如何计算 scikit-learn ML 模型每个样本的二进制对数损失

niu*_*999 3 python numpy machine-learning loss scikit-learn

我正在尝试将二进制日志损失应用于我创建的朴素贝叶斯 ML 模型。我生成了一个分类预测数据集(yNew)和一个概率数据集(probabilityYes),但无法在对数损失函数中成功运行它们。

简单的 sklearn.metrics 函数给出单个对数损失结果 - 不知道如何解释这个

from sklearn.metrics import log_loss
ll = log_loss(yNew, probabilityYes, eps=1e-15)
print(ll)
.0819....
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更复杂的函数为每个“否”返回 2.55,为每个“是”返回 2.50(总共 90 列)-同样,不知道如何解释它

def logloss(yNew,probabilityYes):
epsilon = 1e-15
probabilityYes = sp.maximum(epsilon, probabilityYes)
probabilityYes = sp.minimum(1-epsilon, probabilityYes)

#compute logloss function (vectorised)
ll = sum(yNew*sp.log(probabilityYes) +
            sp.subtract(1,yNew)*sp.log(sp.subtract(1,probabilityYes)))
ll = ll * -1.0/len(yNew)
return ll

print(logloss(yNew,probabilityYes))
2.55352047 2.55352047 2.50358354 2.55352047 2.50358354 2.55352047 .....
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des*_*aut 5

以下是计算每个样本损失的方法:

import numpy as np

def logloss(true_label, predicted, eps=1e-15):
  p = np.clip(predicted, eps, 1 - eps)
  if true_label == 1:
    return -np.log(p)
  else:
    return -np.log(1 - p)
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让我们用一些虚拟数据来检查它(我们实际上不需要模型):

predictions = np.array([0.25,0.65,0.2,0.51,
                        0.01,0.1,0.34,0.97])
targets = np.array([1,0,0,0,
                   0,0,0,1])

ll = [logloss(x,y) for (x,y) in zip(targets, predictions)]
ll
# result:
[1.3862943611198906,
 1.0498221244986778,
 0.2231435513142097,
 0.7133498878774648,
 0.01005033585350145,
 0.10536051565782628,
 0.41551544396166595,
 0.030459207484708574]
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从上面的数组中,您应该能够说服自己,预测距离相应的真实标签越远,损失就越大,正如我们直观地预期的那样。

让我们确认一下上面的计算与 scikit-learn 返回的总(平均)损失一致:

from sklearn.metrics import log_loss

ll_sk = log_loss(targets, predictions)
ll_sk
# 0.4917494284709932

np.mean(ll)
# 0.4917494284709932

np.mean(ll) == ll_sk
# True
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代码改编自这里[链接现已失效]。