I'm sure this is something that can be done, just not sure how!
I have a dataset that is around 500 rows(csv) and it shows footballers match stas(e,g passes, shots on target)etc.I have some of their salaries(around 10) and I'n trying to predict their salaries using a linear regression equation.
In the below, if Y is salaries, is there a way on R to essentially autopopulate? what the rest of the salaries might be based on the ten salaries I do have?
lm(y ~ x1 + x2 +x3)
Any help would be much appreciated.
This is what the predict function does.
Note that you don't need to call predict.lm explicitly. Because the result of a call to lm is an object with class "lm", R "knows" to use predict.lm when you call predict on it.
Eg:
lm1 <- lm(y ~ x1 + x2 +x3)
y.fitted <- predict(lm1)
You should also be able to test the predictive accuracy of your model using cross validation with the function cv.lm in the DAAG library. With this function you create test data to test the model which is generated using training data.
Related
I have a response variable (A) which I transformed (logA) and predictor (B) from data (X) which are both continuous. How do I check the linearity between the two variables using Generalized Additive Model (GAM) in R. I use the following code
model <- gamlss(logA ~ pb(B) , data = X, trace = F)
but I am not sure about it, can I add "family=Poisson" in the code when logA is continuous in GLM? Any thoughts on this?
Thanks in advance
If your dependent variable is a count variable, you can use family=PO() without log transformation. With family=PO() a log link is already applied to transform the variable. See help page for gamlss family and also vignette on count regression section 2.1.
So it will go like:
library(gamlss)
fit = gamlss(gear ~ pb(mpg),data=mtcars,family=PO())
You can see that the predictions are log transformed and you need to take the exponential:
with(mtcars,plot(mpg,gear))
points(mtcars$mpg,exp(predict(fit,what="mu")),col="blue",pch=20)
I've run a simple model using orm (i.e. reg <- orm(formula = y ~ x)) and I'm having trouble understanding how to get predicted values for Y. I've never worked with models that use multiple intercepts. I want to know for each and every value of Y in my dataset what the predicted value from the model would be. I tried predict(reg, type="mean") and this produced values that are close to the predicted values from an OLS regression, but I'm not sure if this is what I want. I really just want something analogous to OLS where you can obtain the E(Y) given a set of predictors. If possible, please provide code I can run to do this with a very short explanation.
Is there a way to get R to run all possible models (with all combinations of variables in a dataset) to produce the best/most accurate linear model and then output that model?
I feel like there is a way to do this, but I am having a hard time finding the information.
There are numerous ways this could be achieved, but for a simple way of doing this I would suggest that you have a look at the glmulti package, which is described in detail in this paper:
glmulti: An R Package for Easy Automated Model Selection with (Generalized) Linear Models
Alternatively, very simple example of the model selection as available on the Quick-R website:
# Stepwise Regression
library(MASS)
fit <- lm(y~x1+x2+x3,data=mydata)
step <- stepAIC(fit, direction="both")
step$anova # display results
Or to simplify even more, you can do more manual model comparison:
fit1 <- lm(y ~ x1 + x2 + x3 + x4, data=mydata)
fit2 <- lm(y ~ x1 + x2, data=mydata)
anova(fit1, fit2)
This should get you started. Although you should read my comment from above. This should build you a model based on all the data in your dataset and then compare all of the models with AIC and BIC.
# create a NULL vector called model so we have something to add our layers to
model=NULL
# create a vector of the dataframe column names used to build the formula
vars = names(data)
# remove variable names you don’t want to use (at least
# the response variable (if its in the first column)
vars = vars[-1]
# the combn function will run every different combination of variables and then run the glm
for(i in 1:length(vars)){
xx = combn(vars,i)
if(is.null(dim(xx))){
fla = paste("y ~", paste(xx, collapse="+"))
model[[length(model)+1]]=glm(as.formula(fla),data=data)
} else {
for(j in 1:dim(xx)[2]){
fla = paste("y ~", paste(xx[1:dim(xx)[1],j], collapse="+"))
model[[length(model)+1]]=glm(as.formula(fla),data=data)
}
}
}
# see how many models were build using the loop above
length(model)
# create a vector to extract AIC and BIC values from the model variable
AICs = NULL
BICs = NULL
for(i in 1:length(model)){
AICs[i] = AIC(model[[i]])
BICs[i] = BIC(model[[i]])
}
#see which models were chosen as best by both methods
which(AICs==min(AICs))
which(BICs==min(BICs))
I ended up running forwards, backwards, and stepwise procedures on data to select models and then comparing them based on AIC, BIC, and adj. R-sq. This method seemed most efficient. However, when I received the actual data to be used (the program I was writing was for business purposes), I was told to only model each explanatory variable against the response, so I was able to just call lm(response ~ explanatory) for each variable in question, since the analysis we ended up using it for wasn't worried about how they interacted with each other.
This is a very old question, but for those who are still encountering this discussion - the package olsrr and specifically the function ols_step_all_possible exhaustively produces an ols model for all possible subsets of variables, based on an lm object (such that by feeding it with a full model you will get all possible combinations), and returns a dataframe with R squared, adjusted R squared, aic, bic, etc. for all the models. This is very helpful in finding the best predictors but it is also very much time consuming.
see https://olsrr.rsquaredacademy.com/reference/ols_step_all_possible.html
I do not recommend just "cherry picking" the best performing model, rather I would actually look at the output and choose carefully for the most reasonable outcome. In case you would want to immediately get the best performing model (by some criteria, say number of predictors and R2) you may write a function that saves the dataframe, arranges it by number of predictors and orders it by descending R2 and spits out the top result.
The dredge() function in R also accomplishes this.
I'm working with a Probit model, and would like to calculate the probabilities from my model for each observation in my DF. I know I can calculate this using the formula, however I am wondering if there is a quick way to output the probabilities and append them to my DF.
I am running the following model:
attach(non.part.2)
y <- cbind(E)
x1 <- cbind(tech.ems, med.com, tech.nonemerg)
probit <- glm(y ~ x1, family = binomial (link = "probit"))
summary(probit)
I am running several models, so it would be nice to be able to have R spit out the probabilities and allow me to name them in my DF(non.part.2) - something like p_x1 - so that I can run summary stats on the various models later.
Any help is much appreciated!
The following should work.
non.part2$p_x1 <- predict(probit, yourDataToPredictOn, type = "response")
I am using glm() function in R with link= log to fit my model. I read on various websites that fitted() returns the value which we can compare with the original data as compared to the predict().
I am facing some problem while fitting the model.
data<-read.csv("training.csv")
data$X2 <- as.Date(data$X2, format="%m/%d/%Y")
data$X3 <- as.Date(data$X3, format="%m/%d/%Y")
data_subset <- subset(...)
attach(data_subset)
#define variable
Y<-cbind(Y)
X<-cbind(X1,X2,X3,X4,X5,X6,X7,X8,X9,X10,X11,X12,X14)
# correlation among variables
cor(Y,X)
model <- glm(Y ~ X , data_subset,family=Gamma(link="log"))
summary(model)
detach(data_subset)
validation_data<-read.csv("validation.csv")
validation_data$X2 <- as.Date(validation_data$X2, format="%m/%d/%Y")
validation_data$X3 <- as.Date(validation_data$X3, format="%m/%d/%Y")
attach(validation_data)
predicted_valid<-predict(model, newdata=validation_data)
I am not sure how does predict work with gamma log link. I want to transform the predicted values so that it can be compared with the original data. Can someone please help me.
Add type="response" to your predict call, to get predictions on the response scale. See ?predict.glm.
predict(model, newdata=*, type="response")
Looks to me like fitted doesn't work the way you seem to think it does.
You probably want to use predict there, since you seem to want to pass it data.
see ?fitted vs ?predict