Here is a different view from the other answers that I think will be helpful to others. The goal here is to allow you to compute these metrics using basic laws of probability.
First, it helps to understand what a confusion matrix is telling us in general. Let $Y$ represent a class label and $\hat Y$ represent a class prediction. In the binary case, let the two possible values for $Y$ and $\hat Y$ be $0$ and $1$, which represent the classes. Next, suppose that the confusion matrix for $Y$ and $\hat Y$ is:
|
$\hat Y = 0$ |
$\hat Y = 1$ |
$Y = 0$ |
10 |
20 |
$Y = 1$ |
30 |
40 |
With hindsight, let us normalize the rows and columns of this confusion matrix, such that the sum of all elements of the confusion matrix is $1$. Currently, the sum of all elements of the confusion matrix is $10 + 20 + 30 + 40 = 100$, which is our normalization factor. After dividing the elements of the confusion matrix by the normalization factor, we get the following normalized confusion matrix:
|
$\hat Y = 0$ |
$\hat Y = 1$ |
$Y = 0$ |
$\frac{1}{10}$ |
$\frac{2}{10}$ |
$Y = 1$ |
$\frac{3}{10}$ |
$\frac{4}{10}$ |
With this formulation of the confusion matrix, we can interpret $Y$ and $\hat Y$ slightly differently. We can interpret them as jointly Bernoulli (binary) random variables, where their normalized confusion matrix represents their joint probability mass function. When we interpret $Y$ and $\hat Y$ this way, the definitions of precision and recall are much easier to remember using Bayes' rule and the law of total probability:
\begin{align}
\text{Precision} &= P(Y = 1 \mid \hat Y = 1) = \frac{P(Y = 1 , \hat Y = 1)}{P(Y = 1 , \hat Y = 1) + P(Y = 0 , \hat Y = 1)} \\
\text{Recall} &= P(\hat Y = 1 \mid Y = 1) = \frac{P(Y = 1 , \hat Y = 1)}{P(Y = 1 , \hat Y = 1) + P(Y = 1 , \hat Y = 0)}
\end{align}
How do we determine these probabilities? We can estimate them using the normalized confusion matrix. From the table above, we see that
\begin{align}
P(Y = 0 , \hat Y = 0) &\approx \frac{1}{10} \\
P(Y = 0 , \hat Y = 1) &\approx \frac{2}{10} \\
P(Y = 1 , \hat Y = 0) &\approx \frac{3}{10} \\
P(Y = 1 , \hat Y = 1) &\approx \frac{4}{10}
\end{align}
Therefore, the precision and recall for this specific example are
\begin{align}
\text{Precision} &= P(Y = 1 \mid \hat Y = 1) = \frac{\frac{4}{10}}{\frac{4}{10} + \frac{2}{10}} = \frac{4}{4 + 2} = \frac{2}{3} \\
\text{Recall} &= P(\hat Y = 1 \mid Y = 1) = \frac{\frac{4}{10}}{\frac{4}{10} + \frac{3}{10}} = \frac{4}{4 + 3} = \frac{4}{7}
\end{align}
Note that, from the calculations above, we didn't really need to normalize the confusion matrix before computing the precision and recall. The reason for this is that, because of Bayes' rule, we end up dividing one value that is normalized by another value that is normalized, which means that the normalization factor can be cancelled out.
A nice thing about this interpretation is that it can be generalized to confusion matrices of any size. In the case where there are more than 2 classes, $Y$ and $\hat Y$ are no longer considered to be jointly Bernoulli, but rather jointly categorical. Moreover, we would need to specify which class we are computing the precision and recall for. In fact, the definitions above may be interpreted as the precision and recall for class $1$. We can also compute the precision and recall for class $0$, but these have different names in the literature.