Classification of variables with logistic regression model – GLM
Hello everyone, one of the great dilemmas in all data analysis is to know which are the variables that most influence us in a certain condition.
To know this, a tool that we can use is a logistic regression model: glm.
Logistic regression allows us to know which are the variables that most influence the result of the variable analyzed; therefore, in this post we will make a complete logistic regression model step by step.
The first step is to read the file that we want to analyze, in our case this file is in an Excel format so we have to install the readxl library in R to read it.
The formula used in R to perform a logistic regression is glm , before using this function we have to divide the data set in two; train formed by 80% of the total data and in test formed by the remaining 20%. This division is done by a simple random sampling method (more, no_mas).
In the function glm we have to introduce the variable that we want to analyze (in our case Churn) and the rest of variables that influence this variable.
set.seed(515616)
mas < sample(seq(1:nrow(datos)), ceiling(nrow(datos)*0.8))
no_mas < which(!1:nrow(datos) %in% mas)
train < datos[mas,]
test < datos[no_mas,]
modelo < glm(Churn ~ ., family = binomial(link = 'logit'), data = train)
summary(modelo)
Call:
glm(formula = Churn ~ ., family = binomial(link = "logit"), data = train)
Deviance Residuals:
Min 1Q Median 3Q Max
2.119 0.520 0.336 0.188 3.061
Coefficients:
Estimate Std. Error z value Pr(>z)
(Intercept) 5.262230 0.823289 6.392 1.64e10 ***
`Day Mins` 0.012578 0.001208 10.413 < 2e16 ***
`Eve Mins` 0.006385 0.001285 4.970 6.69e07 ***
`Night Mins` 0.002036 0.001250 1.629 0.1034
`Intl Mins` 0.098624 0.023041 4.280 1.87e05 ***
`CustServ Calls` 0.435844 0.044124 9.878 < 2e16 ***
Age 0.067486 0.005186 13.013 < 2e16 ***
`Day Calls` 0.006751 0.003126 2.159 0.0308 *
`Eve Calls` 0.001641 0.003128 0.525 0.5998
`Night Calls` 0.002334 0.003184 0.733 0.4635

Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 2180.4 on 2666 degrees of freedom
Residual deviance: 1692.2 on 2657 degrees of freedom
AIC: 1712.2
Number of Fisher Scoring iterations: 6
In the results obtained in the regression we have to focus mainly on two aspects:

 Coeficiente (estimate)
 Pvalor (Pr(>z)).
Day Mins
, Eve Mins
, Intl Mins
, CustServ Calls
, Day Calls
and Age because they all have a pvalue less than 0.1.Night Mins
, Eve Calls
y Night Calls
because they have a pvalue greater than 0.1.exp(coefficients(modelo))
(Intercept) `Day Mins` `Eve Mins` `Night Mins` `Intl Mins` `CustServ Calls`
0.005183732 1.012657675 1.006405441 1.002037919 1.103651137 1.546267924
Age `Day Calls` `Eve Calls` `Night Calls`
0.934741078 1.006774202 1.001642462 1.002336852

 Day Mins: Increase by one unit the variable causes an increase of 0.012% the chances of leaving the company. & nbsp;

 Eve Mins: Increase by one unit the variable causes an increase of 0.006% the chances of leaving the company. & nbsp;

 Night Mins: Increase by one unit the variable causes an increase of 0.002% the chances of leaving the company.

 Intl Mins: Increasing by one unit the variable causes an increase of 0.103% the chances of leaving the company. & nbsp;

 CustServ Calls: Increasing by one unit the variable causes an increase of 0.546% the chances of leaving the company. & nbsp;

 Age: Increasing the variable by one unit causes a 0.07% decrease in the chances of leaving the company. & nbsp;

 Day Calls: Increasing by one unit the variable causes an increase of 0.006% the chances of leaving the company. & nbsp;

 Eve Calls: Increasing by one unit the variable causes an increase of 0.0016% the chances of leaving the company. & nbsp;
 Night Calls: Increasing by one unit the variable causes an increase of 0.0023% the chances of leaving the company.
To know the quality of the model we use the AUC indicator; which can be defined as the probability that a classifier will order a positive instance randomly chosen higher than a negative one. The levels of this indicator are:

 [0.5, 0.6): Bad test. & nbsp;

 [0.6, 0.75): Regular test. & nbsp;

 [0.75, 0.9): Good test. & nbsp;

 [0.9, 0.97): Very good test.
 [0.97, 1): Excellent test.
library(ROCR)
p < predict(modelo, test, type = "response") pr < prediction(p, test$Churn) prf < performance(pr, measure = "tpr", x.measure = "fpr") plot(prf)
auc < performance(pr,measure = "auc") auc < auc@y.values[[1]] auc
In our case we find an auc = 0.797253: so we can say that our test is good. & nbsp;
We also find the ROC curve which is the representation of true positives against false positives, that is, taking 20% of the sample we would reach 60% of the population.
Greetings. & Nbsp;