def calc_chiSquare(sampleSet, feature, target):
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'''
计算某个特征每种属性值的卡方统计量
params:
sampleSet: 样本集
feature: 目标特征
target: 目标Y值 (0或1) Y值为二分类变量
return:
卡方统计量dataframe
feature: 特征名称
act_target_cnt: 实际坏样本数
expected_target_cnt：期望坏样本数
chi_square：卡方统计量
'''
# 计算样本期望频率
target_cnt = sampleSet[target].sum()
sample_cnt = len(sampleSet[target])
expected_ratio = target_cnt * 1.0/sample_cnt
# 对变量按属性值从大到小排序
df = sampleSet[[feature, target]]
col_value = list(set(df[feature]))
# 计算每一个属性值对应的卡方统计量等信息
chi_list = []; target_list = []; expected_target_list = []
for value in col_value:
df_target_cnt = df.loc[df[feature] == value, target].sum()
df_cnt = len(df.loc[df[feature] == value, target])
expected_target_cnt = df_cnt * expected_ratio
chi_square = (df_target_cnt - expected_target_cnt)**2 / expected_target_cnt
chi_list.append(chi_square)
target_list.append(df_target_cnt)
expected_target_list.append(expected_target_cnt)
# 结果输出到dataframe, 对应字段为特征属性值, 卡方统计量, 实际坏样本量, 期望坏样本量
chi_stats = pd.DataFrame({feature:col_value, 'chi_square':chi_list,
'act_target_cnt':target_list, 'expected_target_cnt':expected_target_list})
return chi_stats[[feature, 'act_target_cnt', 'expected_target_cnt', 'chi_square']]
def chiMerge_maxInterval(chi_stats, feature, maxInterval=5):
'''
卡方分箱合并--最大区间限制法
params:
chi_stats: 卡方统计量dataframe
feature: 目标特征
maxInterval：最大分箱数阈值
return:
卡方合并结果dataframe, 特征分割split_list
'''
group_cnt = len(chi_stats)
split_list = [chi_stats[feature].min()]
# 如果变量区间超过最大分箱限制，则根据合并原则进行合并
while(group_cnt maxInterval):
min_index = chi_stats[chi_stats['chi_square']==chi_stats['chi_square'].min()].index.tolist()[0]
# 如果分箱区间在最前,则向下合并
if min_index == 0:
chi_stats = merge_chiSquare(chi_stats, min_index+1, min_index)
# 如果分箱区间在最后，则向上合并
elif min_index == group_cnt-1:
chi_stats = merge_chiSquare(chi_stats, min_index-1, min_index)
# 如果分箱区间在中间，则判断与其相邻的最小卡方的区间，然后进行合并
else:
if chi_stats.loc[min_index-1, 'chi_square'] chi_stats.loc[min_index+1, 'chi_square']:
chi_stats = merge_chiSquare(chi_stats, min_index, min_index+1)
else:
chi_stats = merge_chiSquare(chi_stats, min_index-1, min_index)
group_cnt = len(chi_stats)
chiMerge_result = chi_stats
split_list.extend(chiMerge_result[feature].tolist())
return chiMerge_result, split_list
def chiMerge_minChiSquare(chi_stats, feature, dfree=4, cf=0.1, maxInterval=5):
'''
卡方分箱合并--卡方阈值法
params:
chi_stats: 卡方统计量dataframe
feature: 目标特征
maxInterval: 最大分箱数阈值, default 5
dfree: 自由度, 最大分箱数-1, default 4
cf: 显著性水平, default 10%
return:
卡方合并结果dataframe, 特征分割split_list
'''
threshold = get_chiSquare_distuibution(dfree, cf)
min_chiSquare = chi_stats['chi_square'].min()
group_cnt = len(chi_stats)
split_list = [chi_stats[feature].min()]
# 如果变量区间的最小卡方值小于阈值，则继续合并直到最小值大于等于阈值
while(min_chiSquare threshold and group_cnt maxInterval):
min_index = chi_stats[chi_stats['chi_square']==chi_stats['chi_square'].min()].index.tolist()[0]
# 如果分箱区间在最前,则向下合并
if min_index == 0:
chi_stats = merge_chiSquare(chi_stats, min_index+1, min_index)
# 如果分箱区间在最后，则向上合并
elif min_index == group_cnt-1:
chi_stats = merge_chiSquare(chi_stats, min_index-1, min_index)
# 如果分箱区间在中间，则判断与其相邻的最小卡方的区间，然后进行合并
else:
if chi_stats.loc[min_index-1, 'chi_square'] chi_stats.loc[min_index+1, 'chi_square']:
chi_stats = merge_chiSquare(chi_stats, min_index, min_index+1)
else:
chi_stats = merge_chiSquare(chi_stats, min_index-1, min_index)
min_chiSquare = chi_stats['chi_square'].min()
group_cnt = len(chi_stats)
chiMerge_result = chi_stats
split_list.extend(chiMerge_result[feature].tolist())
return chiMerge_result, split_list
def get_chiSquare_distuibution(dfree=4, cf=0.1):
'''
根据自由度和置信度得到卡方分布和阈值
params:
dfree: 自由度, 最大分箱数-1, default 4
cf: 显著性水平, default 10%
return:
卡方阈值
'''
percents = [0.95, 0.90, 0.5, 0.1, 0.05, 0.025, 0.01, 0.005]
df = pd.DataFrame(np.array([chi2.isf(percents, df=i) for i in range(1, 30)]))
df.columns = percents
df.index = df.index+1
# 显示小数点后面数字
pd.set_option('precision', 3)
return df.loc[dfree, cf]
def merge_chiSquare(chi_result, index, mergeIndex, a = 'expected_target_cnt',
b = 'act_target_cnt', c = 'chi_square'):
'''
params:
chi_result: 待合并卡方数据集
index: 合并后的序列号
mergeIndex: 需合并的区间序号
a, b, c: 指定合并字段
return:
分箱合并后的卡方dataframe
'''
chi_result.loc[mergeIndex, a] = chi_result.loc[mergeIndex, a] + chi_result.loc[index, a]
chi_result.loc[mergeIndex, b] = chi_result.loc[mergeIndex, b] + chi_result.loc[index, b]
chi_result.loc[mergeIndex, c] = (chi_result.loc[mergeIndex, b] - chi_result.loc[mergeIndex, a])**2 /chi_result.loc[mergeIndex, a]
chi_result = chi_result.drop([index])
chi_result = chi_result.reset_index(drop=True)
return chi_result
for col in bin_col:
chi_stats = calc_chiSquare(exp_f_data_label_dr, col, 'label')
chiMerge_result, split_list = chiMerge_maxInterval(chi_stats, col, maxInterval=5)
print(col, 'feature maybe split like this:', split_list)
2018.08.02
R语言中有smbining可以进行最优分箱，python中分箱如果既要考虑箱体个数，分箱后信息量大小，也要考虑单调性等其他因素。
这里给出一种简单的通过IV值来选择如果分箱的方法。
下面是按照分位数来分的，还可以按照卡房分箱，决策树分箱等。
参照toad(由厚本金融开发的较标准的评分卡开发开源包)的分箱方式。
R包有 smbinning CRAN - Package smbinning
SAS中 这本 信用风险评分卡研究 (豆瓣) P140 有提及 SAS 实现自动分箱的宏，SAS代码在书本的附录。
还有这里讲解了决策树的三个处理方法，自动分箱的原理基本就是利用决策树做单变量的分支 Decision Tree Algorithms ！
Monotonic Binning with Python
Monotonic binning is a data preparation technique widely used in scorecard development and is usually implemented with SAS. Below is an attempt to do the monotonic binning with python.
Python Code:
# import packages
import pandas as pd
import numpy as np
import scipy.stats.stats as stats
# import data
data = pd.read_csv("/home/liuwensui/Documents/data/accepts.csv", sep = ",", header = 0)
# define a binning function
def mono_bin(Y, X, n = 20):
# fill missings with median
X2 = X.fillna(np.median(X))
r = 0
while np.abs(r) 1:
d1 = pd.DataFrame({"X": X2, "Y": Y, "Bucket": pd.qcut(X2, n)})
d2 = d1.groupby('Bucket', as_index = True)
r, p = stats.spearmanr(d2.mean().X, d2.mean().Y)
n = n - 1
d3 = pd.DataFrame(d2.min().X, columns = ['min_' + X.name])
d3['max_' + X.name] = d2.max().X
d3[Y.name] = d2.sum().Y
d3['total'] = d2.count().Y
d3[Y.name + '_rate'] = d2.mean().Y
d4 = (d3.sort_index(by = 'min_' + X.name)).reset_index(drop = True)
print "=" * 60
print d4
mono_bin(data.bad, data.ltv)
mono_bin(data.bad, data.bureau_score)
mono_bin(data.bad, data.age_oldest_tr)
mono_bin(data.bad, data.tot_tr)
mono_bin(data.bad, data.tot_income)
Output:
============================================================
min_ltv max_ltv bad total bad_rate
0 0 83 88 884 0.099548
1 84 92 137 905 0.151381
2 93 98 175 851 0.205640
3 99 102 173 814 0.212531
4 103 108 194 821 0.236297
5 109 116 194 769 0.252276
6 117 176 235 793 0.296343
============================================================
min_bureau_score max_bureau_score bad total bad_rate
0 443 630 325 747 0.435074
1 631 655 242 721 0.335645
2 656 676 173 721 0.239945
3 677 698 245 1059 0.231350
4 699 709 64 427 0.149883
5 710 732 73 712 0.102528
6 733 763 53 731 0.072503
7 764 848 21 719 0.029207
============================================================
min_age_oldest_tr max_age_oldest_tr bad total bad_rate
0 1 59 319 987 0.323202
1 60 108 235 975 0.241026
2 109 142 282 1199 0.235196
3 143 171 142 730 0.194521
4 172 250 125 976 0.128074
5 251 588 93 970 0.095876
============================================================
min_tot_tr max_tot_tr bad total bad_rate
0 0 8 378 1351 0.279793
1 9 13 247 1025 0.240976
2 14 18 240 1185 0.202532
3 19 25 165 1126 0.146536
4 26 77 166 1150 0.144348
============================================================
min_tot_income max_tot_income bad total bad_rate
0 0.00 2000.00 323 1217 0.265407
1 2002.00 2916.67 259 1153 0.224631
2 2919.00 4000.00 226 1150 0.196522
3 4001.00 5833.33 231 1186 0.194772
4 5833.34 8147166.66 157 1131 0.138815
list =[None,None,None,None,"a","b","c",None,"d",12,None,2,4,5,4] list = list[4:] len(list)11 list['a', 'b', 'c', None, 'd', 12, None, 2, 4, 5, 4]#如果你的list 格式是相同的 比如前面4个都是None，这个格式是固定的，那么切片很容易解决
