you will see an assignment:
del <- dqstep(x, f, sens = 0.01)
You can see that this calls the function dqstep, which is reflected in your warning.
You can see this function by typing dqstep into the command line of R again. In reading through this function, also long but not so tedious, there is this section of boolean logic:
if (r < 0 | is.na(r) | b == 0) {
warning("oops: unable to find stepsize, use default")
cat("problem with ", x$label[i], "\n")
break
}
This is the culprit, it returns the message you are getting. The line right above it spells out how r is calculated. You are feeding this function your default x from the prior function plus a sensitivity equations (which I assume dfp generates, it is huge and ugly, so I did not untangle all of it). When the previous nested function returns either an r value lower than Zero, and r value of NA or a b value of ZERO, that message is displayed.
The second error tells you that it was likely b==0 because b is in the denominator and it returned and infinity value, so NO STEP SIZE IS RETURNED FROM THIS NESTED FUNCTION to the variable del in dfp.
The step is fed into THIS equation:
h <- logit.hessian(x, f, del, dapprox = FALSE, nfcn)
which you can look into by typing logit.hessian into the R commandline.
When you do, you see that del is a step size in a logit scale, with a default value of del=rep(0.002, length(x$est))...which the function set for you because running the function dqstep returned no value.
So, you now get to decide if using that step size in the calculation of your confidence interval seems right or if there is a problem with your data which needs resolving to make this work better for you.
When I ran it, line by line, I got this message:
Error in if (denom <= 0) { : missing value where TRUE/FALSE needed
at this line of code:
r <- dfp(x,f=nlogf(x))
Which makes me think I was correct.
That is how I chase down issues I have with messages from packages when I get a message like yours.
I am currently writing a program in R to find solutions of a general polynomial difference equation using Picard's method.
For an insight in the mathematics behind it (as math mode isn't available here):
https://math.stackexchange.com/questions/2064669/picard-iterations-for-general-polynomials/2064732
Now since then I've been trying to work with the Ryacas package for integration. However I ran into trouble trying to work with the combination of expression and integration function.
library(Ryacas)
degrees = 3
a = c(3,5,4,6)
x0 = -1
maxIterations(10)
iteration = vector('expression', length = maxIterations)
iteration[1] = x0
for(i in 2:maxIterations){
for(i in 1:degrees){
exp1 = expression( a[i] * iteration[i-1] ^ i)
}
iteration[i] = x0 + Integrate(exp1, t)
}
but this results in
"Error in paste("(", ..., ")") :
cannot coerce type 'closure' to vector of type 'character'"
and exp1 = expression(a[j] * iteration[i-1]^j) instead of an actual expression as I tried to achieve. Is there anyway I can make sure R reads this as a real expression (i.e. for example 3 * ( x0 ) ^ j for i = 2)?
Thanks in advance!
Edit:
I also found the Subst() function, and currently trying to see if anything is fixable using it. Now I am mainly struggling to actually set up an expression for m coefficients of a, as I can't find a way to create e.g. a for loop in the expression() command.
I'm relatively new in R and I would appreciated if you could take a look at the following code. I'm trying to estimate the shape parameter of the Frechet distribution (or inverse weibull) using mmedist (I tried also the fitdist that calls for mmedist) but it seems that I get the following error :
Error in mmedist(data, distname, start = start, fix.arg = fix.arg, ...) :
the empirical moment function must be defined.
The code that I use is the below:
require(actuar)
library(fitdistrplus)
library(MASS)
#values
n=100
scale = 1
shape=3
# simulate a sample
data_fre = rinvweibull(n, shape, scale)
memp=minvweibull(c(1,2), shape=3, rate=1, scale=1)
# estimating the parameters
para_lm = mmedist(data_fre,"invweibull",start=c(shape=3,scale=1),order=c(1,2),memp = "memp")
Please note that I tried many times en-changing the code in order to see if my mistake was in syntax but I always get the same error.
I'm aware of the paradigm in the documentation. I've tried that as well but with no luck. Please note that in order for the method to work the order of the moment must be smaller than the shape parameter (i.e. shape).
The example is the following:
require(actuar)
#simulate a sample
x4 <- rpareto(1000, 6, 2)
#empirical raw moment
memp <- function(x, order)
ifelse(order == 1, mean(x), sum(x^order)/length(x))
#fit
mmedist(x4, "pareto", order=c(1, 2), memp="memp",
start=c(shape=10, scale=10), lower=1, upper=Inf)
Thank you in advance for any help.
You will need to make non-trivial changes to the source of mmedist -- I recommend that you copy out the code, and make your own function foo_mmedist.
The first change you need to make is on line 94 of mmedist:
if (!exists("memp", mode = "function"))
That line checks whether "memp" is a function that exists, as opposed to whether the argument that you have actually passed exists as a function.
if (!exists(as.character(expression(memp)), mode = "function"))
The second, as I have already noted, relates to the fact that the optim routine actually calls funobj which calls DIFF2, which calls (see line 112) the user-supplied memp function, minvweibull in your case with two arguments -- obs, which resolves to data and order, but since minvweibull does not take data as the first argument, this fails.
This is expected, as the help page tells you:
memp A function implementing empirical moments, raw or centered but
has to be consistent with distr argument. This function must have
two arguments : as a first one the numeric vector of the data and as a
second the order of the moment returned by the function.
How can you fix this? Pass the function moment from the moments package. Here is complete code (assuming that you have made the change above, and created a new function called foo_mmedist):
# values
n = 100
scale = 1
shape = 3
# simulate a sample
data_fre = rinvweibull(n, shape, scale)
# estimating the parameters
para_lm = foo_mmedist(data_fre, "invweibull",
start= c(shape=5,scale=2), order=c(1, 2), memp = moment)
You can check that optimization has occurred as expected:
> para_lm$estimate
shape scale
2.490816 1.004128
Note however, that this actually reduces to a crude way of doing overdetermined method of moments, and am not sure that this is theoretically appropriate.