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$linkfun, and the emmeans package knows to look for the other stuff when the response transformation is named linkfun.
BTW, please break the habit of wrapping emmeans() in as.data.frame(). That renders invisible some important annotations, and also disables the possibility of following up with contrasts and comparisons. If you think you want to see more precision than is shown, you can precede the call with emm_options(opt.digits = FALSE); but really, you are kidding yourself if you think those extra digits give you useful information.
Related
I'm trying to use gee to model counts of an outcome with a population offset.I have models with interaction terms and am trying to use the all effects package to summarize parameter estimates and odds ratios (ORs).
When I compute ORs by hand, I'm not sure why its not matching the output I get from the effects::allEffects() function. The data can't be shared but the model is
mdl <- geeglm(count~age+gender+age:gender+offset(log(totalpop)),
family="poisson", corstr="exchangeable", id=geo,
waves=year, data=df)
I use the below code to compute stuff manually. log_OR sums the interaction terms without intercepts added to parameter. log_odds sums the parameters with intercept. The code is taken from here.
tibble(
variables = names(coef(mdl)),
log_OR = c(...),
log_odds = c(...),
OR = exp(log_OR),
odds = exp(log_odds),
probability = odds / (1 + odds)
) %>%
mutate_if(is.numeric, ~round(., 5)) %>%
knitr::kable()
I then compare my manual calculations to the output of allEffects below. They don't match. Can someone help me see what I am doing wrong?
result <- allEffects(mdl)
allEffects(mdl) %>% summary()
variable <- result[["age:gender"]][["x"]]
Prob <- result$`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).
The toy model below stands in for one with a bunch more variables, transforms, lags, etc. Assume I got that stuff right.
My data is ordered in time, but is now formatted as an R time series, because I need to exclude certain periods, etc. I'd rather not make it a time series for this reason, because I think it would be easy to muck up, but if I need to, or it greatly simplifies the estimating process, I'd like to just use an integer sequence, such as index. below, to represent time if that is allowed.
My problem is a simple one (I hope). I would like to use the first part of my data to estimate the coefficients of the model. Then I want to use those estimates, and not estimates from a sliding window, to do one-ahead forecasts for each of the remaining values of that data. The idea is that the formula is applied with a sliding window even though it is not estimated with one. Obviously I could retype the model with coefficients included and then get what I want in multiple ways, with base R sapply, with tidyverse dplyr::mutate or purrr::map_dbl, etc. But I am morally certain there is some standard way of pulling the formula out of the lm object and then wielding it as one desires, that I just haven't been able to find. Example:
set.seed(1)
x1 <- 1:20
y1 <- 2 + x1 + lag(x1) + rnorm(20)
index. <- x1
data. <- tibble(index., x1, y1)
mod_eq <- y1 ~ x1 + lag(x1)
lm_obj <- lm(mod_eq, data.[1:15,])
and I want something along the lines of:
my_forecast_values <- apply_eq_to_data(eq = get_estimated_equation(lm_obj), my_data = data.[16:20])
and the lag shouldn't give me an error.
Also, this is not part of my question per se, but I could use a pointer to a nice tutorial on using R formulas and the standard estimation output objects produced by lm, glm, nls and the like. Not the statistics, just the programming.
The common way to use the coefficients is by calling the predict(), coefficients(), or summary() function on the model object for what it is worth. You might try the ?predict.lm() documentation for details on formula.
A simple example:
data.$lagx <- dplyr::lag(data.$x1, 1) #create lag variable
lm_obj1 <- lm(data=data.[2:15,], y1 ~ x1 + lagx) #create model object
data.$pred1 <- predict(lm_obj1, newdata=data.[16,20]) #predict new data; needs to have same column headings
I am trying to convert Absorbance (Abs) values to Concentration (ng/mL), based on an established linear model & standard curve. I planned to do this by using the predict() function. I am having trouble getting predict() to return the desired results. Here is a sample of my code:
Standards<-data.frame(ng_mL=c(0,0.4,1,4),
Abs550nm=c(1.7535,1.5896,1.4285,0.9362))
LM.2<-lm(log(Standards[['Abs550nm']])~Standards[['ng_mL']])
Abs<-c(1.7812,1.7309,1.3537,1.6757,1.7409,1.7875,1.7533,1.8169,1.753,1.6721,1.7036,1.6707,
0.3903,0.3362,0.2886,0.281,0.3596,0.4122,0.218,0.2331,1.3292,1.2734)
predict(object=LM.2,
newdata=data.frame(Concentration=Abs[1]))#using Abs[1] as an example, but I eventually want predictions for all values in Abs
Running that last lines gives this output:
> predict(object=LM.2,
+ newdata=data.frame(Concentration=Abs[1]))
1 2 3 4
0.5338437 0.4731341 0.3820697 -0.0732525
Warning message:
'newdata' had 1 row but variables found have 4 rows
This does not seem to be the output I want. I am trying to get a single predicted Concentration value for each Absorbance (Abs) entry. It would be nice to be able to predict all of the entries at once and add them to an existing data frame, but I can't even get it to give me a single value correctly. I've read many threads on here, webpages found on Google, and all of the help files, and for the life of me I cannot understand what is going on with this function. Any help would be appreciated, thanks.
You must have a variable in newdata that has the same name as that used in the model formula used to fit the model initially.
You have two errors:
You don't use a variable in newdata with the same name as the covariate used to fit the model, and
You make the problem much more difficult to resolve because you abuse the formula interface.
Don't fit your model like this:
mod <- lm(log(Standards[['Abs550nm']])~Standards[['ng_mL']])
fit your model like this
mod <- lm(log(Abs550nm) ~ ng_mL, data = standards)
Isn't that some much more readable?
To predict you would need a data frame with a variable ng_mL:
predict(mod, newdata = data.frame(ng_mL = c(0.5, 1.2)))
Now you may have a third error. You appear to be trying to predict with new values of Absorbance, but the way you fitted the model, Absorbance is the response variable. You would need to supply new values for ng_mL.
The behaviour you are seeing is what happens when R can't find a correctly-named variable in newdata; it returns the fitted values from the model or the predictions at the observed data.
This makes me think you have the formula back to front. Did you mean:
mod2 <- lm(ng_mL ~ log(Abs550nm), data = standards)
?? In which case, you'd need
predict(mod2, newdata = data.frame(Abs550nm = c(1.7812,1.7309)))
say. Note you don't need to include the log() bit in the name. R recognises that as a function and applies to the variable Abs550nm for you.
If the model really is log(Abs550nm) ~ ng_mL and you want to find values of ng_mL for new values of Abs550nm you'll need to invert the fitted model in some way.
I have created a loop to fit a non-linear model to six data points by participants (each participant has 6 data points). The first model is a one parameter model. Here is the code for that model that works great. The time variable is defined. The participant variable is the id variable. The data is in long form (one row for each datapoint of each participant).
Here is the loop code with 1 parameter that works:
1_p_model <- dlply(discounting_long, .(Participant), function(discounting_long) {wrapnls(indiff ~ 1/(1+k*time), data = discounting_long, start = c(k=0))})
However, when I try to fit a two parameter model, I get this error "Error: singular gradient matrix at initial parameter estimates" while still using the wrapnls function. I realize that the model is likely over parameterized, that is why I am trying to use wrapnls instead of just nls (or nlsList). Some in my field insist on seeing both model fits. I thought that the wrapnls model avoids the problem of 0 or near-0 residuals. Here is my code that does not work. The start values and limits are standard in the field for this model.
2_p_model <- dlply(discounting_long, .(Participant), function(discounting_long) {nlxb(indiff ~ 1/(1+k*time^s), data = discounting_long, lower = c (s = 0), start = c(k=0, s=.99), upper = c(s=1))})
I realize that I could use nlxb (which does give me the correct parameter values for each participant) but that function does not give predictive values or residuals of each data point (at least I don't think it does) which I would like to compute AIC values.
I am also open to other solutions for running a loop through the data by participants.
You mention at the end that 'nlxb doesn't give you residuals', but it does. If your result from your call to nlxbis called fit then the residuals are in fit$resid. So you can get the fitted values using just by adding them to the original data. Honestly I don't know why nlxb hasn't been made to work with the predict() function, but at least there's a way to get the predicted values.
Using predict() one can obtain the predicted value of the dependent variable (y) for a certain value of the independent variable (x) for a given model. Is there any function that predicts x for a given y?
For example:
kalythos <- data.frame(x = c(20,35,45,55,70),
n = rep(50,5), y = c(6,17,26,37,44))
kalythos$Ymat <- cbind(kalythos$y, kalythos$n - kalythos$y)
model <- glm(Ymat ~ x, family = binomial, data = kalythos)
If we want to know the predicted value of the model for x=50:
predict(model, data.frame(x=50), type = "response")
I want to know which x makes y=30, for example.
Saw the previous answer is deleted. In your case, given n=50 and the model is binomial, you would calculate x given y using:
f <- function (y,m) {
(logit(y/50) - coef(m)[["(Intercept)"]]) / coef(m)[["x"]]
}
> f(30,model)
[1] 48.59833
But when doing so, you better consult a statistician to show you how to calculate the inverse prediction interval. And please, take VitoshKa's considerations into account.
Came across this old thread but thought I would add some other info. Package MASS has function dose.p for logit/probit models. SE is via delta method.
> dose.p(model,p=.6)
Dose SE
p = 0.6: 48.59833 1.944772
Fitting the inverse model (x~y) would not makes sense here because, as #VitoshKa says, we assume x is fixed and y (the 0/1 response) is random. Besides, if the data weren’t grouped you’d have only 2 values of the explanatory variable: 0 and 1. But even though we assume x is fixed it still makes sense to calculate a confidence interval for the dose x for a given p, contrary to what #VitoshKa says. Just as we can reparameterize the model in terms of ED50, we can do so for ED60 or any other quantile. Parameters are fixed, but we still calculate CI's for them.
The chemcal package has an inverse.predict() function, which works for fits of the form y ~ x and y ~ x - 1
You just have to rearrange the regression equation, but as the comments above state this may prove tricky and not necessarily have a meaningful interpretation.
However, for the case you presented you can use:
(1/coef(model)[2])*(model$family$linkfun(30/50)-coef(model)[1])
Note I did the division by the x coefficient first to allow the name attribute to be correct.
For just a quick view (without intervals and considering additional issues) you could use the TkPredict function in the TeachingDemos package. It does not do this directly, but allows you to dynamically change the x value(s) and see what the predicted y-value is, so it would be fairly simple to move x until the desired Y is found (for given values of additional x's), this will also show possibly problems with multiple x's that would work for the same y.