I have been foll0wing the logistic regression chapter in Hands on Programing with R. As I started all the codes were working fine but then I retarted my R session and when I run this code
tidy(model1)
it throws this error message.
`Error in UseMethod("tidy") :
no applicable method for 'tidy' applied to an object of class "c('glm', 'lm')"`
so here are my codes up to where it throws the error
library(dplyr)
library(ggplot2)
library(rsample)
library(modeldata) #contains the attrition dataset
library(caret)
library(vip)
df <- attrition |>
mutate_if(is.ordered, factor, ordered=F)
#create training(70%) and test(30% sets )
set.seed(191) # for reproducibility
churn_split <- initial_split(df, prop =0.7, strata ='Attrition')
churn_train <- training(churn_split)
churn_test <- testing(churn_split)
#model simple logistic regression (use 1 variable for prediction)
model1 <- glm(Attrition ~ MonthlyIncome,
family ='binomial',
data = churn_train)
model2 <- glm(Attrition ~ OverTime,
family = 'binomial',
data = churn_train)
summary(model1)
exp(coef(model1))
all these codes work fine
and this tidy(model1) was also working until I restarted R studio. I want to know if I did something or the function is just messing with me and how do I fix it
I tried running your code and it gave me the same error; but after reloading the broom package, it did create a tibble of model1 and model2:
set.seed(123) # for reproducibility
churn_split <- initial_split(df, prop = .7, strata = "Attrition")
churn_train <- training(churn_split)
churn_test <- testing(churn_split)
model1 <- glm(Attrition ~ MonthlyIncome, family = "binomial", data = churn_train) # prob. of attrition on income
model2 <- glm(Attrition ~ OverTime, family = "binomial", data = churn_train) # prob. of attrition on overtime
broom::tidy(model1)
A tibble: 2 × 5
term estimate std.error statistic p.value
<chr> <dbl> <dbl> <dbl> <dbl>
1 (Intercept) -0.886 0.157 -5.64 0.0000000174
2 MonthlyIncome -0.000139 0.0000272 -5.10 0.000000344
broom::tidy(model2)
# A tibble: 2 × 5
term estimate std.error statistic p.value
<chr> <dbl> <dbl> <dbl> <dbl>
1 (Intercept) -2.13 0.119 -17.9 1.46e-71
2 OverTimeYes 1.29 0.176 7.35 2.01e-13
There might be an issue with the package or the combination of loaded packages in your R environment that makes R get confused by the class of your model "c("glm","lm")".
To my knowledge, the tidy function in the broom package works by using a mechanism to identify the specific class of your model object. Based on the object's class, it will select the right tidying method (i.e, tidy.glm or tidy.lm) and transform it into a tidy data frame.
R might be trying to figure out the class of your model but stopping after seeing the model's class doesn't pass the tidy.glm checks. (You can type broom:::tidy.glm to look at the function arguments)
You'll see that a model object of class glm will display the following results:
summary(model1)
Call:
glm(formula = Attrition ~ MonthlyIncome, family = "binomial", # identifies the model
data = churn_train)
Coefficients:
Estimate Std. Error z value Pr(>|z|) # calculates p-value.
(Intercept) -8.861e-01 1.572e-01 -5.636 1.74e-08 ***
MonthlyIncome -1.386e-04 2.719e-05 -5.098 3.44e-07 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1) # Provides a dispersion parameter like `lm` objects
Null deviance: 905.68 on 1027 degrees of freedom
Residual deviance: 870.83 on 1026 degrees of freedom
AIC: 874.83 # Provides a statistic (AIC instead of the R-squared and the F-statistic)
Number of Fisher Scoring iterations: 5
So, in theory, tidy(model1) should work whether your input is a glm or lm model. If there is some sort of issue with your packages, R might be throwing an error thinking there are no tidy functions that take objects of class c("glm","lm").
It seemed like reloading the package is a simple solution that fixes the issue.