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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}}](img/56d5623e5f8790533a9269fc91ab6b8f.png "Rendered by QuickLaTeX.com").

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

![ \hat{f}\epsilon \left \{ \mathbb{R} \right \}^{n_{samples} * n_{labels}} ](img/15ec92b058d97f59955337e42cc6eef6.png "Rendered by QuickLaTeX.com")

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}} ](img/4f4eb9c1c3bd88ee41a4916912a0b776.png "Rendered by QuickLaTeX.com")

where,

![ L_{ij} = \left \{ k\colon y_{ik} =1, \hat{f_{ik}}\geq\hat{f_{ij}} \right \} ](img/2c2ad7c723844e2c194dde604a506dd0.png "Rendered by QuickLaTeX.com")

and

![ rank_{ij} = \left | \left \{ k \colon \hat{f_{ik}}\geq\hat{f_{ij}} \right \}\right | ](img/b9f2dc9057cd88f06be76aa07814cb01.png "Rendered by QuickLaTeX.com")

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 和的位置。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 ](img/84443dcdbf887545c86c039fbe05c4a4.png "Rendered by QuickLaTeX.com")

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 ](img/2a88a7647782df56030bfd8ea24dad1d.png "Rendered by QuickLaTeX.com")

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 ](img/c44688df1775590156520a4d380d1d36.png "Rendered by QuickLaTeX.com")

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 ](img/4935425695c0392dd7ef93acad93936a.png "Rendered by QuickLaTeX.com")
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