I am using sklearn
to compute macro f1
score and I doubt if there are any bugs in the code. Here is an example (label 0
is ignored):
from sklearn.metrics import f1_score, precision_recall_fscore_support
y_true = [1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4]
y_pred = [1, 1, 1, 0, 0, 2, 2, 3, 3, 3, 4, 3, 4, 3]
p_macro, r_macro, f_macro, support_macro \
= precision_recall_fscore_support(y_true=y_true, y_pred=y_pred, labels=[1, 2, 3, 4], average='macro')
p_micro, r_micro, f_micro, support_micro\
= precision_recall_fscore_support(y_true=y_true, y_pred=y_pred, labels=[1, 2, 3, 4], average='micro')
def f(p, r):
return 2*p*r/(p+r)
my_f_macro = f(p_macro, r_macro)
my_f_micro = f(p_micro, r_micro)
print('my f macro {}'.format(my_f_macro))
print('my f micro {}'.format(my_f_micro))
print('macro: p {}, r {}, f1 {}'.format(p_macro, r_macro, f_macro))
print('micro: p {}, r {}, f1 {}'.format(p_micro, r_micro, f_micro))
The output:
my f macro 0.6361290322580646
my f micro 0.6153846153846153
macro: p 0.725, r 0.5666666666666667, f1 0.6041666666666666
micro: p 0.6666666666666666, r 0.5714285714285714, f1 0.6153846153846153
As you can see, sklearn
gives 0.6041666666666666
for macro f1
. However, it does not equal to 2*0.725*0.566666666/(0.725+0.566666666)
, where 0.725
and 0.566666666
are macro precision
and macro recall
computed by sklearn
.