火车测试按特定班级计数分开

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闭合 1年前。

Answer 1

您不应因失衡而影响火车测试分裂。火车测试拆分必须对应于实际测试分布。如果您的问题不平衡 - 您的测试套装也应该是！

您可以更改的是是您使用和/或培训制度的度量，例如：

• 使用平衡准确性而不是准确的评估准确性
• 班级重量重量在培训期间，

这两种技术在技术上都将以同样重要的方式对待课程的效果相同，但是您不必以不同的方式“拆分”。

而且，如果您真的坚持以一种奇怪的方式拆分数据，那就可以手工

import numpy as np

def odd_split(X, y, minority_class=1, minority_test_size=0.1):
  minority_indices = np.where(y==minority_class)[0]
  majority_indices = np.where(y!=minority_class)[0]
    
  n = max(1, int(minority_test_size* len(minority_indices)))
  selected = np.random.choice(range(len(minority_indices)), n, replace=False)
  test_minority_indices = minority_indices[selected]
  assert (y[test_minority_indices] == minority_class).all()
  
  selected = np.random.choice(range(len(majority_indices)), n, replace=False)
  test_majority_indices = majority_indices[selected]
  assert (y[test_majority_indices ] != minority_class).all()
  
  test_indices = np.concatenate((test_minority_indices, test_majority_indices))
  train_indices = np.array([i for i in range(len(y)) if i not in set(test_indices)])
  
  return X[train_indices], y[train_indices], X[test_indices], y[test_indices]
  

from collections import Counter  
X = np.random.normal(size=(1000, 2))  
y = np.random.choice([0, 1], p=[0.9, 0.1], size=1000)
print('Whole', Counter(y))

X_train, y_train, X_test, y_test = odd_split(X, y)
print('Train', Counter(y_train))
print('Test', Counter(y_test))

做

Whole Counter({0: 886, 1: 114})
Train Counter({0: 875, 1: 103})
Test Counter({1: 11, 0: 11})