library(survival)
library(survminer)
library(dplyr)
ovarian=ovarian
ovarian$weighting = sample(1:100,26,replace=T)
fitWEIGHT <- coxph(Surv(futime, fustat) ~ age + rx,data=ovarian,weight=weighting)
fitNOWEIGHT <- coxph(Surv(futime, fustat) ~ age + rx,data=ovarian)
In this example above the value of the R-Squared for fitWEIGHT equals to 1. However the same model without fake sample weights has R-Squared equals to less than half (0.5). Why is this happening?
Weighting here is effectively repeating the observations. You're calculating weights with a perfectly distributed random sample ovarian$weighting = sample(1:100,26,replace=T) that's distributed across your underlying data set. So re-observing each sets of data points according to the normally distributed weights is likely biasing the function to ensure perfect correlation between your dependent and independent variables. It's probably not perfectly perfectly correlated, but the 1:100 range is likely blowing it out beyond the default number of significant digits and so it rounds to 1. If you change the sample to 1:10 or 40:50 or something it would likely continue to push the correlation bias but to reduce the r2 to nearly-1 instead of rounded-to-1 value that you're seeing now under the current weighting strategy.
For additional discussion on weights for this function see below. To ensure that the weights you're specifying are the types of weights you're expecting for this analysis. It's really weighting the observation count (ie, a form of over/re-sampling the observation you're assigning the weight to). https://www.rdocumentation.org/packages/survival/versions/2.43-3/topics/coxph
Where it states:
Case Weights Case weights are treated as replication weights, i.e., a
case weight of 2 is equivalent to having 2 copies of that subject's
observation. When computers were much smaller grouping like subjects
together was a common trick to used to conserve memory. Setting all
weights to 2 for instance will give the same coefficient estimate but
halve the variance. When the Efron approximation for ties (default) is
employed replication of the data will not give exactly the same
coefficients as the weights option, and in this case the weighted fit
is arguably the correct one.
When the model includes a cluster term or the robust=TRUE option the
computed variance treats any weights as sampling weights; setting all
weights to 2 will in this case give the same variance as weights of 1.
Related
I have a response Y that is a percentage ranging between 0-1. My data is nested by taxonomy or evolutionary relationship say phylum/genus/family/species and I have one continuous covariate temp and one categorial covariate fac with levels fac1 & fac2.
I am interested in estimating:
is there a difference in Y between fac1 and fac2 (intercept) and how much variance is explained by that
does each level of fac responds differently in regard to temp (linearly so slope)
is there a difference in Y for each level of my taxonomy and how much variance is explained by those (see varcomp)
does each level of my taxonomy responds differently in regard to temp (linearly so slope)
A brute force idea would be to split my data into the lowest taxonomy here species, do a linear beta regression for each species i as betareg(Y(i)~temp) . Then extract slope and intercepts for each speies and group them to a higher taxonomic level per fac and compare the distribution of slopes (intercepts) say, via Kullback-Leibler divergence to a distribution that I get when bootstrapping my Y values. Or compare the distribution of slopes (or interepts) just between taxonomic levels or my factor fac respectively.Or just compare mean slopes and intercepts between taxonomy levels or my factor levels.
Not sure is this is a good idea. And also not sure of how to answer the question of how many variance is explained by my taxonomy level, like in nested random mixed effect models.
Another option may be just those mixed models, but how can I include all the aspects I want to test in one model
say I could use the "gamlss" package to do:
library(gamlss)
model<-gamlss(Y~temp*fac+re(random=~1|phylum/genus/family/species),family=BE)
But here I see no way to incorporate a random slope or can I do:
model<-gamlss(Y~re(random=~temp*fac|phylum/genus/family/species),family=BE)
but the internal call to lme has some trouble with that and guess this is not the right notation anyways.
Is there any way to achive what I want to test, not necessarily with gamlss but any other package that inlcuded nested structures and beta regressions?
Thanks!
In glmmTMB, if you have no exact 0 or 1 values in your response, something like this should work:
library(glmmTMB)
glmmTMB(Y ~ temp*fac + (1 + temp | phylum/genus/family/species),
data = ...,
family = beta_family)
if you have zero values, you will need to do something . For example, you can add a zero-inflation term in glmmTMB; brms can handle zero-one-inflated Beta responses; you can "squeeze" the 0/1 values in a little bit (see the appendix of Smithson and Verkuilen's paper on Beta regression). If you have only a few 0/1 values it won't matter very much what you do. If you have a lot, you'll need to spend some serious time thinking about what they mean, which will influence how you handle them. Do they represent censoring (i.e. values that aren't exactly 0/1 but are too close to the borders to measure the difference)? Are they a qualitatively different response? etc. ...)
As I said in my comment, computing variance components for GLMMs is pretty tricky - there's not necessarily an easy decomposition, e.g. see here. However, you can compute the variances of intercept and slope at each taxonomic level and compare them (and you can use the standard deviations to compare with the magnitudes of the fixed effects ...)
The model given here might be pretty demanding, depending on the size of your phylogeny - for example, you might not have enough replication at the phylum level (in which case you could fit the model ~ temp*(fac + phylum) + (1 + temp | phylum:(genus/family/species)), i.e. pull out the phylum effects as fixed effects).
This is assuming that you're willing to assume that the effects of fac, and its interaction with temp, do not vary across the phylogeny ...
Suppose I have a following formula for a mixed effects model:
Performance ~ 1 + WorkingHours + Tenure + (1 + WorkingHours + Tenure || JobClass)
then I can specify priors for fixed slopes and fixed intercept as:
prior = normal(c(mu1,mu2), c(sd1,sd2), autoscale = FALSE)
prior_intercept = normal(mean, scale, autoscale = FALSE)
But how do I specify the priors for random slopes and intercept using
prior_covariance = decov(regularization, concentration, shape, scale)
(or)
lkj(regularization, scale, df)
if I know the variance between the slopes and intercepts and the correlation between them.
I am unable to understand how to specify the parameters for the above mixed effects formula.
Because you're working in a Bayesian model, you aren't going to specify the correlations or variances. You're going to specify a likelihood distribution of covariance matrices (by way of the correlation matrix and vector of variances) by giving the values for a few parameters.
The regularization parameter is a positive real value that determines how likely things are to be correlated. A value of 1 is sort of the "anything's possible" option (this is the default). Values greater than 1 mean that you believe there are few, if any, correlations. Values less than 1 mean you believe there is a lot of correlation.
The scale parameter is related to the sum of the variances. In particular, the scale parameter is equal to the square root of the average variance.
The concentration parameter is used to control how the total variance is distributed among the different variables. A value of 1 is saying you don't have an expectation. Larger values say that you believe that the variables have similar proportions of the total variance. Values between 0 and 1 mean that you think there are dissimilar contributions.
The shape parameter is used for a Gamma distribution that acts as a prior on the scale.
Then, finally, df is your prior degrees of freedom.
So, decov and lkj are each giving you a different way to express your expectations about properties of the covariance matrix, but they won't let you specify which specific variables you believe to be correlated with which other specific variables. It should decide that as part of the model fitting process.
This is all from the rstanarm documentation
I have the following data:
# actual value:
a <- c(26.77814,29.34224,10.39203,29.66659,20.79306,20.73860,22.71488,29.93678,10.14384,32.63233,24.82544,38.14778,25.12343,23.07767,14.60789)
# predicted value
p <- c(27.238142,27.492240,13.542026,32.266587,20.473063,20.508603,21.414882,28.536775,18.313844,32.082333,24.545438,30.877776,25.703430,22.397666,15.627892)
I already calculated MSE and RMSE for these two, but they're asking for AUC and ROC curve. How can I calculate it from this data using R? I thought AUC is for classification problems, was I mistaken? Can we still calculate AUC for numeric values like above?
Question:
I thought AUC is for classification problems, was I mistaken?
You are not mistaken. The area under the receiver operating characteristic curve can't be computed for two numeric vectors like in your example. It's used to determine how well your binary classifier stands up to a gold standard binary classifier. You need a vector of cases vs. controls, or levels for the a vector that put each value in one of two categories.
Here's an example of how you'd do this with the pROC package:
library(pROC)
# actual value
a <- c(26.77814,29.34224,10.39203,29.66659,20.79306,20.73860,22.71488,29.93678,10.14384,32.63233,24.82544,38.14778,25.12343,23.07767,14.60789)
# predicted value
p <- c(27.238142,27.492240,13.542026,32.266587,20.473063,20.508603,21.414882,28.536775,18.313844,32.082333,24.545438,30.877776,25.703430,22.397666,15.627892)
df <- data.frame(a = a, p = p)
# order the data frame according to the actual values
odf <- df[order(df$a),]
# convert the actual values to an ordered binary classification
odf$a <- odf$a > 12 # arbitrarily decided to use 12 as the threshold
# construct the roc object
roc_obj <- roc(odf$a, odf$p)
auc(roc_obj)
# Area under the curve: 0.9615
Here, we have arbitrarily decided that threshold for the gold standard (a) is 12. If that's the case, than observations that have a lower value than 12 are controls. The prediction (p) classifies very well, with an AUC of 0.9615. We don't have to decide on the threshold for our prediction classifier in order to determine the AUC, because it's independent of the threshold decision. We can slide up and down depending on whether it's more important to find cases or to not misclassify a control.
Important Note
I completely made up the threshold for the gold standard classifier. If you choose a different threshold (for the gold standard), you'll get a different AUC. For example, if we chose 28, the AUC would be 1. The AUC is independent of the threshold for the predictor, but absolutely depends on the threshold for the gold standard.
EDIT
To clarify the above note, which was apparently misunderstood, you were not mistaken. This kind of analysis is for classification problems. You cannot use it here without more information. In order to do it, you need a threshold for your a vector, which you don't have. You CAN'T make one up and expect to get a non made up result for the AUC. Because the AUC depends on the threshold for the gold standard classifier, if you just make up the threshold, as we did in the exercise above, you are also just making up the AUC.
I want to exam which variable impacts most on the outcome, in my data, which is the stock yield. My data is like below.
And my code is also attached.
library(randomForest)
require(data.table)
data = fread("C:/stockcrazy.csv")
PEratio <- data$offeringPE/data$industryPE
data_update <- data.frame(data,PEratio)
train <- data_update[1:47,]
test <- data_update[48:57,]
For the above subset data set train and test, I am not sure if I need to do a cross validation on this data. And I don't know how to do it.
data.model <- randomForest(yield ~ offerings + offerprice + PEratio + count + bingo
+ purchase , data=train, importance=TRUE)
par(mfrow = c(1, 1))
varImpPlot(data.model, n.var = 6, main = "Random Forests: Top 6 Important Variables")
importance(data.model)
plot(data.model)
model.pred <- predict(data.model, newdata=test)
model.pred
d <- data.frame(test,model.pred)
I am sure not sure if the result of IncMSE is good or bad. Can someone interpret this?
Additionally, I found the predicted values of the test data is not a good prediction of the real data. So how can I improve this?
Let's see. Let's start with %IncMSE:
I found this really good answer on cross validated about %IncMSE which I quote:
if a predictor is important in your current model, then assigning
other values for that predictor randomly but 'realistically' (i.e.:
permuting this predictor's values over your dataset), should have a
negative influence on prediction, i.e.: using the same model to
predict from data that is the same except for the one variable, should
give worse predictions.
So, you take a predictive measure (MSE) with the original dataset and
then with the 'permuted' dataset, and you compare them somehow. One
way, particularly since we expect the original MSE to always be
smaller, the difference can be taken. Finally, for making the values
comparable over variables, these are scaled.
This means that in your case the most important variable is purchase i.e. when the variable purchase was permuted (i.e. the order of the values randomly changed) the resulting model was 12% worse than having the variable in its original order in terms of calculating the mean square error. The MSE was 12% higher using a permuted purchase variable meaning that the this variable is the most important. Variable importance is just a measure of how important your predictor variables were in the model you used. In your case purchase was the most important and P/E ratio was the least (for those 6 variables). This is not something you can interpret as good or bad because it doesn't show you how well the model fits unseen data. I hope this is clear now.
For the cross-validation:
You do not need to do a cross validation during the training phase because it happens automatically. Approximately, 2/3 of the records are used for the creation of a tree and the 1/3 that is left out (out-of-bag data) is used to assess the tree afterwards (the R squared for the tree is computed using the oob data)
As for the improvement of the model:
By showing just the 10 first lines of the predicted and the actual values of yield, you cannot make a safe decision on whether the model is good or bad. What you need is a test of fitness. The most common one is the R squared. It is simplistic but for comparing models and getting a first opinion about your model it does its job. This is calculated by the model for every tree that you make and can be accessed by data.model$rsq. This ranges from 0 to 1 with 1 being the perfect model and 0 showing really poor fit ( it can sometimes even take negative values which shows a bad fit). If your rsq is bad then you can try the following to improve your model although it is not certain that you will get the results you wish for:
Calibrate your trees in a different way. Change the number of trees grown and prune the trees by specifying a big nodesize number. (here you use the default 500 trees and a nodesize of 5 which might overfit your model.)
Increase the number of variables if possible.
Choose a different model. There are cases were a random Forest would not work well
I'm running a regression on census data where my dependent variable is life expectancy and I have eight independent variables. The data is aggregated be cities, so I have many thousand observations.
My model is somewhat heteroscedastic though. I want to run a weighted least-squares where each observation is weighted by the city’s population. In this case, it would mean that I want to weight the observations by the inverse of the square root of the population. It’s unclear to me, however, what would be the best syntax. Currently, I have:
Model=lm(…,weights=(1/population))
Is that correct? Or should it be:
Model=lm(…,weights=(1/sqrt(population)))
(I found this question here: Weighted Least Squares - R but it does not clarify how R interprets the weights argument.)
From ?lm: "weights: an optional vector of weights to be used in the fitting process. Should be NULL or a numeric vector. If non-NULL, weighted least squares is used with weights weights (that is, minimizing sum(w*e^2)); otherwise ordinary least squares is used." R doesn't do any further interpretation of the weights argument.
So, if what you want to minimize is the sum of (the squared distance from each point to the fit line * 1/sqrt(population) then you want ...weights=(1/sqrt(population)). If you want to minimize the sum of (the squared distance from each point to the fit line * 1/population) then you want ...weights=1/population.
As to which of those is most appropriate... that's a question for CrossValidated!
To answer your question, Lucas, I think you want weights=(1/population). R parameterizes the weights as inversely proportional to the variances, so specifying the weights this way amounts to assuming that the variance of the error term is proportional to the population of the city, which is a common assumption in this setting.
But check the assumption! If the variance of the error term is indeed proportional to the population size, then if you divide each residual by the square root of its corresponding sample size, the residuals should have constant variance. Remember, dividing a random variable by a constant results in the variance being divided by the square of that constant.
Here's how you can check this: Obtain residuals from the regression by
residuals = lm(..., weights = 1/population)$residuals
Then divide the residuals by the square roots of the population variances:
standardized_residuals = residuals/sqrt(population)
Then compare the sample variance among the residuals corresponding to the bottom half of population sizes:
variance1 = var(standardized_residuals[population < median(population)])
to the sample variance among the residuals corresponding to the upper half of population sizes:
variance2 = var(standardized_residuals[population > median(population)])
If these two numbers, variance1 and variance2 are similar, then you're doing something right. If they are drastically different, then maybe your assumption is violated.