Reusing a Model Built in R

RModels

R Problem Overview


When building a model in R, how do you save the model specifications such that you can reuse it on new data? Let's say I build a logistic regression on historical data but won't have new observations until next month. What's the best approach?

Things that I have considered:

  • Saving the model object and loading in a new session
  • I know that some models can be exported with PMML, but haven't really seen anything about importing PMML

Simply, I am trying to get a sense of what you do when you need to use your model in a new session.

Thanks in advance.

R Solutions


Solution 1 - R

Reusing a model to predict for new observations

If the model is not computationally costly, I tend to document the entire model building process in an R script that I rerun when needed. If a random element is involved in the model fitting, I make sure to set a known random seed.

If the model is computationally costly to compute, then I still use a script as above, but save out the model objects using save() into and rda object. I then tend to modify the script such that if the saved object exists, load it, or if not, refit the model, using a simple if()...else clause wrapped around the relevant parts of the code.

When loading your saved model object, be sure to reload any required packages, although in your case if the logit model were fit via glm() there will not be any additional packages to load beyond R.

Here is an example:

> set.seed(345)
> df <- data.frame(x = rnorm(20))
> df <- transform(df, y = 5 + (2.3 * x) + rnorm(20))
> ## model
> m1 <- lm(y ~ x, data = df)
> ## save this model
> save(m1, file = "my_model1.rda")
> 
> ## a month later, new observations are available: 
> newdf <- data.frame(x = rnorm(20))
> ## load the model
> load("my_model1.rda")
> ## predict for the new `x`s in `newdf`
> predict(m1, newdata = newdf)
        1         2         3         4         5         6 
6.1370366 6.5631503 2.9808845 5.2464261 4.6651015 3.4475255 
        7         8         9        10        11        12 
6.7961764 5.3592901 3.3691800 9.2506653 4.7562096 3.9067537 
       13        14        15        16        17        18 
2.0423691 2.4764664 3.7308918 6.9999064 2.0081902 0.3256407 
       19        20 
5.4247548 2.6906722 

If wanting to automate this, then I would probably do the following in a script:

## data
df <- data.frame(x = rnorm(20))
df <- transform(df, y = 5 + (2.3 * x) + rnorm(20))

## check if model exists? If not, refit:
if(file.exists("my_model1.rda")) {
    ## load model
    load("my_model1.rda")
} else {
    ## (re)fit the model
    m1 <- lm(y ~ x, data = df)
}

## predict for new observations
## new observations
newdf <- data.frame(x = rnorm(20))
## predict
predict(m1, newdata = newdf)

Of course, the data generation code would be replaced by code loading your actual data.

Updating a previously fitted model with new observations

If you want to refit the model using additional new observations. Then update() is a useful function. All it does is refit the model with one or more of the model arguments updated. If you want to include new observations in the data used to fit the model, add the new observations to the data frame passed to argument 'data', and then do the following:

m2 <- update(m1, . ~ ., data = df)

where m1 is the original, saved model fit, . ~ . is the model formula changes, which in this case means include all existing variables on both the left and right hand sides of ~ (in other words, make no changes to the model formula), and df is the data frame used to fit the original model, expanded to include the newly available observations.

Here is a working example:

> set.seed(123)
> df <- data.frame(x = rnorm(20))
> df <- transform(df, y = 5 + (2.3 * x) + rnorm(20))
> ## model
> m1 <- lm(y ~ x, data = df)
> m1

Call:
lm(formula = y ~ x, data = df)

Coefficients:
(Intercept)            x  
      4.960        2.222  

> 
> ## new observations
> newdf <- data.frame(x = rnorm(20))
> newdf <- transform(newdf, y = 5 + (2.3 * x) + rnorm(20))
> ## add on to df
> df <- rbind(df, newdf)
> 
> ## update model fit
> m2 <- update(m1, . ~ ., data = df)
> m2

Call:
lm(formula = y ~ x, data = df)

Coefficients:
(Intercept)            x  
      4.928        2.187

Other have mentioned in comments formula(), which extracts the formula from a fitted model:

> formula(m1)
y ~ x
> ## which can be used to set-up a new model call
> ## so an alternative to update() above is:
> m3 <- lm(formula(m1), data = df)

However, if the model fitting involves additional arguments, like 'family', or 'subset' arguments in more complex model fitting functions. If update() methods are available for your model fitting function (which they are for many common fitting functions, like glm()), it provides a simpler way to update a model fit than extracting and reusing the model formula.

If you intend to do all the modelling and future prediction in R, there doesn't really seem much point in abstracting the model out via PMML or similar.

Solution 2 - R

If you use the same name of the dataframe and variables, you can (at least for lm() and glm() ) use the function update on the saved model :

Df <- data.frame(X=1:10,Y=(1:10)+rnorm(10))

model <- lm(Y~X,data=Df)
model

Df <- rbind(Df,data.frame(X=2:11,Y=(10:1)+rnorm(10)))

update(model)

This is off course without any preparation of the data and so forth. It just reuses the model specifications set. Be aware that if you change the contrasts in the meantime, the new model gets updated with the new contrasts, not the old.

So the use of a script is in most cases the better answer. One could include all steps in a convenience function that just takes the dataframe, so you can source the script and then use the function on any new dataset. See also the answer of Gavin for that.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionBtibert3View Question on Stackoverflow
Solution 1 - RGavin SimpsonView Answer on Stackoverflow
Solution 2 - RJoris MeysView Answer on Stackoverflow