在 scikit-learn 中获得 multi-label 预测的准确性

Question

在multilabel classification setting, sklearn.metrics.accuracy_score中只计算子集准确度 (3)：即为样本预测的标签集必须与[中相应的标签集完全匹配=51=].

这种计算准确度的方法有时被命名为 精确匹配率 (1):

在scikit-learn中有没有其他典型的计算精度的方法，即

（如 (1) 和 (2) 中所定义，并不太明确地称为 汉明得分 (4)（因为它与汉明损失密切相关） , 或 label-based 准确性) ？

(1) Sorower, Mohammad S. "A literature survey on algorithms for multi-label learning." 俄勒冈州立大学，科瓦利斯 (2010)。

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

(3) Ghamrawi、Nadia 和 Andrew McCallum。 "Collective multi-label classification." 第 14 届 ACM 信息与知识管理国际会议论文集。美国计算机学会，2005 年。

(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.

Answer 1

你可以自己写一个版本，这里是一个不考虑weight和normalize的例子。

import numpy as np

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

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

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
    
    '''
    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, y_pred, normalize=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

Answer 2

一个简单的汇总函数：

import numpy as np

def hamming_score(y_true, y_pred):
    return (
        (y_true & y_pred).sum(axis=1) / (y_true | y_pred).sum(axis=1)
    ).mean()


hamming_score(y_true, y_pred)
# 0.375

在 scikit-learn 中获得 multi-label 预测的准确性

Getting the accuracy for multi-label prediction in scikit-learn

python

classification

scikit-learn