信用评分模型开发-基于R语言（9）
2019-05-06 本文已影响37人
Liam_ml
scorecard 信用评分包
scorecard包是在R中提供了一个完整的信用评分模型开发的解决方案。本节会对这一部分内容做一个详细的讲解
首先是现在安装:
install.packages(scorecard)
require(scorecard)
split_df 划分数据集
这个函数是用于划分数据集，使用方法如下:
split_df(dt, y = NULL, ratio = 0.7, seed = 618)
- y表示样本的标签，或者说是因变量
- ratio 代表的是训练集合与测试集合的比例
data(germancredit)
# Example I
dt_list = split_df(germancredit, y="creditability")
train = dt_list[[1]]
test = dt_list[[2]]
dim(germancredit)
## [1] 1000 21
dim(train)
## [1] 681 21
dim(test)
## [1] 319 21
IV 计算信息值
使用这个函数来计算特征的IV值，用于特征选择作为参考，使用方法如下：
iv(dt, y, x = NULL, positive = "bad|1", order = TRUE)
x表示因变量，默认表示计算所有的自变量的IV
order 表示根据IV进行排序
data(germancredit)
# information values
info_value = iv(germancredit, y = "creditability")
info_value
## variable info_value
## 1: status.of.existing.checking.account 6.660115e-01
## 2: duration.in.month 3.345035e-01
## 3: credit.history 2.932335e-01
## 4: age.in.years 2.596514e-01
## 5: savings.account.and.bonds 1.960096e-01
## 6: purpose 1.691951e-01
## 7: property 1.126383e-01
## 8: present.employment.since 8.643363e-02
## 9: housing 8.329343e-02
## 10: other.installment.plans 5.761454e-02
## 11: foreign.worker 4.387741e-02
## 12: personal.status.and.sex 4.268938e-02
## 13: credit.amount 3.895727e-02
## 14: other.debtors.or.guarantors 3.201932e-02
## 15: installment.rate.in.percentage.of.disposable.income 2.632209e-02
## 16: number.of.existing.credits.at.this.bank 1.326652e-02
## 17: job 8.762766e-03
## 18: telephone 6.377605e-03
## 19: present.residence.since 3.588773e-03
## 20: number.of.people.being.liable.to.provide.maintenance.for 4.339223e-05
var_filter 筛选变量
通过设定标准，使用这个函数可以通过特定的标准，信息值，缺失率，筛选特征，使用方法如下:
var_filter(dt, y, x = NULL, iv_limit = 0.02, missing_limit = 0.95,
identical_limit = 0.95, var_rm = NULL, var_kp = NULL,
return_rm_reason = FALSE, positive = "bad|1")
- iv_limit 表示信息值超过多少，才保留此特征，默认是大于0.02
- missing_limit 表示保留某个缺失率以下的特征，默认是0.95
- identical_limit 表示，如果某个特征的值一样的比例小于某个比例，则保留，默认是0.95
data(germancredit)
# variable filter
dt_sel = var_filter(germancredit, y = "creditability")
names(dt_sel)
## [1] "status.of.existing.checking.account"
## [2] "duration.in.month"
## [3] "credit.history"
## [4] "purpose"
## [5] "credit.amount"
## [6] "savings.account.and.bonds"
## [7] "present.employment.since"
## [8] "installment.rate.in.percentage.of.disposable.income"
## [9] "personal.status.and.sex"
## [10] "other.debtors.or.guarantors"
## [11] "property"
## [12] "age.in.years"
## [13] "other.installment.plans"
## [14] "housing"
## [15] "creditability"
woebin 进行WOE变换
使用这个函数进行WOE进行连续变量WOE分箱，使用方法如下：
woebin(dt, y, x = NULL, var_skip = NULL, breaks_list = NULL,
special_values = NULL, stop_limit = 0.1, count_distr_limit = 0.05,
bin_num_limit = 8, positive = "bad|1", no_cores = NULL,
print_step = 0L, method = "tree", save_breaks_list = NULL,
ignore_const_cols = TRUE, ignore_datetime_cols = TRUE,
check_cate_num = TRUE, replace_blank_na = TRUE, ...)
- method 是使用分箱的方法，默认是使用决策树
- breaks list 可以制定自己的分箱规则
- stop_limit 如果使用树方法，当信息值增益比小于stop_limit时停止分箱分段; 如果使用chimerge方法，当最小卡方大于’qchisq（1-stoplimit，1）’时停止合并。 可接受的范围：0-0.5; 默认值为0.1。
bins2_tree = woebin(germancredit, y="creditability", method="tree")
bins2_tree$status.of.existing.checking.account
## variable
## 1: status.of.existing.checking.account
## 2: status.of.existing.checking.account
## 3: status.of.existing.checking.account
## bin count
## 1: ... < 0 DM%,%0 <= ... < 200 DM 543
## 2: ... >= 200 DM / salary assignments for at least 1 year 63
## 3: no checking account 394
## count_distr good bad badprob woe bin_iv total_iv
## 1: 0.543 303 240 0.4419890 0.6142040 0.225500603 0.639372
## 2: 0.063 49 14 0.2222222 -0.4054651 0.009460853 0.639372
## 3: 0.394 348 46 0.1167513 -1.1762632 0.404410499 0.639372
## breaks
## 1: ... < 0 DM%,%0 <= ... < 200 DM
## 2: ... >= 200 DM / salary assignments for at least 1 year
## 3: no checking account
## is_special_values
## 1: FALSE
## 2: FALSE
## 3: FALSE
woebin_ply 将原始数据转换成为WOE数据
WOE分箱的具体划分规则指定好了，使用woebin_ply将原始数据转化成为WOE数据，使用方法如下:
woebin_ply(dt, bins, no_cores = NULL, print_step = 0L,
replace_blank_na = TRUE, ...)
- dt 是原始数据
- bins 是woebin 的返回结果
dt_woe = woebin_ply(germancredit, bins=bins2_tree)
head(dt_woe)
## creditability status.of.existing.checking.account_woe
## 1: good 0.614204
## 2: bad 0.614204
## 3: good -1.176263
## 4: good 0.614204
## 5: bad 0.614204
## 6: good -1.176263
## duration.in.month_woe credit.history_woe purpose_woe credit.amount_woe
## 1: -1.3121864 -0.73374058 -0.4100628 0.03366128
## 2: 1.1349799 0.08831862 -0.4100628 0.39053946
## 3: -0.3466246 -0.73374058 0.2799201 -0.25830746
## 4: 0.5245245 0.08831862 0.2799201 0.39053946
## 5: 0.1086883 0.08515781 0.2799201 0.39053946
## 6: 0.5245245 0.08831862 0.2799201 0.39053946
## savings.account.and.bonds_woe present.employment.since_woe
## 1: -0.7621401 -0.23556607
## 2: 0.2713578 0.03210325
## 3: 0.2713578 -0.39441527
## 4: 0.2713578 -0.39441527
## 5: 0.2713578 0.03210325
## 6: -0.7621401 0.03210325
## installment.rate.in.percentage.of.disposable.income_woe
## 1: 0.1039609
## 2: -0.1554665
## 3: -0.1554665
## 4: -0.1554665
## 5: 0.1039609
## 6: -0.1554665
## personal.status.and.sex_woe other.debtors.or.guarantors_woe
## 1: -0.1655476 0.02797385
## 2: 0.2646926 0.02797385
## 3: -0.1655476 0.02797385
## 4: -0.1655476 -0.58778666
## 5: -0.1655476 0.02797385
## 6: -0.1655476 0.02797385
## present.residence.since_woe property_woe age.in.years_woe
## 1: -0.01359409 -0.46103496 -0.2123715
## 2: 0.07015071 -0.46103496 0.5288441
## 3: -0.01359409 -0.46103496 -0.2123715
## 4: -0.01359409 0.02857337 -0.2123715
## 5: -0.01359409 0.58608236 -0.2123715
## 6: -0.01359409 0.58608236 -0.8724881
## other.installment.plans_woe housing_woe
## 1: -0.1211786 -0.1941560
## 2: -0.1211786 -0.1941560
## 3: -0.1211786 -0.1941560
## 4: -0.1211786 0.4726044
## 5: -0.1211786 0.4726044
## 6: -0.1211786 0.4726044
## number.of.existing.credits.at.this.bank_woe job_woe
## 1: -0.1347806 -0.02278003
## 2: 0.0748775 -0.02278003
## 3: 0.0748775 -0.07847162
## 4: 0.0748775 -0.02278003
## 5: -0.1347806 -0.02278003
## 6: 0.0748775 -0.07847162
## number.of.people.being.liable.to.provide.maintenance.for_woe
## 1: 0
## 2: 0
## 3: 0
## 4: 0
## 5: 0
## 6: 0
## telephone_woe foreign.worker_woe
## 1: -0.09863759 0
## 2: 0.06469132 0
## 3: 0.06469132 0
## 4: 0.06469132 0
## 5: 0.06469132 0
## 6: -0.09863759 0
scorecard 构建评分卡
使用scorecard通过模型和woebin的结果构建出评分卡规则，使用方法如下：
scorecard(bins, model, points0 = 600, odds0 = 1/19, pdo = 50,
basepoints_eq0 = FALSE)
- bins 是woebin的返回结果
- model 是构建好的逻辑回归模型
- odds 见第七章
- pdo 见第七章
dt_woe$creditability <- as.character(dt_woe$creditability)
dt_woe$creditability[as.character(dt_woe$creditability)=='good']=0
dt_woe$creditability[as.character(dt_woe$creditability)=='bad']=1
dt_woe$creditability <- as.factor(dt_woe$creditability)
l <- glm(creditability~.,data = dt_woe,family = binomial())
l <- step(l)
score <- scorecard(bins = bins2_tree,model = l)
score$status.of.existing.checking.account
## variable
## 1: status.of.existing.checking.account
## 2: status.of.existing.checking.account
## 3: status.of.existing.checking.account
## bin count
## 1: ... < 0 DM%,%0 <= ... < 200 DM 543
## 2: ... >= 200 DM / salary assignments for at least 1 year 63
## 3: no checking account 394
## count_distr good bad badprob woe bin_iv total_iv
## 1: 0.543 303 240 0.4419890 0.6142040 0.225500603 0.639372
## 2: 0.063 49 14 0.2222222 -0.4054651 0.009460853 0.639372
## 3: 0.394 348 46 0.1167513 -1.1762632 0.404410499 0.639372
## breaks
## 1: ... < 0 DM%,%0 <= ... < 200 DM
## 2: ... >= 200 DM / salary assignments for at least 1 year
## 3: no checking account
## is_special_values points
## 1: FALSE -36
## 2: FALSE 24
## 3: FALSE 68
scorecard_ply
将一个新用户的原始数据获取这个用户的分数，使用方法如下：
scorecard_ply(dt, card, only_total_score = TRUE, print_step = 0L,
replace_blank_na = TRUE, var_kp = NULL)
- dt 训练模型的原始数据集
- 使用scorecard 建立起来的评分卡规则
resutl <- scorecard_ply(dt = germancredit,card = score)
resutl
## score
## 1: 648
## 2: 314
## 3: 638
## 4: 439
## 5: 310
## ---
## 996: 535
## 997: 462
## 998: 559
## 999: 342
## 1000: 402
