I have produced a linear data set and have used lm() to fit a model to that dataset. I am now trying to find the MSE using mse()
I know the formula for MSE but I'm trying to use this function. What would be the proper way to do so? I have looked at the documentation, but I'm either dumb or it's just worded for people who actually know what they're doing.
library(hydroGOF)
x.linear <- seq(0, 200, by=1) # x data
error.linear <- rnorm(n=length(x.linear), mean=0, sd=1) # Error (0, 1)
y.linear <- x.linear + error.linear # y data
training.data <- data.frame(x.linear, y.linear)
training.model <- lm(training.data)
training.mse <- mse(training.model, training.data)
plot(training.data)
mse() needs two data frames. I'm not sure how to get a data frame out of lm(). Am I even on the right track to finding a proper MSE for my data?
Try this:
mean((training.data - predict(training.model))^2)
#[1] 0.4467098
You can also use below mentioned code which is very clean to get mean square error
install.packages("Metrics")
library(Metrics)
mse(actual, predicted)
The first data set on which is actual one : training.data
The second argument is the one which you will predict like :
pd <- predict(training.model , training.data)
mse(training.data$,pd)
Seems you have not done prediction yet so first predict the data based on your model and then calculate mse
You can use the residual component from lm model output to find mse in this manner :
mse = mean(training.model$residuals^2)
Note: if you come from another program (like SAS)
they get the mean using the sum and the degrees of freedom of the residual. I recommend doing the same if you want a more accurate estimate of the error.
mse = sum(training.model$residuals^2)/training.model$df.residual
I found this while trying to figure out why mean(my_model$residuals^2) was different in R than the MSE in SAS.
Related
I have a linear model in R of the form
lm(log(num_encounters) ~ log(distance)*sampling_effort, data=df)
I want to interpret the residuals but get them back on the scale of num_encounters. I have seen residuals.lm(x, type="working") and residuals.lm(x, type="response") but I'm not sure about the values returned by them. Do I for instance still need to use exp() to get the residual values back on the num_encounters scale? Or are they already on that scale? I want to plot these absolute values back, both in a histogram and in a raster map afterwards.
EDIT:
Basically my confusion is that the following code results in 3 different histograms, while I was expecting the first 2 to be identical.
df$predicted <- exp(predict(x, newdata=df))
histogram(df$num_encounters-df$predicted)
histogram(exp(residuals(x, type="response")))
histogram(residuals(x, type="response"))
I want to interpret the residuals but get them back on the scale of
num_encounters.
You can easily calculate them:
mod <- lm(log(num_encounters) ~ log(distance)*sampling_effort, data=df)
res <- df$num_encounters - exp(predict(mod))
In addition what #Roland suggests, which indeed is correct and works, the problem with my confusion was just basic high-school logarithm algebra.
Indeed the absolute response residuals (on the scale of the original dependent variable) can be calculated as #Roland says with
mod <- lm(log(num_encounters) ~ log(distance)*sampling_effort, data=df)
res <- df$num_encounters - exp(predict(mod))
If you want to calculate them from the model residuals, you need to keep logarithm substraction rules into account.
log(a)-log(b)=log(a/b)
The residual is calculated from the original model. So in my case, the model predicts log(num_encounters). So the residual is log(observed)-log(predicted).
What I was trying to do was
exp(resid) = exp(log(obs)-log(pred)) = exp(log(obs/pred)) = obs/pred
which is clearly not the number I was looking for. To get the absolute response residual from the model response residual, this is what I needed.
obs-obs/exp(resid)
So in R code, this is what you could also do:
mod <- lm(log(num_encounters) ~ log(distance)*sampling_effort, data=df)
abs_resid <- df$num_encounters - df$num_encounters/exp(residuals(mod, type="response"))
This resulted in the same number as with the method described by #Roland which is much easier of course. But at least I got my brain lined up again.
I am trying to fit a curve to a set of data points but did not succeed. So I ask you.
plot(time,val) # look at data
exponential.model <- lm(log(val)~ a) # compute model
fit <- exp(predict(exponential.model,list(Time=time))) # create the fitted curve
plot(time,val)#plot it again
lines(time, fit,lwd=2) # show the fitted line
My only problem is, that my data contains negative values and so log(val) produces a lot of NA making the model computation crash.
I know that my data does not necessarily look like exponential , but I want to see the fit anyway. I also used another program which shows me val=27.1331*exp(-time/2.88031) is a nice fit but I do not know, what I am doing wrong.
I want to compute it with R.
I had the idea to shift data so no negative values remain, but result is poor and quite sure wrong.
plot(time,val+20) # look at data
exponential.model <- lm(log(val+20)~ a) # compute model
fit <- exp(predict(exponential.model,list(Time=time))) # create the fitted curve
plot(time,val)#plot it again
lines(time, fit-20,lwd=2) # show the (BAD) fitted line
Thank you!
I figured some things out and have a satisfying solution.
exponential.model <- lm(log(val)~ a) # compute model
The log(val) term is trying to rescale the values, so a linear model can be applied. Since this not possible to my values, you have to use a non-linear model (nls).
exponential.model <- nls(val ~ a*exp(b*time), start=c(b=-0.1,h=30))
This worked fine for me.
satisfying fit
First of all, I thank you all beforehand for reading this.
I am trying to fit a Standardized T-Student Distribution (i.e. a T-Student with standard deviation = 1) on a series of data; that is: I want to estimate the degrees of freedom via Maximum Likelihood Estimation.
An example of what I need to achieve can be found in the following (simple) Excel file I made:
https://www.dropbox.com/s/6wv6egzurxh4zap/Excel%20Implementation%20Example.xlsx?dl=0
Inside the Excel file, I have an image that contains the formula corresponding to the calculation of the loglikelihood function for the Standardized T Student Distribution. The formula was extracted from a Finance book (Elements of Financial Risk Management - by Peter Christoffersen).
So far, I have tried this with R:
copula.data <- read.csv(file.choose(),header = TRUE)
z1 <- copula.data[,1]
library(fitdistrplus)
ft1 = fitdist(z1, "t", method = "mle", start = 10)
df1=ft1$estimate[1]
df1
logLik(ft1)
df1 yields the number: 13.11855278779897
logLike(ft1) yields the number: -3600.2918050056487
However, the Excel file yields degrees of freedom of: 8.2962365022727, and a log-likelihood of: -3588.8879 (which is the right answer).
Note: the .csv file that my code reads is the following:
https://www.dropbox.com/s/nnh2jgq4fl6cm12/Data%20for%20T%20Copula.csv?dl=0
Any ideas? Thank you people!
The formula from your spreadsheet (with n, x substituted for the df parameter and the data)
=GAMMALN((n+1)/2)-GAMMALN(n/2)-LN(PI())/2-LN(n-2)/2-1/2*(1+n)*LN(1+x^2/(n-2))
or, exponentiating,
Gamma((n+1)/2) / (sqrt((n-2) pi) Gamma(n/2)) (1+x^2/(n-2))^-((n+1)/2)
?dt gives
f(x) = Gamma((n+1)/2) / (sqrt(n pi) Gamma(n/2)) (1 + x^2/n)^-((n+1)/2)
So the difference lies in those n-2 values in two places in the formula. I don't have enough context to see why the author is defining the t distribution in that different way; there may be some good reason ...
Looking at the negative log-likelihood curve directly, it certainly seems as though the fitdistrplus answer is agreeing with the direct calculation. (It would be very surprising if there were a bug in the dt() function, R's distribution functions are very broadly used and thoroughly tested.)
LL <- function(p,data=z1) {
-sum(dt(data,df=p,log=TRUE))
}
pvec <- seq(6,20,by=0.05)
Lvec <- sapply(pvec,LL)
par(las=1,bty="l")
plot(pvec,Lvec,type="l",
xlab="df parameter",ylab="negative log-likelihood")
## superimpose fitdistr results ...
abline(v=coef(ft1),lty=2)
abline(h=-logLik(ft1),lty=2)
Unless there's something else you're not telling us about the problem definition, it seems to me that R is getting the right answer. (The mean and sd of the data you gave were not exactly equal to 0 and 1 respectively, but they were close; centering and scaling gave an even larger value for the parameter.)
I am trying to fit t-distributions to my data but am unable to do so. My first try was
fitdistr(myData, "t")
There are 41 warnings, all saying that NaNs are produced. I don't know how, logarithms seem to be involved. So I adjusted my data somewhat so that all data is >0, but I still have the same problem (9 fewer warnings though...). Same problem with sstdFit(), produces NaNs.
So instead I try with fitdist which I've seen on stackoverflow and CrossValidated:
fitdist(myData, "t")
I then get
Error in mledist(data, distname, start, fix.arg, ...) :
'start' must be defined as a named list for this distribution
What does this mean? I tried looking into the documentation but that told me nothing. I just want to possibly fit a t-distribution, this is so frustrating :P
Thanks!
Start is the initial guess for the parameters of your distribution. There are logs involved because it is using maximum likelihood and hence log-likelihoods.
library(fitdistrplus)
dat <- rt(100, df=10)
fit <- fitdist(dat, "t", start=list(df=2))
I think it's worth adding that in most cases, using the fitdistrplus package to fit a t-distribution to real data will lead to a very bad fit, which is actually quite misleading. This is because the default t-distribution functions in R are used, and they don't support shifting or scaling. That is, if your data has a mean other than 0, or is scaled in some way, then the fitdist function will simply lead to a bad fit.
In real life, if data fits a t-distribution, it is usually shifted (i.e. has a mean other than 0) and / or scaled. Let's generate some data like that:
data = 1.5*rt(10000,df=5) + 0.5
Given this data has been sampled from the t-distribution with 5 degrees of freedom, you'd think that trying to fit a t-distribution to this should work quite nicely. But actually, here is the result. It estimates a df of 2, and provides a bad fit as shown in the qq plot.
> fit_bad <- fitdist(data,"t",start=list(df=3))
> fit_bad
Fitting of the distribution ' t ' by maximum likelihood
Parameters:
estimate Std. Error
df 2.050967 0.04301357
> qqcomp(list(fit_bad)) # generates plot to show fit
When you fit to a t-distribution you want to not only estimate the degrees of freedom, but also a mean and scaling parameter.
The metRology package provides a version of the t-distribution called t.scaled that has a mean and sd parameter in addition to the df parameter [metRology]. Now let's fit it again:
> library("metRology")
> fit_good <- fitdist(data,"t.scaled",
start=list(df=3,mean=mean(data),sd=sd(data)))
> fit_good
Fitting of the distribution ' t.scaled ' by maximum likelihood
Parameters:
estimate Std. Error
df 4.9732159 0.24849246
mean 0.4945922 0.01716461
sd 1.4860637 0.01828821
> qqcomp(list(fit_good)) # generates plot to show fit
Much better :-) The parameters are very close to how we generated the data in the first place! And the QQ plot shows a much nicer fit.
Apologies if this is better suited in CrossValidated.
I am fitting GAM models to binomial data using the mgcv package in R. One of the covariates is periodic, so I am specifying the bs = "cc" cyclic cubic spline. I am doing this in a cross validation framework, but when I go to fit my holdout data using the predict function I get the following error:
Error in pred.mat(x, object$xp, object$BD) :
can't predict outside range of knots with periodic smoother
Here is some code that should replicate the error:
# generate data:
x <- runif(100,min=-pi,max=pi)
linPred <- 2*cos(x) # value of the linear predictor
theta <- 1 / (1 + exp(-linPred)) #
y <- rbinom(100,1,theta)
plot(x,theta)
df <- data.frame(x=x,y=y)
# fit gam with periodic smoother:
gamFit <- gam(y ~ s(x,bs="cc",k=5),data=df,family=binomial())
summary(gamFit)
plot(gamFit)
# predict y values for new data:
x.2 <- runif(100,min=-pi,max=pi)
df.2 <- data.frame(x=x.2)
predict(gamFit,newdata=df.2)
Any suggestions on where I'm going wrong would be greatly appreciated. Maybe manually specifying knots to fall on -pi and pi?
I did not get an error on the first run but I did replicate the error on the second try. Perhaps you need to use set.seed(123) #{no error} and set.seed(223) #{produces error}. to see if that creates partial success. I think you are just seeing some variation with a relatively small number of points in your derivation and validation datasets. 100 points for GAM fit is not particularly "generous".
Looking at the gamFit object it appears that the range of the knots is encoded in gamFit$smooth[[1]]['xp'], so this should restrict your inputs to the proper range:
x.2 <- runif(100,min=-pi,max=pi);
x.2 <- x.2[findInterval(x.2, range(gamFit$smooth[[1]]['xp']) )== 1]
# Removes the errors in all the situations I tested
# There were three points outside the range in the set.seed(223) case
The problem is that your test set contains values that were not in the range of your training set. Since you used a spline, knots were created at the minimum and maximum value of x, and your fitted function is not defined outside of that range. So, when you test the model, you should exclude those points that are outside the range. Here is how you would exclude the points in the test set:
set.seed(2)
... <Your code>
predict(gamFit,newdata=df.2[df.2$x>=min(df$x) & df.2$x<=max(df$x),,drop=F])
Or, you could specify the "outer" knot points in the model to the min and max of your whole data. I don't know how to do that offhand.