I am running an ARIMA model using the fable package. Just curious to find out whether there is a way to specify the order of the model (e.g specifying an order of 2,1,1) when using the ARIMA function in the package as opposed to using the optimal lags which are specified automatically?
Also trying to figure out the best way to add a vector as a dummy variable in order to control for a structural break for the first observation
I've used the built in dataset in the package (tourism)
library(fable)
fit <- tourism %>% slice(tail(row_number(), 10)) %>%
model(arima = ARIMA(Trips))
TIA!
If you check out the help page for this ARIMA() function, it says how to specify p, d, and q, and it provides a nice example:
# Manual ARIMA specification
USAccDeaths %>%
as_tsibble() %>%
model(arima = ARIMA(log(value) ~ 0 + pdq(0, 1, 1) + PDQ(0, 1, 1))) %>%
report()
You can specify numbers (or ranges of numbers that it can use to determine which one is the best fit) for p, d, and q in the right-hand side of the formula argument within the ARIMA() function. pdq() is for non-seasonal components and PDQ() is for seasonal components.
It also says:
To force a nonseasonal fit, specify PDQ(0, 0, 0) in the RHS of the model formula.
I'm not exactly sure what you're trying to accomplish with the dummy variable. Could you provide a reproducible example?
Related
I had to transform a variable response (e.g. Variable 1) to fulfil the assumptions of linear models in lmer using an approach suggested here https://www.r-bloggers.com/2020/01/a-guide-to-data-transformation/ for heavy-tailed data and demonstrated below:
TransformVariable1 <- sqrt(abs(Variable1 - median(Variable1))
I then fit the data to the following example model:
fit <- lmer(TransformVariable1 ~ x + y + (1|z), data = dataframe)
Next, I update the reference grid to account for the transformation as suggested here Specifying that model is logit transformed to plot backtransformed trends:
rg <- update(ref_grid(fit), tran = "TransformVariable1")
Neverthess, the emmeans are not back transformed to the original scale after using the following command:
fitemm <- as.data.frame(emmeans(rg, ~ x + y, type = "response"))
My question is: How can I back transform the emmeans to the original scale?
Thank you in advance.
There are two major problems here.
The lesser of them is in specifying tran. You need to either specify one of a handful of known transformations, such as "log", or a list with the needed functions to undo the transformation and implement the delta method. See the help for make.link, make.tran, and vignette("transformations", "emmeans").
The much more serious issue is that the transformation used here is not a monotone function, so it is impossible to back-transform the results. Each transformed response value corresponds to two possible values on either side of the median of the original variable. The model we have here does not estimate effects on the given variable, but rather effects on the dispersion of that variable. It's like trying to use the speedometer as a substitute for a navigation system.
I would suggest using a different model, or at least a different response variable.
A possible remedy
Looking again at this, I wonder if what was meant was the symmetric square-root transformation -- what is shown multiplied by sign(Variable1 - median(Variable1)). This transformation is available in emmeans::make.tran(). You will need to re-fit the model.
What I suggest is creating the transformation object first, then using it throughout:
require(lme4)
requre(emmeans)
symsqrt <- make.tran("sympower", param = c(0.5, median(Variable1)))
fit <- with(symsqrt,
lmer(linkfun(Variable1) ~ x + y + (1|z), data = dataframe)
)
emmeans(fit, ~ x + y, type = "response")
symsqrt comprises a list of functions needed to implement the transformation. The transformation itself is symsqrt$linkfun, and the emmeans package knows to look for the other stuff when the response transformation is named linkfun.
BTW, please break the habit of wrapping emmeans() in as.data.frame(). That renders invisible some important annotations, and also disables the possibility of following up with contrasts and comparisons. If you think you want to see more precision than is shown, you can precede the call with emm_options(opt.digits = FALSE); but really, you are kidding yourself if you think those extra digits give you useful information.
I'm trying to use gee to model counts of an outcome with a population offset.I have models with interaction terms and am trying to use the all effects package to summarize parameter estimates and odds ratios (ORs).
When I compute ORs by hand, I'm not sure why its not matching the output I get from the effects::allEffects() function. The data can't be shared but the model is
mdl <- geeglm(count~age+gender+age:gender+offset(log(totalpop)),
family="poisson", corstr="exchangeable", id=geo,
waves=year, data=df)
I use the below code to compute stuff manually. log_OR sums the interaction terms without intercepts added to parameter. log_odds sums the parameters with intercept. The code is taken from here.
tibble(
variables = names(coef(mdl)),
log_OR = c(...),
log_odds = c(...),
OR = exp(log_OR),
odds = exp(log_odds),
probability = odds / (1 + odds)
) %>%
mutate_if(is.numeric, ~round(., 5)) %>%
knitr::kable()
I then compare my manual calculations to the output of allEffects below. They don't match. Can someone help me see what I am doing wrong?
result <- allEffects(mdl)
allEffects(mdl) %>% summary()
variable <- result[["age:gender"]][["x"]]
Prob <- result$`age:gender`$fit
Prob_upper <- result$`age:gender`$upper
Prob_lower <- result$`age:gender`$lower
model_Est <- data.frame("Est"=Prob, "CI Lower"= Prob_lower,
"CI Upper"= Prob_upper)
model_Prob <- exp(model_Est)
model_est <- data.frame("Variable"=variable, model_est)
model_OR <- data.frame("Variable"=variable, model_OR)
You haven't given us very much to go on, but the cause is almost certainly that the offset isn't being dealt with properly. (The first thing I would try is running the model without the offset to see if the results from effects and your by-hand calculations match: that's not the model you want, but it will confirm that the problem is with the offsets.)
?effects says:
offset a function to be applied to the offset values (if
there is an offset) in a linear or generalized linear
model, or a mixed-effects model fit by ‘lmer’ or ‘glmer’;
or a numeric value, to which the offset will be set. The
default is the ‘mean’ function, and thus the offset will
be set to its mean; in the case of ‘"svyglm"’ objects,
the default is to use the survey-design weighted mean.
Note: Only offsets defined by the ‘offset’ argument to
‘lm’, ‘glm’, ‘svyglm’, ‘lmer’, or ‘glmer’ will be handled
correctly; use of the ‘offset’ function in the model
formula is not supported.
(emphasis added)
methods("effects") lists only effects.glm and effects.lm, which suggests that the model is being treated as a glm (i.e., there is no specialized method for GEE models). So, this suggests:
(1) you need to include offset= as a separate argument in your model.
(2) when doing your hand calculation, make sure the value of the offset is set to the mean value across all observations (unless you choose to use the offset= argument to effects/allEffects to change the default summary function).
As McFadden (1978) showed, if the number of alternatives in a multinomial logit model is so large that computation becomes impossible, it is still feasible to obtain consistent estimates by randomly subsetting the alternatives, so that the estimated probabilities for each individual are based on the chosen alternative and C other randomly selected alternatives. In this case, the size of the subset of alternatives is C+1 for each individual.
My question is about the implementation of this algorithm in R. Is it already embedded in any multinomial logit package? If not - which seems likely based on what I know so far - how would one go about including the procedure in pre-existing packages without recoding extensively?
Not sure whether the question is more about doing the sampling of alternatives or the estimation of MNL models after sampling of alternatives. To my knowledge, there are no R packages that do sampling of alternatives (the former) so far, but the latter is possible with existing packages such as mlogit. I believe the reason is that the sampling process varies depending on how your data is organized, but it is relatively easy to do with a bit of your own code. Below is code adapted from what I used for this paper.
library(tidyverse)
# create artificial data
set.seed(6)
# data frame of choser id and chosen alt_id
id_alt <- data.frame(
id = 1:1000,
alt_chosen = sample(1:30, 1)
)
# data frame for universal choice set, with an alt-specific attributes (alt_x2)
alts <- data.frame(
alt_id = 1:30,
alt_x2 = runif(30)
)
# conduct sampling of 9 non-chosen alternatives
id_alt <- id_alt %>%
mutate(.alts_all =list(alts$alt_id),
# use weights to avoid including chosen alternative in sample
.alts_wtg = map2(.alts_all, alt_chosen, ~ifelse(.x==.y, 0, 1)),
.alts_nonch = map2(.alts_all, .alts_wtg, ~sample(.x, size=9, prob=.y)),
# combine chosen & sampled non-chosen alts
alt_id = map2(alt_chosen, .alts_nonch, c)
)
# unnest above data.frame to create a long format data frame
# with rows varying by choser id and alt_id
id_alt_lf <- id_alt %>%
select(-starts_with(".")) %>%
unnest(alt_id)
# join long format df with alts to get alt-specific attributes
id_alt_lf <- id_alt_lf %>%
left_join(alts, by="alt_id") %>%
mutate(chosen=ifelse(alt_chosen==alt_id, 1, 0))
require(mlogit)
# convert to mlogit data frame before estimating
id_alt_mldf <- mlogit.data(id_alt_lf,
choice="chosen",
chid.var="id",
alt.var="alt_id", shape="long")
mlogit( chosen ~ 0 + alt_x2, id_alt_mldf) %>%
summary()
It is, of course, possible without using the purrr::map functions, by using apply variants or looping through each row of id_alt.
Sampling of alternatives is not currently implemented in the mlogit package. As stated previously, the solution is to generate a data.frame with a subset of alternatives and then using mlogit (and importantly to use a formula with no intercepts). Note that mlogit can deal with unbalanced data, ie the number of alternatives doesn't have to be the same for all the choice situations.
My recommendation would be to review the mlogit package.
Vignette:
https://cran.r-project.org/web/packages/mlogit/vignettes/mlogit2.pdf
the package has a set of example exercises that (in my opinion) are worth looking at:
https://cran.r-project.org/web/packages/mlogit/vignettes/Exercises.pdf
You may also want to take a look at the gmnl package (I have not used it)
https://cran.r-project.org/web/packages/gmnl/index.html
Multinomial Logit Models with Continuous and Discrete Individual Heterogeneity in R: The gmnl Package
Mauricio Sarrias' (Author) gmnl Web page
Question: What specific problem(s) are you trying to apply a multinomial logit model too? Suitably intrigued.
Aside from the above question, I hope the above points you in the right direction.
I have this data set.
wbh
I wanted to use the R package glmnet to determine which predictors would be useful in predicting fertility. However, I have been unable to do so, most likely due to not having a full understanding of the package. The fertility variable is SP.DYN.TFRT.IN. I want to see which predictors in the data set give the most predictive power for fertility. I wanted to use LASSO or ridge regression to shrink the number of coefficients, and I know this package can do that. I'm just having some trouble implementing it.
I know there are no code snippets which I apologize for but I am rather lost on how I would code this out.
Any advice is appreciated.
Thank you for reading
Here is an example on how to run glmnet:
library(glmnet)
library(tidyverse)
df is the data set your provided.
select y variable:
y <- df$SP.DYN.TFRT.IN
select numerical variables:
df %>%
select(-SP.DYN.TFRT.IN, -region, -country.code) %>%
as.matrix() -> x
select factor variables and convert to dummy variables:
df %>%
select(region, country.code) %>%
model.matrix( ~ .-1, .) -> x_train
run model(s), several parameters here can be tweaked I suggest checking the documentation. Here I just run 5-fold cross validation to determine the best lambda
cv_fit <- cv.glmnet(x, y, nfolds = 5) #just with numeric variables
cv_fit_2 <- cv.glmnet(cbind(x ,x_train), y, nfolds = 5) #both factor and numeric variables
par(mfrow = c(2,1))
plot(cv_fit)
plot(cv_fit_2)
best lambda:
cv_fit$lambda[which.min(cv_fit$cvm)]
coefficients at best lambda
coef(cv_fit, s = cv_fit$lambda[which.min(cv_fit$cvm)])
equivalent to:
coef(cv_fit, s = "lambda.min")
after running coef(cv_fit, s = "lambda.min") all features with - in the resulting table are dropped from the model. This situation corresponds to the left lambda depicted with the left vertical dashed line on the plots.
I suggest reading the linked documentation - elastic nets are quite easy to grasp if you know a bit of linear regression and the package is quite intuitive. I also suggest reading ISLR, at least the part with L1 / L2 regularization. and these videos: 1, 2, 3 4, 5, 6, first three are about estimating model performance via test error and the last three are about the question at hand. This one is how to implement these models in R. By the way these guys on the videos invented LASSO and made glment.
Also check the glmnetUtils library which provides a formula interface and other nice things like in built mixing parameter (alpha) selection. Here is the vignette.
I ran a three way repeated measures ANOVA with ezANOVA.
anova_1<-ezANOVA(data = main_data, dv = .(rt), wid.(id),
within = .(A,B,C), type = 3, detailed = TRUE)
I'm trying to see what's going on with the residuals via a qqplot but I don't know how to get to them or if they'r even there. With my lme models I simply extract them from the model
main_data$model_residuals <- as.numeric(residuals(model_1))
and plot them
residuals_qq<-ggplot(main_data, aes(sample = main_data$model_residuals)) +
stat_qq(color="black", alpha=1, size =2) +
geom_abline(intercept = mean(main_data$model_residuals), slope = sd(main_data$model_residuals))
I'd like to use ggplot since I'd like to keep a sense of consistency in my graphing.
EDIT
Maybe I wasn't clear in what I'm trying to do. With lme models I can simply create the variable model_residuals from the residuals object in the main_data data.frame that then contains the residuals I plot in ggplot. I want to know if something similar is possible for the residuals in ezAnova or if there is a way I can get hold of the residuals for my ANOVA.
I had the same trouble with ezANOVA. The solution I went for was to switch to ez.glm (from the afex package). Both ezANOVA and ez.glm wrap a function from a different package, so you should get the same results.
This would look like this for your example:
anova_1<-ez.glm("id", "rt", main_data, within=c("A","B","C"), return="full")
nice.anova(anova_1$Anova) # show the ANOVA table like ezANOVA does.
Then you can pull out the lm object and get your residuals in the usual way:
residuals(anova_1$lm)
Hope that helps.
Edit: A few changes to make it work with the last version
anova_1<-aov_ez("id", "rt", main_data, within=c("A","B","C"))
print(m1)
print(m1$Anova)
summary(m1$Anova)
summary(m1)
Then you can pull out the lm object and get your residuals in the usual way:
residuals(anova_1$lm)
A quite old post I know, but it's possible to use ggplot to plot the residuals after modeling your data with ez package by using this function:
proj(ez_outcome$aov)[[3]][, "Residuals"]
then:
qplot(proj(ez_outcome$aov)[[3]][, "Residuals"])
Hope it helps.
Also potentially adding to an old post, but I butted up against this problem as well and as this is the first thing that pops up when searching for this question I thought I might add how I got around it.
I found that if you include the return_aov = TRUE argument in the ezANOVA setup, then the residuals are in there, but ezANOVA partitions them up in the resulting list it produces within each main and interaction effect, similar to how base aov() does if you include an Error term for subject ID as in this case.
These can be pulled out into their own list with purrr by mapping the residual function over this aov sublist in ezANOVA, rather than the main output. So from the question example, it becomes:
anova_1 <- ezANOVA(data = main_data, dv = .(rt), wid = .(id),
within = .(A,B,C), type = 3, detailed = TRUE, return_aov = TRUE)
ezanova_residuals <- purrr::map(anova_1$aov, residuals)
This will produce a list where each entry is the residuals from a part of the ezANOVA model for effects and interactions, i.e. $(Intercept), $id, id:a, id:b, id:a:b etc.
I found it useful to then stitch these together in a tibble using enframe and unnest (as the list components will probably be different lengths) to very quickly get them in a long format, that can then be plotted or tested:
ezanova_residuals_tbl <- enframe(ezanova_residuals) %>% unnest
hist(ezanova_residuals_tbl$value)
shapiro.test(ezanova_residuals_tbl$value)
I've not used this myself but the mapping idea also works for the coefficients and fitted.values functions to pull them out of the ezANOVA results, if needed. They might come out in some odd formats and need some extra manipulation afterwards though.