I have a glm model and I want to select the name of the variable whose coefficient has the highest p-value. I know how to find the highest p-value, and I know how to get the number of the variable (in the order in which it appears in the model), but I don't know how to actually get the variable name. The reason I would like to do this is that I want to create a loop that on each iteration removes the variable with the least significant coefficient and reruns the model. I would do that manually, but I just have way too many variables.
The following sample code could be of help. This code outputs the column name corresponding to Maximum P value (also, ignores the intercept)
mydata <- read.csv("http://www.ats.ucla.edu/stat/data/binary.csv")
mydata$rank <- factor(mydata$rank)
mylogit <- glm(admit ~ gre + gpa + rank, data = mydata, family = "binomial")
coefficients <- coef(summary(mylogit))
maxPColumn <- rownames(coefficients)[2:nrow(coefficients)][which.max(coefficients[2:nrow(coefficients),4])]
maxPColumn
Related
I need to find a regression in R which has the form of
lm(Binary_value ~ Age, data=dataframe)
But my age variable starts at 15 yrs old so I'm not interested in ages that are less than 14. How can I specify that I only want my regression to be accurate at the age point of 15 and not worry about smaller values? I tried it this way:
lm(Binary_value ~ Age, data=dataframe)
But I get nonsense results for higher ages.
First things first, remember that R is case-sensitive, so the function would look like lm, not LM. I edited your question to fix that. Second, a regression only includes data that is available for prediction. It will not magically make up 14 data points if they are not already present, so there is no issue there. However, the regression line will not map to just => 15 years old because it uses the model coefficients to draw an intercept. An example below with fake data:
#### Create Fake Data ####
set.seed(123)
x <- 15:100 # use these numbers for age
age <- sample(x, # using x
size=1000, # sample 1000 times
replace=T) # sample with replacement
outcome <- age * .60 + rnorm(n=1000,sd=15) # make fake outcome variable
df <- data.frame(age,outcome)
#### Fit Data ####
fit <- lm(outcome ~ age, data = df)
summary(fit)
plot(age,outcome)
abline(fit,
col = "red")
You will see that the regression line, despite only including 15, will still draw to the left where there is no data. This is because the intercept is a conditional value based on the coefficients.
P.S. I used a normal Gaussian regression for this example because you used the lm function in your question, but included a binary response. For a logistic regression, the rationale would be the same, but it would use glm instead.
I am doing a multi-level meta-analysis. Many studies have several subgroups. When I make a forest plot studies are presented as subgroups. There are 60 of them, however, I would like to plot studies according to the study, then it would be 25 studies and it would be more appropriate. Does anyone have an idea how to do this forest plot?
I did it this way:
full.model <- rma.mv(yi = yi,
V = vi,
slab = Author,
data = df,
random = ~ 1 | Author/Study,
test = "t",
method = "REML")
forest(full.model)
It is not clear to me if you want to aggregate to the Author level or to the Study level. If there are multiple rows of data for particular studies, then the model isn't really complete and you would want to add another random intercept for the level of the estimates within studies. Essentially, the lowest random effect should have as many values for nlvls in the output as there are estimates (k).
Let's first tackle the case where we have a multilevel structure with two levels, studies and multiple estimates within studies (for some technical reasons, some might call this a three-level model, but let's not get into this). I will use a fully reproducible example for illustration purposes, using the dat.konstantopoulos2011 dataset, where we have districts and schools within districts. We fit a multilevel model of the type as you have with:
library(metafor)
dat <- dat.konstantopoulos2011
res <- rma.mv(yi, vi, random = ~ 1 | district/school, data=dat)
res
We can aggregate the estimates to the district level using the aggregate() function, specifying the marginal var-cov matrix of the estimates from the model to account for their non-independence (note that this makes use of aggregate.escalc() which only works with escalc objects, so if it is not, you need to convert the dataset to one - see help(aggregate.escalc) for details):
agg <- aggregate(dat, cluster=dat$district, V=vcov(res, type="obs"))
agg
You will find that if you then fit an equal-effects model to these estimates based on the aggregated data that the results are identical to what you obtained from the multilevel model (we use an equal-effects model since the heterogeneity accounted for by the multilevel model is already encapsulated in vcov(res, type="obs")):
rma(yi, vi, method="EE", data=agg)
So, we can now use these aggregated values in a forest plot:
with(agg, forest(yi, vi, slab=district))
My guess based on your description is that you actually have an additional level that you should include in the model and that you want to aggregate to the intermediate level. This is a tad more complicated, since aggregate() isn't meant for that. Just for illustration purposes, say we use year as another (higher) level and I will mess a bit with the data so that all three variance components are non-zero (again, just for illustration purposes):
dat$yi[dat$year == 1976] <- dat$yi[dat$year == 1976] + 0.8
res <- rma.mv(yi, vi, random = ~ 1 | year/district/school, data=dat)
res
Now instead of aggregate(), we can accomplish the same thing by using a multivariate model, including the intermediate level as a factor and using again vcov(res, type="obs") as the var-cov matrix:
agg <- rma.mv(yi, V=vcov(res, type="obs"), mods = ~ 0 + factor(district), data=dat)
agg
Now the model coefficients of this model are the aggregated values and the var-cov matrix of the model coefficients is the var-cov matrix of these aggregated values:
coef(agg)
vcov(agg)
They are not all independent (since we haven't aggregated to the highest level), so if we want to check that we can obtain the same results as from the multilevel model, we must account for this dependency:
rma.mv(coef(agg), V=vcov(agg), method="EE")
Again, exactly the same results. So now we use these coefficients and the diagonal from vcov(agg) as their sampling variances in the forest plot:
forest(coef(agg), diag(vcov(agg)), slab=names(coef(agg)))
The forest plot cannot indicate the dependency that still remains in these values, so if one were to meta-analyze these aggregated values using only diag(vcov(agg)) as their sampling variances, the results would not be identical to what you get from the full multilevel model. But there isn't really a way around that and the plot is just a visualization of the aggregated estimates and the CIs shown are correct.
You need to specify your own grouping in a new column of data and use this as the new random effect:
df$study_group <- c(1,1,1,2,2,3,4,5,5,5) # example
full.model <- rma.mv(yi = yi,
V = vi,
slab = Author,
data = df,
random = ~ 1 | study_group,
test = "t",
method = "REML")
forest(full.model)
I'm learning to implement robust glms in R, but can't figure out why I am unable to get glmrob to predict values from my regression models when I have a model where some columns are dropped due to co-linearity. Specifically when I use the predict function to predict values from a glmrob, it always gives NA for all values. I don't observe this when predicting values from the same data & model using glm. It doesn't seem to matter what data I use -- as long as there is a NA coefficient in the fitted model (and the NA isn't the last coefficient in the coefficient vector), the predict does not work.
This behavior holds for all datasets and models I have tried where an internal column is dropped due to co-linearity. I include a fake data set where two columns are dropped from the model, which gives two NAs in the coefficient list. Both glm and glmrob give nearly identical coefficients, yet predict only works with the glm model. So my question is: what don't I understand about robust regression that would prevent my glmrob models from generating predicted values?
library(robustbase)
#Make fake data with two categorial predictors
df <- data.frame("category" = rep(c("A","B","C"),each=6))
df$location <- rep(1:6,each=3)
val <- rep(c(500,50,5000),each=6)+rep(c(50,100,25,200,100,1),each=3)
df$value <- rpois(NROW(df),val)
#note that predict works if we omit the newdata parameter. However I need the newdata param
#so I use the original dataframe here as a stand-in.
mod <- glm(val ~ category + as.factor(location), data=df, family=poisson)
predict(mod, newdata=df) # works fine
mod <- glmrob(val ~ category + as.factor(location), data=df, family=poisson)
predict(mod, newdata=df) #predicts NA for all values
I've been digging into this and have concluded that the problem does not lie in my understanding of robust regression, but rather the problem lies with a bug in the robustbase package. The predict.lmrob function does not correctly pick the necessary coefficients from the model before the prediction. It needs to pick the first x non-NA coefficients (where x=rank of the model matrix). Instead it merely picks the first x coefficients without checking if they are NA. This explains why this problem only surfaces for models where the NA isn't the last coefficient in the coefficient vector.
To fix this, I copied the predict.lmrob source using:
getAnywhere(predict.lmrob)
and created my own replacement function. In this function I made a single modification to the code:
...
p <- object$rank
if (is.null(p)) {
df <- Inf
p <- sum(!is.na(coef(object)))
#piv <- seq_len(p) # old code
piv <- which(!is.na(coef(object))) # new code
}
else {
p1 <- seq_len(p)
piv <- if (p)
qr(object)$pivot[p1]
}
...
I've run a few hundred datasets using this change and it has worked well.
I have some data with some missing values for one variable, and I want to be able to create (random) predictions for what these could be. Here's my first thought:
# miss indicates where the observations with missing response are
library(MASS)
model <- glm.nb(data[-miss,4] ~ ., data=data[-miss,-4])
predict(model, newdata=data[miss,-4])
However, if I repeat the last line, it gives the same answers over and over - it appears to give the predicted mean of responses given that data and the model. I want a random prediction which incorporates variance i.e. a random draw from the distribution of the response of an observation with such predictors under the given model.
It could have something to do with the pred.var argument, but I'm unsure how to use that.
Suppose we have data like this:
set.seed(101)
dd <- data.frame(x=(1:20)*0.1)
dd$y <- rnbinom(20,mu=exp(dd$x),size=1)
## make some missing values
miss <- c(2,3,5)
dd$y[miss] <- NA
Now fit a model:
m1 <- MASS::glm.nb(y~x,dd,na.action=na.exclude)
Now use predictions from that model to get the expected mean value and rnbinom to generate the random values ...
p <- predict(m1,newdata=dd,type="response")
randvals <- rnbinom(length(p),mu=p,size=m1$theta)
(This gives random values for every element, not just the missing ones, but obviously you can pick out just the ones you want ...) It would be nice if the simulate method did this, but it's not quite flexible enough ...
I want to obtain the equations of the probability functions represented by plotmo (R). This is the equations of the model when varying one or two predictors while holding the other predictors constant in the mean value. I want an easy way to obtain the mathematical equation because a must to make to many models with different variables.
if my model is like this:
glm(formula = pres_aus ~ pH_sp + Annual_prec + I(pH_sp^2) + I(Annual_prec^2), family = binomial(link = "logit"), data = puntos_calibrado)
how can i make it?
No data example provided, so no testing done, but couldn't you just skip the construction of a symbolic expression and do something along the lines of:
model.matrix(data.frame(one=1, dat) ) %*% coef(mdl.fit)
# where mdl.fit is returned from glm()
In a sense this is the R matrix representation of the formula: sum( beta_i*X_1). If you want to specify a mean value for a particular column then just pull that dataframe apart and use only parts of it for a calculation. So for the first column held at the mean:
model.matrix(data.frame(one=1, mn1 =mean(dat[[1]]), dat[-1]) ) %*%
coef(mdl.fit)