python - 如何计算每个 scikit-learn ML 模型样本的二进制对数损失

Question

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

简单的 sklearn.metrics 函数给出一个单一的对数损失结果——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....

更复杂的函数为每个 NO 返回 2.55 的值，为每个 YES 返回 2.50 的值(总共 90 列)- 同样，不知道如何解释这个

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 .....

Answer 1

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)

让我们用一些虚拟数据检查一下(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]

从上面的数组中，您应该能够说服自己，预测与相应的真实标签越远，损失就越大，正如我们直觉上所期望的那样。

让我们确认上面的计算与 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

代码改编自here [链接现已失效]。