I have a matrix Expr with rows representing variables and columns samples.
I have a categorical vector called groups (containing either "A","B", or "C")
I want to test which of variables 'Expr' can be explained by the fact that the sample belong to a group.
My strategy would be modelling the problem with a generalized additive model (with a negative binomial distribution).
And then I want use a likelihood ratio test in a variable wise way to get a p value for each variable.
I do:
require(VGAM)
m <- vgam(Expr ~ group, family=negbinomial)
m_alternative <- vgam(Expr ~ 1, family=negbinomial)
and then:
lr <- lrtest(m, m_alternative)
The last step is wrong because it is testing the overall likelihood ratio of the two model not the variable wise.
Instead of a single p value I would like to get a vector of the p-values for every variable.
How should I do it?
(I am very new to R, so forgive me my stupidity)
It sounds like you want to use Expr as your predictors It think you may have your formula backwards. The response should be on the left, so I guess that's groups in your case.
If Expr is a data.frame, you can do regression on all variables with
m <- vgam(group ~ ., Expr, family=negbinomial)
If class(Expr)=="matrix", then
m <- vgam(group ~ Expr, family=negbinomial)
probably should work, but you may just get slightly odd looking coefficient labels.
Related
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.$pred1 <- predict(lm_obj1, newdata=data.[16,20]) #predict new data; needs to have same column headings
I would like to know how I can loop a regression n times, and in each time with a different set of variables, extract a data.frame where each column is a regression and each row represent a variable.
In my case I have a data.frame of:
dt_deals <- data.frame(Premium=c(1,3,4,5),Liquidity=c(0.2,0.3,1.5,0.8),Leverage=c(1,3,0.5,0.7))
But I have another explanatory dummy variable called hubris, that is the product of a binomial distribution, with 0.25 of mean. Like that:
n <- 10
hubris_dataset <- data.frame(replicate(n, rbinom(4,1,0.25))
In this sense, what I need is to make n simulation of hubris, so I can, make n regression each one with a different set of random binomial distribution and the output of each distribution needs to be put in a data.frame
So far I could reach this:
# define n as the number of simulations i want
n=10
# define beta as a data.frame to put every coefficient from the lm regression
beta=NULL
for(i in 1:n) {
dt_deals2 <- dt_deals
beta[[i]] <- coef(lm(dt_deals$Premium ~ dt_deals$Liquidity + dt_deals$Leverage + hubris_dataset[,i], data=dt_deals2))
beta <- cbind(reg$coefficients)
}
But this way it only generate the first set of coefficient, and doesn't make another ten columns for the data.frame.
#jogo give an idea to change the for-loop method and use sapply, and change the object beta to list(). This was the result:
beta <- sapply(1:n, function(i) coef(lm(Premium ~ Liquidity +Leverage+ hubris_dataset[,i], data=dt_deals2)))
And it worked
My problem is with the predict() function, its structure, and plotting the predictions.
Using the predictions coming from my model, I would like to visualize how my significant factors (and their interaction) affect the probability of my response variable.
My model:
m1 <-glm ( mating ~ behv * pop +
I(behv^2) * pop + condition,
data=data1, family=binomial(logit))
mating: individual has mated or not (factor, binomial: 0,1)
pop: population (factor, 4 levels)
behv: behaviour (numeric, scaled & centered)
condition: relative fat content (numeric, scaled & centered)
Significant effects after running the glm:
pop1
condition
behv*pop2
behv^2*pop1
Although I have read the help pages, previous answers to similar questions, tutorials etc., I couldn't figure out how to structure the newdata= part in the predict() function. The effects I want to visualise (given above) might give a clue of what I want: For the "behv*pop2" interaction, for example, I would like to get a graph that shows how the behaviour of individuals from population-2 can influence whether they will mate or not (probability from 0 to 1).
Really the only thing that predict expects is that the names of the columns in newdata exactly match the column names used in the formula. And you must have values for each of your predictors. Here's some sample data.
#sample data
set.seed(16)
data <- data.frame(
mating=sample(0:1, 200, replace=T),
pop=sample(letters[1:4], 200, replace=T),
behv = scale(rpois(200,10)),
condition = scale(rnorm(200,5))
)
data1<-data[1:150,] #for model fitting
data2<-data[51:200,-1] #for predicting
Then this will fit the model using data1 and predict into data2
model<-glm ( mating ~ behv * pop +
I(behv^2) * pop + condition,
data=data1,
family=binomial(logit))
predict(model, newdata=data2, type="response")
Using type="response" will give you the predicted probabilities.
Now to make predictions, you don't have to use a subset from the exact same data.frame. You can create a new one to investigate a particular range of values (just make sure the column names match up. So in order to explore behv*pop2 (or behv*popb in my sample data), I might create a data.frame like this
popbbehv<-data.frame(
pop="b",
behv=seq(from=min(data$behv), to=max(data$behv), length.out=100),
condition = mean(data$condition)
)
Here I fix pop="b" so i'm only looking at the pop, and since I have to supply condition as well, i fix that at the mean of the original data. (I could have just put in 0 since the data is centered and scaled.) Now I specify a range of behv values i'm interested in. Here i just took the range of the original data and split it into 100 regions. This will give me enough points to plot. So again i use predict to get
popbbehvpred<-predict(model, newdata=popbbehv, type="response")
and then I can plot that with
plot(popbbehvpred~behv, popbbehv, type="l")
Although nothing is significant in my fake data, we can see that higher behavior values seem to result in less mating for population B.
In modeling it is helpful to do univariate regressions of a dependent on an independent in linear, quadratic, cubic and quaternary (?) forms to see which captures the basic shape of the statistical data. I'm a fairly new R programmer and need some help.
Here's pseudocode:
for i in 1:ncol(data)
data[,ncol(data) + i] <- data[, i]^2 # create squared term
data[,ncol(data) + i] <- data[, i]^3 # create cubed term
...and similarly for cubed and fourth power terms
# now do four regressions starting with linear and including one higher order term each time and display for each i the form of regression that has the highest adj R2.
lm(y ~ data[,i], ... )
# retrieve R2 and save indexed for linear case in vector in row i
lm(y tilda data[,i], data[,ncol(data) + i, ...]
# retrieve R2 and save...
Result is a dataframe indexed by i with column name in data of original x variable and results for each of the four regressions (all run with an intercept term).
Ordinarily we do this by looking at plots but where you have 800 variables that is not feasible.
If you really want to help out write code to automatically insert the required number of exponentiated variables into data.
And this doesn't even take care of the kinky variables that come clumped up in a couple of clusters or only relevant at one value, etc.
I'd say the best way to do this is by using the polynomial function in R, poly(). Imagine you have an independent numeric variable, x, and a numeric response variable, y.
models=list()
for (i in 1:4)
models[[i]]=lm(y~poly(x,i),raw=TRUE)
The raw=TRUE part ensures that the model uses the raw polynomials, and not the orthogonal polynomials.
When you want to get one of the models, just type in models[[1]] or models[[2]], etc.
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.