I've got some coefficients for a logit model set by a non-r user. I'd like to import those coefficients into r and generate some goodness of fit estimates on the same dataset (ROC and confusion matrix) vs my own model. My first thought was to coerce the coefficients into an existing GLM object using something like
summary(fit)$coefficients[,1] <- y
or
summary(fit)$coefficients <- x
where y and x are matrices containing the coefficients I'm trying to use to predict and fit is a previously created dummy glm object fit to the dataset. Of course, this gives me only errors.
Is there any way to pass an arbitrary coefficient vector to the predict() function or to specify coefficients in a model? Can I somehow force this by passing a vector into the offset argument in GLM? Thanks
Edit: As mentioned in the comments, there's not much statistical basis for using the arbitrary coefficients. I have a business partner who believes he/she 'knows' the right coefficients and I'm trying to quantify the loss of predictive power based on those estimates versus the coefficients generated by a proper model.
Edit2: Per BondedDust's answer, I was able to coerce the coefficients, however wasn't able to clear the error messages that predict() returned due to the coercion, it would appear that predict.lm, which is called by predict, also looks at the rank of the coefficients and that is causing the error.
The predict() function in R is used to predict the values based on the input data. All the modeling aspects in the R program will make use of the predict() function in its own way, but note that the functionality of the predict() function remains the same irrespective of the case.
The predict() function can be used to predict the probability that the market will go up, given values of the predictors. The type="response" option tells R to output probabilities of the form P(Y = 1|X) , as opposed to other information such as the logit .
How to Use a Linear Regression Model to Calculate a Predicted Response Value. Step 1: Identify the independent variable x . Step 2: Calculate the predicted response value ^y by plugging in the given x value into the least-squares linear regression line ^y(x)=ax+b y ^ ( x ) = a x + b .
This is not an answer to your posted question - which BondedDust answered - but describes an alternate way in calculating the predicted probabilities yourself which might help in this case.
# Use the mtcars dataset for a minimum worked example
data(mtcars)
# Run a logistic regression and get predictions
mod <- glm(vs ~ mpg + factor(gear) + factor(am), mtcars, family="binomial")
p1 <- predict(mod, type="response")
# Calculate predicted probabilities manually
m <- model.matrix(~ mpg + factor(gear) + factor(am), mtcars)[,]
p2 <- coef(mod) %*% t(m)
p2 <- plogis(p2)
all(p1 == p2)
#identical(as.numeric(p1), as.numeric(p2))
You can replace coef(mod)
with the vector of coefficients given to you. model.matrix
will generate the dummy variables required for the calculation - check that the ordering is the same as that of the coefficient vector.
If you follow the code through predict.glm
which passes the object to predict.lm
, it appears that the node of the model list that needs to be altered is indeed fit$coefficients
. However, altering the summary()
-object will have no effect. The [['coefficients']]
in the glm and lm objects are not matrices with columns: 'Estimate', 'Std. Error', 't value', 'Pr(>|t|)' such as produced by summary
, but rather just a vector of coefficients.
fit$coefficients <- y
newpred <- predict(fit)
You might make a copy and work on it if you will need any further use of fit
.
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