I'd like to use the ols() (ordinary least squares) function from the rms package to do a multivariate linear regression, but I would not like it to calculate the intercept. Using lm() the syntax would be like:
model <- lm(formula = z ~ 0 + x + y, data = myData)
where the 0 stops it from calculating an intercept, and only two coefficients are returned, on for x and the other for y. How do I do this when using ols()?
Trying
model <- ols(formula = z ~ 0 + x + y, data = myData)
did not work, it still returns an intercept and a coefficient each for x and y.
Here is a link to a csv file
It has five columns. For this example, can only use the first three columns:
model <- ols(formula = CorrEn ~ intEn_anti_ncp + intEn_par_ncp, data = ccd)
Thanks!
rms::ols uses rms:::Design instead of model.frame.default. Design is called with the default of intercept = 1, so there is no (obvious) way to specify that there is no intercept. I assume there is a good reason for this, but you can try changing ols using trace.
Related
I have generated randomly a dataset that has been split in two (L and I).
First I run the regression on L using all the covariates.
After defining the set of variables that are significantly different form zero I want to run the regression on I using this set of variables.
reg_L = lm(y ~ ., data = data)
S_hat = as.data.frame(round(summary(reg_L)$coefficients[,"Pr(>|t|)"], 3)<0.05)
S_hat_L = rownames(which(S_hat==TRUE, arr.ind = TRUE))
Therefore here I want to run the new model that doesn't work only due to a problem in the specification of the variable x.
What am I doing wrong?
# Using the I proportion to construct the p-values
x = noquote(paste(S_hat_L, collapse = " + "))
reg_I = lm(y ~ x, data = data)
summary(reg_I)
A simpler way than trying to manipulate a formula programmatically would be to remove the unwanted predictors from the data:
wanted <- summary(fit)$coefficients[,"Pr(>|t|)"] < 0.05
reduced.data <- data[, wanted]
reg_S <- lm(y ~ ., data=reduced.data)
Note however, that it is more robust with respect to out-of-sample performance to reduce variables with the LASSO. This will yield a model that has some coefficients set to zero, but the other coefficients are adjusted in such a way that the uot-of-sample performance will be better.
When we plot a GAM model using the mgcv package with isotropic smoothers, we have a contour plot that looks something like this:
x axis for one predictor,
y axis for another predictor,
the main is a function s(x1, x2) (isotropic smother).
Suppose that in this model we have many other isotropic smoothers like:
y ~ s(x1, x2) + s(x3, x4) + s(x5, x6)
My doubts are: when interpreting the contour plot for s(x1, x2), what happens to the others isotropic smoothers? Are they "fixed at their medians"? Can we interpret a s(x1, x2) plot separately?
Because this model is additive in the functions you can interpret the functions (the separate s() terms) separately, but not necessarily as separate effects of covariates on the response. In your case there is no overlap between the covariates in each of the bivariate smooths, so you can also interpret them as the effects of the covariates on the response separately from the other smoothers.
All of the smooth functions are typically subject to a sum to zero constraint to allow the model constant term (the intercept) to be an identifiable parameter. As such, the 0 line in each plot is the value of the model constant term (on the scale of the link function or linear predictor).
The plots shown in the output from plot.gam(model) are partial effects plots or partial plots. You can essentially ignore the other terms if you are interested in understanding the effect of that term on the response as a function of the covariates for the term.
If you have other terms in the model that might include one or more covariates in another terms, and you want to look at how the response changes as you vary that term or coavriate, then you should predict from the model over the range of the variables you are interested in, whilst holding the other variables at some representation values, say their means or medians.
For example if you had
model <- gam(y ~ s(x, z) + s(x, v), data = foo, method = 'REML')
and you want to know how the response varied as a function of x only, you would fix z and v at representative values and then predict over a range of values for x:
newdf <- with(foo, expand.grid(x = seq(min(x), max(x), length = 100),
z = median(z)
v = median(v)))
newdf <- cbind(newdf, fit = predict(model, newdata = newdf, type = 'response'))
plot(fit ~ x, data = newdf, type = 'l')
Also, see ?vis.gam in the mgcv package as a means of preparing plots like this but where it does the hard work.
I am currently trying to fit a polynomial model to measurement data using lm().
fit_poly4 <- lm(y ~ poly(x, degree = 4, raw = T), weights = w)
with x as independent, y as dependent variable and w = 1/variance of the measurements.
I want to try a polynomial with given coefficients instead of the ones determined by R. Specifically I want my polynomial to be
y = -3,3583*x^4 + 43*x^3 - 191,14*x^2 + 328,2*x - 137,7
I tried to enter it as
fit_poly4 <- lm(y ~ 328.2*x-191.14*I(x^2)+43*I(x^3)-3.3583*I(x^4)-137.3,
weights = w)
but this just returns an error:
Error in terms.formula(formula, data = data) : invalid model formula in ExtractVars
Is there a way to determine the coefficients in lm() and how would one do this?
I'm not sure why you want to do this, but you can use an offset term:
set.seed(101)
dd <- data.frame(x=rnorm(1000),y=rnorm(1000), w = rlnorm(1000))
fit_poly4 <- lm(y ~
-1 + offset(328.2*x-191.14*I(x^2)+43*I(x^3)-3.3583*I(x^4)-137.3),
data=dd,
weights = w)
the -1 suppresses the usual intercept term.
I am supposed to find the intercept term using Ridge Regression model.
"Use Ridge Regression with alpha = 0 to and lambda = 0, divorce as the response and all the other variables as predictors."
I know I'm supposed to convert my data to matrix mode and then transform it to fit the glmnet function. I've converted my response to matrix mode, but I'm not sure how to convert all my predictors into matrix mode, too.
set.seed(100)
require(faraway)
require(leaps)
require(glmnet)
mydata = divusa
mymodel = lm(divorce ~ year + unemployed + femlab + marriage + birth +
military, data=mydata)
summary(mymodel)
.
.
.
y = model.matrix(divorce~.,mydata)
Can anyone help with the code for my x variable? I'm very new to R and finding it very hard to understand it.
Your y = model.matrix(divorce~.,mydata) actually created your predictor matrix (usually called X). Try
X = model.matrix(divorce~.,mydata)
y = mydata$divorce
glmnet(X,y)
glmnet(X,y,alpha=0,lambda=0)
I think if you set lambda=0 you're actually doing ordinary regression (i.e., you're setting the penalty to zero, so ridge -> OLS).
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.