数据抽样平衡方法重写

之前在R里面可以通过调用Rose这个package调用数据平衡函数,这边用python改写了一下,也算是自我学习了。

R:

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# 设定工作目录
setwd(path)
# 安装包
install.packages("ROSE")
library(ROSE)
# 检查数据
data(hacide)
table(hacide.train$cls)
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980 20


过抽样实现:

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data_balanced_over <- ovun.sample(cls ~ ., data = hacide.train, method = "over",N = 1960)$data
table(data_balanced_over$cls)
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980 980

这边需要注意是ovun不是over


欠采样实现:

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data_balanced_under <- ovun.sample(cls ~ ., data = hacide.train, method = "under", N = 40, seed = 1)$data
table(data_balanced_under$cls)
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20 20

这边需要注意的是欠采样是不放回采样,同时对数据信息的损失也是极大的


组合采样实现:

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data_balanced_both <- ovun.sample(cls ~ ., data = hacide.train, method = "both", p=0.5, N=1000, seed = 1)$data
table(data_balanced_both$cls)
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520 480

method的不同值代表着不同的采样方法,p这边是控制正类的占比,seed保证抽取样本的固定,也就是种子值。



在python上,我也没有发现有现成的package可以import,所以就参考了R的实现逻辑重写了一遍,新增了一个分层抽样group_sample,删除了过采样,重写了组合抽样combine_sample,欠抽样under_sample:

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# -*- coding:utf-8 -*-
import pandas as pd
import random as rd
import numpy as np
import math as ma


class sample_s(object):
def __init__(self):
''''this is my pleasure'''

def group_sample(self, data_set, label, percent=0.1):
# 分层抽样
# data_set:数据集
# label:分层变量
# percent:抽样占比
# q:每次抽取是否随机,null为随机
# 抽样根据目标列分层,自动将样本数较多的样本分层按percent抽样,得到目标列样本较多的特征欠抽样数据
x = data_set
y = label
z = percent
diff_case = pd.DataFrame(x[y]).drop_duplicates([y])
result = []
result = pd.DataFrame(result)
for i in range(len(diff_case)):
k = np.array(diff_case)[i]
data_set = x[x[y] == k[0]]
nrow_nb = data_set.iloc[:, 0].count()
data_set.index = range(nrow_nb)
index_id = rd.sample(range(nrow_nb), int(nrow_nb * z))
result = pd.concat([result, data_set.iloc[index_id, :]], axis=0)
new_data = pd.Series(result['label']).value_counts()
new_data = pd.DataFrame(new_data)
new_data.columns = ['cnt']
k1 = pd.DataFrame(new_data.index)
k2 = new_data['cnt']
new_data = pd.concat([k1, k2], axis=1)
new_data.columns = ['id', 'cnt']
max_cnt = max(new_data['cnt'])
k3 = new_data[new_data['cnt'] == max_cnt]['id']
result = result[result[y] == k3[0]]
return result

def under_sample(self, data_set, label, percent=0.1, q=1):
# 欠抽样
# data_set:数据集
# label:抽样标签
# percent:抽样占比
# q:每次抽取是否随机
# 抽样根据目标列分层,自动将样本数较多的样本按percent抽样,得到目标列样本较多特征的欠抽样数据
x = data_set
y = label
z = percent
diff_case = pd.DataFrame(pd.Series(x[y]).value_counts())
diff_case.columns = ['cnt']
k1 = pd.DataFrame(diff_case.index)
k2 = diff_case['cnt']
diff_case = pd.concat([k1, k2], axis=1)
diff_case.columns = ['id', 'cnt']
max_cnt = max(diff_case['cnt'])
k3 = diff_case[diff_case['cnt'] == max_cnt]['id']
new_data = x[x[y] == k3[0]].sample(frac=z, random_state=q, axis=0)
return new_data

def combine_sample(self, data_set, label, number, percent=0.35, q=1):
# 组合抽样
# data_set:数据集
# label:目标列
# number:计划抽取多类及少类样本和
# percent:少类样本占比
# q:每次抽取是否随机
# 设定总的期待样本数量,及少类样本占比,采取多类样本欠抽样,少类样本过抽样的组合形式
x = data_set
y = label
n = number
p = percent
diff_case = pd.DataFrame(pd.Series(x[y]).value_counts())
diff_case.columns = ['cnt']
k1 = pd.DataFrame(diff_case.index)
k2 = diff_case['cnt']
diff_case = pd.concat([k1, k2], axis=1)
diff_case.columns = ['id', 'cnt']
max_cnt = max(diff_case['cnt'])
k3 = diff_case[diff_case['cnt'] == max_cnt]['id']
k4 = diff_case[diff_case['cnt'] != max_cnt]['id']
n1 = p * n
n2 = n - n1
fre1 = n2 / float(x[x[y] == k3[0]]['label'].count())
fre2 = n1 / float(x[x[y] == k4[1]]['label'].count())
fre3 = ma.modf(fre2)
new_data1 = x[x[y] == k3[0]].sample(frac=fre1, random_state=q, axis=0)
new_data2 = x[x[y] == k4[1]].sample(frac=fre3[0], random_state=q, axis=0)
test_data = pd.DataFrame([])
if int(fre3[1]) > 0:
i = 0
while i < (int(fre3[1])):
data = x[x[y] == k4[1]]
test_data = pd.concat([test_data, data], axis=0)
i += 1
result = pd.concat([new_data1, new_data2, test_data], axis=0)
return result

后续使用,只需要复制上述code,存成.py的文件,后续使用的时候:

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#加载函数
import sample_s as sa
#这边可以选择你需要的分层抽样、欠抽样、组合抽样的函数
sample = sa.group_sample()
#直接调用函数即可
new_data3 = sample.combine_sample(data_train, 'label', 60000, 0.4)
#将data_train里面的label保持正样本(少类样本)达到0.4的占比下,总数抽取到60000个样本

其实不是很难的一个过程,只是强化自己对python及R语言的书写方式的记忆,谢谢。

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