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多标签排名指标-标签排名平均精度| ML

原文:https://www.geesforgeks.org/multi label-ranking-metrics-label-ranking-average-precision-ml/

标签排名平均精度(LRAP)衡量预测模型的平均精度,但使用精度召回。它衡量每个样本的标签排名。其值始终大于 0 。这个指标的最佳值是 1 。这个度量与平均精度有关,但是使用了标签排名而不是精度和召回 LRAP 基本上提出了一个问题,对于每个给定的样本,排名较高的标签中有多少百分比是真实标签。 给定一个基本事实标签的二进制指示矩阵

y\epsilon \left { 0, 1 \right }^{n_{samples} * n_{labels}}.

The score associated with each label is denoted by \hat{f} where,

 \hat{f}\epsilon \left { \mathbb{R} \right }^{n_{samples} * n_{labels}}

Then we can calculate LRAP using following formula:

 LRAP(y, \hat{f}) = \dfrac{1}{n_{samples}}*\sum_{i=0}^{n_{samples}-1}\dfrac{1}{\left | y_i \right |_{0}} \sum _{j:y_{ij}=1} \dfrac{\left |L_{ij} \right |}{rank_{ij}}

where,

 L_{ij} = \left { k\colon y_{ik} =1, \hat{f_{ik}}\geq\hat{f_{ij}} \right }

and

 rank_{ij} = \left | \left {  k \colon \hat{f_{ik}}\geq\hat{f_{ij}} \right }\right |

Code : Python code to implement LRAP

# import numpy and scikit-learn libraries
import numpy as np
from sklearn.metrics import label_ranking_average_precision_score

# take sample datasets
y_true = np.array([[1, 0, 0], 
                   [1, 0, 1], 
                   [1, 1, 0]])
y_score = np.array([[0.75, 0.5, 1], 
                    [1, 0.2, 0.1],
                    [0.9, 0.7, 0.6]])

# print the output
print(label_ranking_average_precision_score(
    y_true, y_score))

输出:

0.777

为了理解上面的例子,我们来看三类人(以【1,0,0】为代表),猫(以【0,1,0】为代表),狗(以【0,0,1】为代表)。我们收到了三个样品,如【1,0,0】、【1,0,1】、【1,1,0】。这意味着我们总共有 5 个地面真实标签(人类 3 个,猫 1 个,狗 1 个)。例如,在第一个样本中,只有真正的标签人类在预测标签中获得了 2 的位置。所以,秩= 2。接下来我们需要找出一路上有多少正确的标签。只有一个正确的人类标签,因此分子值为 1。因此分数变为 1/2 = 0.5。 因此,第一个样品的 LRAP 值为:

 LRAP_{1st}=\dfrac{\left ( 0.5 \right )}{1} = 0.5

In the second sample, the first rank prediction is of human, followed by cat and dog. The fraction for the human is 1/1 = 1 and the dog is 2/3 = 0.66 (number of true label ranking along the way/ranking of dog class in the predicted label). LRAP value of 2nd sample is:

 LRAP_{2nd}=\dfrac{\left ( 1+0.66 \ right )}{2} = 0.83

Similarly, for the third sample, the value of fractions for the human class is 1/1 = 1 and the cat class is 2/2 = 1. LRAP value of 3rd sample is:

 LRAP_{3rd}=\dfrac{\left ( 2 \right )}{2} = 1

Therefore total LRAP is the sum of LRAP’s on each sample divided by the number of samples.

 LRAP =  \sum_{i=0}^{n_{samples}-1}\left ( 1 + 0.83 + 0.5 \right ) \ LRAP =  \dfrac{1}{3}\left ( 2.33 \right ) \          \ LRAP =   .77



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