I have a linear mixed model which is run 50 different times in a loop.
Each time the model is run, I want the response variable b to be weighted inversely with the variance. So if the variance of b is small, I want the weighting to be bigger and vice versa. This is a simplified version of the model:
model <- lme(b ~ type, random = ~1|replicate,weights = ~ I(1/b))
Here's the R data files:
b: https://www.dropbox.com/s/ziipdtsih5f0252/b.Rdata?dl=0
type: https://www.dropbox.com/s/90682ewib1lw06e/type.Rdata?dl=0
replicate: https://www.dropbox.com/s/kvrtao5i2g4v3ik/replicate.Rdata?dl=0
I'm trying to do this using the weights option in lme. Right now I have this as:
weights = ~ I(1/b).
But I don't think this is correct....maybe weights = ~ I(1/var(b)) ??
I also want to adjust this slightly as b consists of two types of data specified in the factor variable (of 2 levels) type.
I want to inversely weight the variance of each of these two levels separately. How could I do this?
I'm not sure it makes sense to talk about weighting the response variable in this manner. The descriptions I have found in the R-SIG-mixed-models mailing list refer to using inverse weighting derived from the predictor variables, either the fixed effects or the random effects. The weighting is used in minimizing the deviations of approximation of the model fits to the response. There is a function that returns the fixed effects variance (a sub-class of the varFunc family of functions) and it has a help page (linked from the weights section of the ?gls page):
?varFixed
?varFunc
It requires a formula object as its argument. So my original guess was:
model <- lme(b ~ type, random = ~1|replicate, weights = varFixed( ~type) )
Which you proved incorrect. How about seeing if this works:
model <- lme(b ~ type, random = ~1|replicate, weights = varFixed( ~1| type) )
(My continuing guess is that this weighting is the default situation and specifying these particular weights may not be needed. The inverse nature of the weighting is implied and does not need to be explicitly stated with "1/type". In the case of mixed models the "correct" construction depends on the design and the prior science and none of this has been presented, so this is really only a syntactic comment and not an endorsement of this model. I did not download the files. Seems rather odd to have three separate files and no code for linking them into a dataframe. Generally one would want to have a single data object within which the column names would be used in the formulas of the regression function. (I also suspect this is the default behavior of this function and so my untested prediction is that that you would be getting no change by omitting that 'weights' parameter.)
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 ...
I need to perform glm (poisson) estimations with fixed-effects (say merely unit FE) and several regressors (RHS variables). I have an unbalanced panel dataset where most (~90%) observations have missing values (NA) for some but not all regressors.
fixest::feglm() can handle this and returns my fitted model.
However, to do so, it (and fixest::demean too) removes observations that have at least one regressor missing, before constructing the fixed-effect means.
In my case, I am afraid this implies not using a significant share of available information in the data.
Therefore, I would like to demean my variables by hand, to be able to include as much information as possible in each fixed-effect dimension's mean, and then run feglm on the demeaned data. However, this implies getting negative dependent variable values, which is not compatible with Poisson. If I run feglm with "poisson" family and my manually demeaned data, I (coherently) get: "Negative values of the dependent variable are not allowed for the "poisson" family.". The same error is returned with data demeaned with the fixest::demean function.
Question:
How does feglm handle negative values of the demeaned dependent variable? Is there a way (like some data transformation) to reproduce fepois on a fixed-effect in the formula with fepois on demeaned data and a no fixed-effect formula?
To use the example from fixest::demean documentation (with two-way fixed-effects):
data(trade)
base = trade
base$ln_dist = log(base$dist_km)
base$ln_euros = log(base$Euros)
# We center the two variables ln_dist and ln_euros
# on the factors Origin and Destination
X_demean = demean(X = base[, c("ln_dist", "ln_euros")],
fe = base[, c("Origin", "Destination")])
base[, c("ln_dist_dm", "ln_euros_dm")] = X_demean
and I would like to reproduce
est_fe = fepois(ln_euros ~ ln_dist | Origin + Destination, base)
with
est = fepois(ln_euros_dm ~ ln_dist_dm, base)
I think there are two main problems.
Modelling strategy
In general, it is important to be able to formally describe the estimated model.
In this case it wouldn't be possible to write down the model with a single equation, where the fixed-effects are estimated using all the data and other variables only on the non-missing observations. And if the model is not clear, then... maybe it's not a good model.
On the other hand, if your model is well defined, then removing random observations should not change the expectation of the coefficients, only their variance. So again, if your model is well specified, you shouldn't worry too much.
By suggesting that observations with missing values are relevant to estimate the fixed-effects coefficients (or stated differently, that they are used to demean some variables) you are implying that these observations are not randomly distributed. And now you should worry.
Just using these observations to demean the variables wouldn't remove the bias on the estimated coefficients due to the selection to non-missingness. That's a deeper problem that cannot be removed by technical tricks but rather by a profound understanding of the data.
GLM
There is a misunderstanding with GLM. GLM is a super smart trick to estimate maximum likelihood models with OLS (there's a nice description here). It was developed and used at a time when regular optimization techniques were very expensive in terms of computational time, and it was a way to instead employ well developed and fast OLS techniques to perform equivalent estimations.
GLM is an iterative process where typical OLS estimations are performed at each step, the only changes at each iteration concern the weights associated to each observation. Therefore, since it's a regular OLS process, techniques to perform fast OLS estimations with multiple fixed-effects can be leveraged (as is in the fixest package).
So actually, you could do what you want... but only within the OLS step of the GLM algorithm. By no means you should demean the data before running GLM because, well, it makes no sense (the FWL theorem has absolutely no hold here).
I'm currently trying to use a GAM to calculate a rough estimation of expected goals model based purely on the commentary data from ESPN. However, all the data is either a categorical variable or a logical vector, so I'm not sure if there's a way to smooth, or if I should just use the factor names.
Here are my variables:
shot_where (factor): shot location (e.g. right side of the box)
assist_class (factor): type of assist (cross, through ball, pass)
follow_corner (logical): whether the shot follows a corner
shot_with (factor): right foot, left food, header
follow_set_piece (logical): whether the shot follows a set piece
I think I should just use the formula as just the variable names.
model <- bam(is_goal ~ shot_where + assist_class + follow_set_piece + shot_where + follow_corner + shot_where:shot_with, family = "binomial", method = "REML")
The shot_where and shot_with would incorporate any interactions between these two varaibles.
However, I was told I could smooth factor variables as well using the below structure.
model <- bam(is_goal ~ s(shot_where, bs = 'fs') + s(assist_class, bs = 'fs') + as.logical(follow_set_piece) +
as.logical(follow_corner) + s(shot_with, bs = 'fs'), data = model_data, family = "binomial", method = "REML")
This worked for creating a model, but I want to make sure this is a correct method of building the model. I've yet to see any information on using only factor/logical variables in a GAM model, so I thought it was worth asking.
If you only have categorical covariates then you aren't fitting a GAM, whether you fit the model with gam(), bam(), or something else.
What you are doing when you pass factor variables to s() using the fs basis like this
s(f, bs = 'fs')`
is creating a random intercept for each level of the factor f.
There's no smoothing going on here at all; the model is simply exploiting the equivalence of the Bayesian view of smoothing with random effects.
Given that none of your covariates could reasonably be considered random in the sense of a mixed effects model then the only justification for doing what you're doing might be as a computational trick.
Your first model is just a simple GLM (note the typo in the formula as shot_where is repeated twice in the formula.)
It's not clear to me why you are using bam() to fit this model; you're loosing computational efficiency that bam() provides by using method = 'REML'; it should be 'fREML' for bam() models. But as there is no smoothness selection going on in the first model you'd likely be better off using glm() to fit that model. If the issue is large sample sizes, there are several packages that can fit GLMs to large data, for example biglm and it's bigglm() function.
In the second model there is no smoothing going on but there is penalisation which is shrinking the estimates for the random intercepts toward zero. You're likely to get better performance on big data using the lme4 package or TMB and the glmmTMB package to fit what is a GLMM.
This is more of a theoretical question than about R, but let me provide a brief answer. Essentially, the most flexible model you could estimate would be one where you used the variables as factors. It also produces a model that is reasonably easily interpreted - where each coefficient gives you the expected difference in y between the reference level and the level represented by the dummy regressor.
Smoothing splines try to strike the appropriate bias-variance tradeoff. If you've got lots of data and relatively few categories in the categorical variables, there will be no real loss in efficiency for including all of the dummy regressors representing the categories and the bias will also be as small as possible. To the extent that the smoothing spline model is different from the one treating everything as factors, it is likely inducing bias without a corresponding increase in efficiency. If it were me, I would stick with a model that treats all of the categorical variables as factors.
I want to determine the marginal effects of each dependent variable in a probit regression as follows:
predict the (base) probability with the mean of each variable
for each variable, predict the change in probability compared to the base probability if the variable takes the value of mean + 1x standard deviation of the variable
In one of my regressions, I have a multiplicative variable, as follows:
my_probit <- glm(a ~ b + c + I(b*c), family = binomial(link = "probit"), data=data)
Two questions:
When I determine the marginal effects using the approach above, will the value of the multiplicative term reflect the value of b or c taking the value mean + 1x standard deviation of the variable?
Same question, but with an interaction term (* and no I()) instead of a multiplicative term.
Many thanks
When interpreting the results of models involving interaction terms, the general rule is DO NOT interpret coefficients. The very presence of interactions means that the meaning of coefficients for terms will vary depending on the other variate values being used for prediction. The right way to go about looking at the results is to construct a "prediction grid", i.e. a set of values that are spaced across the range of interest (hopefully within the domain of data support). The two essential functions for this process are expand.grid and predict.
dgrid <- expand.grid(b=fivenum(data$b)[2:4], c=fivenum(data$c)[2:4]
# A grid with the upper and lower hinges and the medians for `a` and `b`.
predict(my_probit, newdata=dgrid)
You may want to have the predictions on a scale other than the default (which is to return the linear predictor), so perhaps this would be easier to interpret if it were:
predict(my_probit, newdata=dgrid, type ="response")
Be sure to read ?predict and ?predict.glm and work with some simple examples to make sure you are getting what you intended.
Predictions from models containing interactions (at least those involving 2 covariates) should be thought of as being surfaces or 2-d manifolds in three dimensions. (And for 3-covariate interactions as being iso-value envelopes.) The reason that non-interaction models can be decomposed into separate term "effects" is that the slopes of the planar prediction surfaces remain constant across all levels of input. Such is not the case with interactions, especially those with multiplicative and non-linear model structures. The graphical tools and insights that one picks up in a differential equations course can be productively applied here.
When using R's rpart function, I can easily fit a model with it. for example:
# Classification Tree with rpart
library(rpart)
# grow tree
fit <- rpart(Kyphosis ~ Age + Number + Start,
method="class", data=kyphosis)
printcp(fit) # display the results
plotcp(fit)
summary(fit) # detailed summary of splits
# plot tree
plot(fit, uniform=TRUE,
main="Classification Tree for Kyphosis")
text(fit, use.n=TRUE, all=TRUE, cex=.8)
My question is -
How can I measure the "importance" of each of my three explanatory variables (Age, Number, Start) to the model?
If this was a regression model, I could have looked at p-values from the "anova" F-test (between lm models with and without the variable). But what is the equivalence of using "anova" on lm to an rpart object?
(I hope I managed to make my question clear)
Thanks.
Of course anova would be impossible, as anova involves calculating the total variation in the response variable and partitioning it into informative components (SSA, SSE). I can't see how one could calculate sum of squares for a categorical variable like Kyphosis.
I think that what you actually talking about is Attribute Selection (or evaluation). I would use the information gain measure for example. I think that this is what is used to select the test attribute at each node in the tree and the attribute with the highest information gain (or greatest entropy reduction) is chosen as the test attribute for the current node. This attribute minimizes the information needed to classify the samples in the resulting partitions.
I am not aware whether there is a method of ranking attributes according to their information gain in R, but I know that there is in WEKA and is named InfoGainAttributeEval It evaluates the worth of an attribute by measuring the information gain with respect to the class. And if you use Ranker as the Search Method, the attributes are ranked by their individual evaluations.
EDIT
I finally found a way to do this in R using Library CORElearn
estInfGain <- attrEval(Kyphosis ~ ., kyphosis, estimator="InfGain")
print(estInfGain)