# Can the F1 score be equal to zero?

As it is mentioned in the F1 score Wikipedia, 'F1 score reaches its best value at 1 (perfect precision and recall) and worst at 0'.

What is the worst condition that was mentioned? Even if we consider the case of: either precision or recall is 0. The whole F1-score value becomes undefined. Because when either precision or recall is to be 0, true postives should be 0. When the true positives value becomes 0, both the precision and recall become 0.

F1 will never be zero, but very near to zero for a bad classifier. If TP or TN is zero then there isn't any need to check F1.

It's a mistake on Wikipedia.

$$F_{1}$$ as the harmonic mean is defined only at positive real numbers. $$PRE$$ or $$REC$$ could be equal 0 in case $$TP=0$$. Which provides to undefined result $$F_1=\frac{0}{0}$$.

• Have you added something to the talk page for that Wikipedia article (to point out mistake)? – Peter Mortensen Apr 11 at 12:14

It can't be exactly zero. We need exactly one (only one) of precision. Or recall to be zero to make f1 = zero, but both have "tp" as the numerator.

#### Will be Nan
y_test = np.array([0,0,1,1])
y_pred = np.array([0,1,0,0])

from sklearn.metrics import confusion_matrix
tn, fp, fn, tp = confusion_matrix(y_test, y_pred).ravel()
precision = tp/(tp+tn)
recall = tp/(tp+fn)
f1 = 2*precision*recall/(precision+recall)
print(f1)


nan

#### Can be ~0 with a specific case, with this data
y_test = np.hstack((np.zeros((1,2)),np.ones((1,1000000))))
y_pred = np.hstack((np.ones((1,1)),np.zeros((1,1000000)),np.ones((1,1))))
tn, fp, fn, tp = confusion_matrix(y_test, y_pred).ravel()
precision = tp/(tp+tn)
recall = tp/(tp+fn)
f1 = 2*precision*recall/(precision+recall)
print('{0:1.6f}'.format(f1))


0.000002