Naïve Bayes Theorem for multiple features

I understand the basic principles for the naïve bayes classification with one feature:

$$P(Class|feature) = (P(f|Class) * P(Class)) / P(f)$$

We have a dataset that has the following attributes/features:

day | outlook | temperature | humidity | wind | play

• Day is just a number (sequence)
• Outlook can be [sunny | overcast | rain]
• Temperature can be [cool | mild | hot]
• Humidity can be [normal | high]
• Wind can be [strong | weak]
• Play is [yes | no]

Now, we have a new instance: today = (sunny, cool, high, strong) and we want to know if we can play outside. This is Bayes classification with multiple features, as you've recognized.

The image below is a slide from my course at uni, however I don't understand anything of it.

Who can explain to work out the above formulas to me like I'm five, maybe with Python code? I'd like to understand how I can do naïve bayes classification for multiple features.

• There is no naive Bayes theorem, there are naive Bayes algorithm and Bayes theorem. What exactly is unclear for you? – Tim Jan 17 '18 at 17:40
• @Tim It's about the Bayes Theorem with multiple features, as the image shows. I'd like an explanation of the steps I should take to do that and maybe some code :) – jbehrens94 Jan 18 '18 at 11:58
• Your image related to naive Bayes algorithm... what exactly is unclear for you in here? Have you seen stats.stackexchange.com/questions/21822/… or stats.stackexchange.com/questions/33185/… ? – Tim Jan 18 '18 at 13:16
• @Tim I understand the algorithm for one feature ( P(Class|feature) = P(feature|Class)*P(Class)/P(feature)). However, when I get multiple features (for example outlook, wind and humidity) I get lost with the formula in the image I included. I just don't understand what to do, how to tackle this. – jbehrens94 Jan 18 '18 at 13:23
• @Tim I added more detail to my question, do you think you could help me out with the question? – jbehrens94 Jan 18 '18 at 14:50

\begin{align} p(C_k, x_1, x_2, ..., x_n) &\propto p(x_1 | C_k) \, p(x_2 | C_k) \dots p(x_n | C_k) \, p(C_k) \\ &= \prod_{i=1}^n p(x_i|C_k) \, p(C_k) \end{align}
This follow from the Bayes theorem and independence. So in your example today = (sunny, cool, high, strong), you look at $p($day = sunny $|$ play = yes $)$, and $p($outlook = cool $|$ play = yes $)$, etc.
• Let me check if I understand correctly: P(Class) multiplied by P(feature 1|Class) x ... x ... P(feature n|Class) for every feature? – jbehrens94 Jan 18 '18 at 15:21