Iso-F1 curve for Precision-Recall Curve

Question

I'm reading through Sklearn's tutorial on computing precision/recall! I came across this curve called "Iso-F1" curve they are plotting: link.

I tried to read their code for generating it, but I can't seem to understand -- the idea seems to be fixing F1 score, generating x points, and then generate y based on the f-score?

for f_score in f_scores:
    x = np.linspace(0.01, 1)
    y = f_score * x / (2 * x - f_score)
    l, = plt.plot(x[y >= 0], y[y >= 0], color='gray', alpha=0.2)
    plt.annotate('f1={0:0.1f}'.format(f_score), xy=(0.9, y[45] + 0.02))

I can't find a lot of information about this curve online, and the only one that seems to be discussing about creating this curve: https://github.com/scikit-learn/scikit-learn/issues/8313

Another question I have is:

For precision-recall curve, if the curve is concave like below, does it mean I have a very good classifier?

Answer 1

By definition, an iso-F₁ curve contains all points in the precision/recall space whose F₁ scores are the same.

We can present as many iso-F₁ curves in the plot of a precision-recall curve as we'd like. E.g., one would contain all points for which F₁ equals 0.2, the second one all points for which F₁ equals 0.4, and so on. In the code snippet, each iteration of the loop plots a single iso-F₁ curve, and in each iteration variable f_score stores the value of F₁ corresponding to the current curve.

A point in the plot has coordinates $x$ and $y$ corresponding to a pair of recall and precision values. In the snippet, the $x$ (recall) coordinates of a curve are stored in x and are calculated with np.linspace(0.01, 1), which gives 50 evenly spaced numbers over the interval $[0.01, 1]$.

For calculating the $y$ (precision) coordinate of a point for given values of recall and F₁, we have to turn to the formula of F₁ score:

$$ \mathrm{F}_{1} = 2 \cdot \frac{\mathrm{precision} \cdot \mathrm{recall}}{\mathrm{precision} + \mathrm{recall}} $$

We can express $\mathrm{precision}$ as:

$$ \mathrm{precision} = \frac{\mathrm{F}_{1} \cdot \mathrm{recall}}{2 \cdot \mathrm{recall} - \mathrm{F}_{1}} $$

y = f_score * x / (2 * x - f_score) corresponds to the equation above. With this we obtain the $y$ coordinates of the points of the iso-F₁ curve. The curve now can be plotted.

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As for the second question: precision-recall curves are usually concave like yours. Your curve seems OK, but one needs to know about the specific task (domain, use-case, baseline solutions etc.) to tell if it's a very good classifier.