我正在使用python和scikit-learn解决多类分类问题。目前,我正在使用 classification_report
函数来评估分类器的性能,获取如下报告:
>>> print(classification_report(y_true, y_pred, target_names=target_names))
precision recall f1-score support
class 0 0.50 1.00 0.67 1
class 1 0.00 0.00 0.00 1
class 2 1.00 0.67 0.80 3
avg / total 0.70 0.60 0.61 5
为了进行进一步的分析,我感兴趣的是获得每个可用类的每个类的 f1 分数。也许是这样的:
>>> print(calculate_f1_score(y_true, y_pred, target_class='class 0'))
0.67
scikit-learn上有类似的东西吗?
取
自f1_score
文档。
from sklearn.metrics import f1_score
y_true = [0, 1, 2, 0, 1, 2]
y_pred = [0, 2, 1, 0, 0, 1]
f1_score(y_true, y_pred, average=None)
输出:
array([ 0.8, 0. , 0. ])
这是每个班级的分数。
我会将f1_score
与labels
参数一起使用
from sklearn.metrics import f1_score
y_true = [0, 1, 2, 0, 1, 2]
y_pred = [0, 2, 1, 0, 0, 1]
labels = [0, 1, 2]
f1_scores = f1_score(y_true, y_pred, average=None, labels=labels)
f1_scores_with_labels = {label:score for label,score in zip(labels, f1_scores)}
输出:
{0: 0.8, 1: 0.0, 2: 0.0}
您只需
要使用pos_label作为参数并分配要打印的类值。
f1_score(ytest, ypred_prob, pos_label=0)# default is pos_label=1
如果您只有混淆矩阵C
,其中行对应于预测,列对应于真值,则可以使用以下函数计算 F1 分数:
def f1(C):
num_classes = np.shape(C)[0]
f1_score = np.zeros(shape=(num_classes,), dtype='float32')
weights = np.sum(C, axis=0)/np.sum(C)
for j in range(num_classes):
tp = np.sum(C[j, j])
fp = np.sum(C[j, np.concatenate((np.arange(0, j), np.arange(j+1, num_classes)))])
fn = np.sum(C[np.concatenate((np.arange(0, j), np.arange(j+1, num_classes))), j])
# tn = np.sum(C[np.concatenate((np.arange(0, j), np.arange(j+1, num_classes))), np.concatenate((np.arange(0, j), np.arange(j+1, num_classes)))])
precision = tp/(tp+fp) if (tp+fp) > 0 else 0
recall = tp/(tp+fn) if (tp+fn) > 0 else 0
f1_score[j] = 2*precision*recall/(precision + recall)*weights[j] if (precision + recall) > 0 else 0
f1_score = np.sum(f1_score)
return f1_score