python – 获得scikit-learn中多标签预测的准确性

在 multilabel classification设置中,sklearn.metrics.accuracy_score仅计运算符集精度(3)：即,为样本预测的标签集必须与y_true中的相应标签集完全匹配.

这种计算精度的方法有时被命名,可能不那么模糊,精确匹配率(1)：

有没有办法让其他典型的方法来计算scikit-learn的准确性,即

(如(1)和(2)中所定义,并且不那么模糊地称为汉明分数(4)(因为它与汉明损失密切相关),或基于标签
准确性)
？

(1)Sorower,Mohammad S.“A literature survey on algorithms for multi-label learning.”俄勒冈州立大学,Corvallis(2010年).

(2)Tsoumakas,Grigorios和Ioannis Katakis. “Multi-label classification: An overview.”信息学系,希腊塞萨洛尼基亚里士多德大学(2006年).

(3)Ghamrawi,Nadia和Andrew McCallum. “Collective multi-label classification.”第14届ACM国际信息与知识管理会议论文集. ACM,2005.

(4)Godbole,Shantanu和Sunita Sarawagi. “Discriminative methods for multi-labeled classification.”知识发现和数据挖掘的进展. Springer Berlin Heidelberg,2004.22-30.

解决方法

您可以自己编写一个版本,这是一个不考虑权重和规范化的示例.

import numpy as np

y_true = np.array([[0,1,0],[0,1],[1,1]])

y_pred = np.array([[0,0]])

def hamming_score(y_true,y_pred,normalize=True,sample_weight=None):
    '''
    Compute the Hamming score (a.k.a. label-based accuracy) for the multi-label case
    https://stackoverflow.com/q/32239577/395857
    '''
    acc_list = []
    for i in range(y_true.shape[0]):
        set_true = set( np.where(y_true[i])[0] )
        set_pred = set( np.where(y_pred[i])[0] )
        #print('\nset_true: {0}'.format(set_true))
        #print('set_pred: {0}'.format(set_pred))
        tmp_a = None
        if len(set_true) == 0 and len(set_pred) == 0:
            tmp_a = 1
        else:
            tmp_a = len(set_true.intersection(set_pred))/\
                    float( len(set_true.union(set_pred)) )
        #print('tmp_a: {0}'.format(tmp_a))
        acc_list.append(tmp_a)
    return np.mean(acc_list)

if __name__ == "__main__":
    print('Hamming score: {0}'.format(hamming_score(y_true,y_pred))) # 0.375 (= (0.5+1+0+0)/4)

    # For comparison sake:
    import sklearn.metrics

    # Subset accuracy
    # 0.25 (= 0+1+0+0 / 4) --> 1 if the prediction for one sample fully matches the gold. 0 otherwise.
    print('Subset accuracy: {0}'.format(sklearn.metrics.accuracy_score(y_true,sample_weight=None)))

    # Hamming loss (smaller is better)
    # $$\text{HammingLoss}(x_i,y_i) = \frac{1}{|D|} \sum_{i=1}^{|D|} \frac{xor(x_i,y_i)}{|L|},$$
    # where
    #  - \\(|D|\\) is the number of samples  
    #  - \\(|L|\\) is the number of labels  
    #  - \\(y_i\\) is the ground truth  
    #  - \\(x_i\\)  is the prediction.  
    # 0.416666666667 (= (1+0+3+1) / (3*4) )
    print('Hamming loss: {0}'.format(sklearn.metrics.hamming_loss(y_true,y_pred)))

输出：

Hamming score: 0.375
Subset accuracy: 0.25
Hamming loss: 0.416666666667

python – 获得scikit-learn中多标签预测的准确性

解决方法

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