Here is my code for building a design matrix for a linear modelling function:
f <- factor(targets$Sample.Name, levels = unique(targets$Sample.Name))
design <- model.matrix(~0 + f)
colnames(design) <- levels(f)
Am not sure how to interpret the formula "~0". I looked up ?lm() and found that if a formula has an implied intercept term one can remove this using either y ~ x - 1 or y ~ 0 + x, but am not sure if this is the same case here.
Yes it's the same case here. You can examine the output from design to check.
Related
In coef(l), where l is a object of class "lm", is (Intercept) always listed first?
R's source code for lm() is not so straightforward. lm() appears to call lm.fit(), which gets coefficients by calling a C function with .Call(C_Cdqrls, x, y, tol, FALSE), which ultimately calls a least squares fitting routine in FORTRAN according to this informative blog post. I'm not really familiar enough with R internals or actual code to do least squares regression to answer my question.
No, only when you have an intercept. Intercept is implicit in formula, but you can specify a model without it using - 1 or 0 +:
x <- rnorm(20)
y <- rnorm(20, 10)
> coef(lm(y ~ x + I(x^2)))
(Intercept) x I(x^2)
10.3035412 -0.1506304 -0.3092836
> coef(lm(y ~ I(x^3) + x - 1))
I(x^3) x
-0.5094851 -0.6598634
The coefficients will be listed in the order they appear in the formula. If there is an intercept, it will be the first. But as in many other situations in R, if you need to obtain the value of a specific component (intercept or any other), it is a good practice to call by it's name. It will return NA if the object don't have it:
intercept <- coef(model)["(Intercept)"]
Is there a way how I can extract coefficients of globally fitted terms in local regression modeling?
Maybe I do misunderstand the role of globally fitted terms in the function loess, but what I would like to have is the following:
# baseline:
x <- sin(seq(0.2,0.6,length.out=100)*pi)
# noise:
x_noise <- rnorm(length(x),0,0.1)
# known structure:
x_1 <- sin(seq(5,20,length.out=100))
# signal:
y <- x + x_1*0.25 + x_noise
# fit loess model:
x_seq <- seq_along(x)
mod <- loess(y ~ x_seq + x_1,parametric="x_1")
The fit is done perfectly, however, how can I extract the estimated value of the globally fitted term x_1 (i.e. some value near 0.25 for the example above)?
Finally, I found a solution to my problem using the function gam from the package gam:
require(gam)
mod2 <- gam(y ~ lo(x_seq,span=0.75,degree=2) + x_1)
However, the fits from the two models are not exactly the same (which might be due to different control settings?)...
I have an x-matrix of 8 columns. I want to run glmnet to do a lasso regression. I know I need to call:
glmnet(x, y, family = "binomial", ...).
However, how do I get x to consider all one way interactions as well? Do I have to manually remake the data frame: if so, is there an easier way? I suppose I was hoping to do something using an R formula.
Yes, there is a convenient way for that. Two steps in it are important.
library(glmnet)
# Sample data
data <- data.frame(matrix(rnorm(9 * 10), ncol = 9))
names(data) <- c(paste0("x", 1:8), "y")
# First step: using .*. for all interactions
f <- as.formula(y ~ .*.)
y <- data$y
# Second step: using model.matrix to take advantage of f
x <- model.matrix(f, data)[, -1]
glmnet(x, y)
f <- as.formula( ~ .^2) should also work for including main effects and all pairwise interactions
I was reading the documentation on R Formula, and trying to figure out how to work with depmix (from the depmixS4 package).
Now, in the documentation of depmixS4, sample formula tends to be something like y ~ 1.
For simple case like y ~ x, it is defining a relationship between input x and output y, so I get that it is similar to y = a * x + b, where a is the slope, and b is the intercept.
If we go back to y ~ 1, the formula is throwing me off. Is it equivalent to y = 1 (a horizontal line at y = 1)?
To add a bit context, if you look at the depmixs4 documentation, there is one example below
depmix(list(rt~1,corr~1),data=speed,nstates=2,family=list(gaussian(),multinomial()))
I think in general, formula that end with ~ 1 is confusing to me. Can any explain what ~ 1 or y ~ 1 mean?
Many of the operators used in model formulae (asterix, plus, caret) in R, have a model-specific meaning and this is one of them: the 'one' symbol indicates an intercept.
In other words, it is the value the dependent variable is expected to have when the independent variables are zero or have no influence. (To use the more common mathematical meaning of model terms, you wrap them in I()). Intercepts are usually assumed so it is most common to see it in the context of explicitly stating a model without an intercept.
Here are two ways of specifying the same model for a linear regression model of y on x. The first has an implicit intercept term, and the second an explicit one:
y ~ x
y ~ 1 + x
Here are ways to give a linear regression of y on x through the origin (that is, without an intercept term):
y ~ 0 + x
y ~ -1 + x
y ~ x - 1
In the specific case you mention ( y ~ 1 ), y is being predicted by no other variable so the natural prediction is the mean of y, as Paul Hiemstra stated:
> data(city)
> r <- lm(x~1, data=city)
> r
Call:
lm(formula = x ~ 1, data = city)
Coefficients:
(Intercept)
97.3
> mean(city$x)
[1] 97.3
And removing the intercept with a -1 leaves you with nothing:
> r <- lm(x ~ -1, data=city)
> r
Call:
lm(formula = x ~ -1, data = city)
No coefficients
formula() is a function for extracting formula out of objects and its help file isn't the best place to read about specifying model formulae in R. I suggest you look at this explanation or Chapter 11 of An Introduction to R.
if your model were of the form y ~ x1 + x2 This (roughly speaking) represents:
y = β0 + β1(x1) + β2(x2)
Which is of course the same as
y = β0(1) + β1(x1) + β2(x2)
There is an implicit +1 in the above formula. So really, the formula above is y ~ 1 + x1 + x2
We could have a very simple formula, whereby y is not dependent on any other variable. This is the formula that you are referencing,
y ~ 1 which roughly would equate to
y = β0(1) = β0
As #Paul points out, when you solve the simple model, you get β0 = mean (y)
Here is an example
# Let's make a small sample data frame
dat <- data.frame(y= (-2):3, x=3:8)
# Create the linear model as above
simpleModel <- lm(y ~ 1, data=dat)
## COMPARE THE COEFFICIENTS OF THE MODEL TO THE MEAN(y)
simpleModel$coef
# (Intercept)
# 0.5
mean(dat$y)
# [1] 0.5
In general such a formula describes the relation between dependent and independent variables in the form of a linear model. The lefthand side are the dependent variables, the right hand side the independent. The independent variables are used to calculate the trend component of the linear model, the residuals are then assumed to have some kind of distribution. When the independent are equal to one ~ 1, the trend component is a single value, e.g. the mean value of the data, i.e. the linear model only has an intercept.
Could anybody explain to me why
simulatedCase <- rbinom(100,1,0.5)
simDf <- data.frame(CASE = simulatedCase)
posterior_m0 <<- MCMClogit(CASE ~ 1, data = simDf, b0 = 0, B0 = 1)
always results in a MCMC acceptance ratio of 0? Any explanation would be greatly appreciated!
I think your problem is the model formula, since logistic regression models have no error term. Thus you model CASE ~ 1 should be replaced by something like CASE ~ x (the predictor variable x is mandatory). Here is your example, modified:
CASE <- rbinom(100,1,0.5)
x <- 1:100
posterior_m0 <- MCMClogit (CASE ~ x, b0 = 0, B0 = 1)
classic_m0 <- glm (CASE ~ x, family=binomial(link="logit"), na.action=na.pass)
So I think your problem is not related to the MCMCpack library (disclaimer: I have never used this package).
For anyone stumbling into this same problem :
It seems that the MCMClogit function cannot handle anything but B0=0 if your model only has an intercept.
If you add a covariate, then you can specify a precision just fine.
I would consider other packages (such as arm or rjags) if you really want to sample from this model. For a list of options available for Bayesian regression, see http://cran.r-project.org/web/views/Bayesian.html