[R] Obtaining predicted probabilities for Logistic regression

[Rui Barradas]

Às 12:13 de 13/07/2024, Christofer Bogaso escreveu:
> Hi,
> 
> I ran below code
> 
> Dat = read.csv('https://raw.githubusercontent.com/sam16tyagi/Machine-Learning-techniques-in-python/master/logistic%20regression%20dataset-Social_Network_Ads.csv')
> head(Dat)
> Model = glm(Purchased ~ Gender, data = Dat, family = binomial())
> head(predict(Model, type="response"))
> My_Predict = 1/(1+exp(-1 * (as.vector(coef(Model))[1] *
> as.vector(coef(Model))[2] * ifelse(Dat['Gender'] == "Male", 1, 0))))
> head(My_Predict)
> 
> However, My_Predict and predict(Model, type="response")) are differing
> when I tried to manually calculate prediction.
> 
> Could you please help to identify what was the mistake I made?
> 
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> and provide commented, minimal, self-contained, reproducible code.
Hello,

Sometimes when there is an error, the best way to correct it is to 
rewrite the offending part of the code.
In your case, after as.vector(coef(Model))[1] you should have a plus sign.



Dat = 
read.csv('https://raw.githubusercontent.com/sam16tyagi/Machine-Learning-techniques-in-python/master/logistic%20regression%20dataset-Social_Network_Ads.csv')
head(Dat)
Model = glm(Purchased ~ Gender, data = Dat, family = binomial())

# use matrix algebra
x <- cbind(1, (Dat$Gender == "Male")) %*% coef(Model)
pred1 <- exp(x)/(1 + exp(x))

# use the fitted line equation
y <- coef(Model)[1L] + coef(Model)[2L] * (Dat$Gender == "Male")
pred2 <- exp(y)/(1 + exp(y))

head(predict(Model, type="response"))
head(pred1) |> c()
head(pred2)


Hope this helps,

Rui Barradas


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