机器学习中难免遇到正负样本不平衡问题，处理办法通常有梁总，一：过采样，增加正样本数据；二：欠采样，减少负样本数据，缺点是会丢失一些重要信息。smote属于过采样。

代码

# from imblearn.over_sampling import BorderlineSMOTE

# from imblearn.over_sampling import SMOTENC

# from imblearn.over_sampling import SVMSMOTE

# from imblearn.over_sampling import KMeansSMOTE

# from imblearn.over_sampling import ADASYN

# from imblearn.over_sampling import RandomOverSampler

import pandas as pd

import numpy as np

from collections import Counter

from imblearn.over_sampling import SMOTE# 使用imlbearn库中上采样方法中的SMOTE接口

import matplotlib.pyplot as plt

# 生成一组0和1比例为9比1的样本，X为特征，y为对应的标签

x1=[np.random.randint(1,31) for i in range(90)]+[np.random.randint(50,81) for i in range(10)]

x2=[np.random.randint(1,31) for i in range(90)]+[np.random.randint(50,81) for i in range(10)]

y=[0 for i in range(90)]+[1 for i in range(10)]

x=pd.DataFrame({'x1':x1,'x2':x2})

y=pd.DataFrame(y)

# 查看所生成的样本类别分布，0和1样本比例9比1，属于类别不平衡数据

print(Counter(list(y[0])))

fig1=plt.figure(1)

plt.scatter(x['x1'],x['x2'])

plt.show

# 定义SMOTE模型，random_state相当于随机数种子的作用

smo = SMOTE(sampling_strategy='auto',random_state=10)

x_smo, y_smo = smo.fit_sample(x, y)

print(Counter(list(y_smo[0])))

fig2=plt.figure(2)

plt.scatter(x_smo['x1'],x_smo['x2'])

plt.show

结果

处理前

处理后

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