贷款利润最大化

数据下载地址 https://www.lendingclub.com/info/download-data.action

import pandas as pd
loans_2007 = pd.read_csv('LoanStats3a.csv', skiprows=1)
half_count = len(loans_2007) / 2
# 去掉没有用的数据
loans_2007 = loans_2007.dropna(thresh=half_count, axis=1)
loans_2007 = loans_2007.drop(['desc', 'url'],axis=1)
loans_2007.to_csv('loans_2007.csv', index=False)

import pandas as pd
loans_2007 = pd.read_csv("loans_2007.csv")
#loans_2007.drop_duplicates() 打印第一行数据和第一列
print(loans_2007.iloc[0])
print(loans_2007.shape[1])

数据预处理 –去掉没用的特征

loans_2007 = loans_2007.drop(["id", "member_id", "funded_amnt", "funded_amnt_inv", "grade", "sub_grade", "emp_title", "issue_d"], axis=1)

loans_2007 = loans_2007.drop(["zip_code", "out_prncp", "out_prncp_inv", "total_pymnt", "total_pymnt_inv", "total_rec_prncp"], axis=1)

loans_2007 = loans_2007.drop(["total_rec_int", "total_rec_late_fee", "recoveries", "collection_recovery_fee", "last_pymnt_d", "last_pymnt_amnt"], axis=1)
print(loans_2007.iloc[0])
print(loans_2007.shape[1])

统计loan_status取值的次数

print(loans_2007['loan_status'].value_counts())

数据预处理 – 拿到最多的两个值作为分类的值–这里做二分类

loans_2007 = loans_2007[(loans_2007['loan_status'] == "Fully Paid") | (loans_2007['loan_status'] == "Charged Off")]

status_replace = {
    "loan_status" : {
        "Fully Paid": 1,
        "Charged Off": 0,
    }
}
# pandas来做label值
loans_2007 = loans_2007.replace(status_replace)

#let's look for any columns that contain only one unique value and remove them
# 拿到所有的列进行清洗工作
orig_columns = loans_2007.columns
drop_columns = []
for col in orig_columns:
# 去掉缺失值，再去计算唯一属性有几个
    col_series = loans_2007[col].dropna().unique()
#如果列的值只有一个 则去掉
    if len(col_series) == 1:
        drop_columns.append(col)
loans_2007 = loans_2007.drop(drop_columns, axis=1)
print(drop_columns)
print loans_2007.shape
loans_2007.to_csv('filtered_loans_2007.csv', index=False)

只有24列了：
[‘initial_list_status’, ‘collections_12_mths_ex_med’, ‘policy_code’, ‘application_type’, ‘acc_now_delinq’, ‘chargeoff_within_12_mths’, ‘delinq_amnt’, ‘tax_liens’]
(39560, 24)
现在数据集里还包含了很多的缺失值、字符、标点符号之类的

import pandas as pd
loans = pd.read_csv('filtered_loans_2007.csv')
# 。isnull是否有缺失值 sum计算一共有多少个缺失值
null_counts = loans.isnull().sum()
print(null_counts)

缺失值较少的就直接去掉对应的样本

统计各类型的特征个数

loans = loans.drop("pub_rec_bankruptcies", axis=1)
loans = loans.dropna(axis=0)
print(loans.dtypes.value_counts())

统计各个object类的数据到底是什么情况，以便进行数据清洗

object_columns_df = loans.select_dtypes(include=["object"])
print(object_columns_df.iloc[0])

cols = ['home_ownership', 'verification_status', 'emp_length', 'term', 'addr_state']
for c in cols:
    print(loans[c].value_counts())

print(loans["purpose"].value_counts())
print(loans["title"].value_counts())

将数据转换为sklearn能处理的数据

mapping_dict = {
    "emp_length": {
        "10+ years": 10,
        "9 years": 9,
        "8 years": 8,
        "7 years": 7,
        "6 years": 6,
        "5 years": 5,
        "4 years": 4,
        "3 years": 3,
        "2 years": 2,
        "1 year": 1,
        "< 1 year": 0,
        "n/a": 0
    }
}
loans = loans.drop(["last_credit_pull_d", "earliest_cr_line", "addr_state", "title"], axis=1)
loans["int_rate"] = loans["int_rate"].str.rstrip("%").astype("float")
loans["revol_util"] = loans["revol_util"].str.rstrip("%").astype("float")
loans = loans.replace(mapping_dict)

由于关系到利润最大化问题，不能再单纯的考虑精度和召回率了，比如FP和FN对公司的损失不是一样大的

cat_columns = ["home_ownership", "verification_status", "emp_length", "purpose", "term"]
dummy_df = pd.get_dummies(loans[cat_columns])
loans = pd.concat([loans, dummy_df], axis=1)
loans = loans.drop(cat_columns, axis=1)
loans = loans.drop("pymnt_plan", axis=1)

loans.to_csv('cleaned_loans2007.csv', index=False)

import pandas as pd
loans = pd.read_csv("cleaned_loans2007.csv")
print(loans.info())

import pandas as pd
# False positives.
fp_filter = (predictions == 1) & (loans["loan_status"] == 0)
fp = len(predictions[fp_filter])

# True positives.
tp_filter = (predictions == 1) & (loans["loan_status"] == 1)
tp = len(predictions[tp_filter])

# False negatives.
fn_filter = (predictions == 0) & (loans["loan_status"] == 1)
fn = len(predictions[fn_filter])

# True negatives
tn_filter = (predictions == 0) & (loans["loan_status"] == 0)
tn = len(predictions[tn_filter])

先用逻辑回归来预测利润最大化问题的表现情况（二分类问题）

from sklearn.linear_model import LogisticRegression
lr = LogisticRegression()
cols = loans.columns
train_cols = cols.drop("loan_status")
features = loans[train_cols]
target = loans["loan_status"]
lr.fit(features, target)
predictions = lr.predict(features)

from sklearn.linear_model import LogisticRegression
from sklearn.cross_validation import cross_val_predict, KFold
lr = LogisticRegression()
kf = KFold(features.shape[0], random_state=1)
predictions = cross_val_predict(lr, features, target, cv=kf)
predictions = pd.Series(predictions)

# False positives.
fp_filter = (predictions == 1) & (loans["loan_status"] == 0)
fp = len(predictions[fp_filter])

# True positives.
tp_filter = (predictions == 1) & (loans["loan_status"] == 1)
tp = len(predictions[tp_filter])

# False negatives.
fn_filter = (predictions == 0) & (loans["loan_status"] == 1)
fn = len(predictions[fn_filter])

# True negatives
tn_filter = (predictions == 0) & (loans["loan_status"] == 0)
tn = len(predictions[tn_filter])

# Rates
tpr = tp / float((tp + fn))
fpr = fp / float((fp + tn))

print(tpr)
print(fpr)
print predictions[:20]

打印预测结果发现，全都是1，全借，证明这是一个废模型–原因是样本不均衡，因为大部分都信用良好
增加权重项–负样本权重加大，正样本权重降低

from sklearn.linear_model import LogisticRegression
from sklearn.cross_validation import cross_val_predict
# 使用sklearn的样本均衡策略  更改正负样本权重
lr = LogisticRegression(class_weight="balanced")
kf = KFold(features.shape[0], random_state=1)
predictions = cross_val_predict(lr, features, target, cv=kf)
predictions = pd.Series(predictions)

# False positives.
fp_filter = (predictions == 1) & (loans["loan_status"] == 0)
fp = len(predictions[fp_filter])

# True positives.
tp_filter = (predictions == 1) & (loans["loan_status"] == 1)
tp = len(predictions[tp_filter])

# False negatives.
fn_filter = (predictions == 0) & (loans["loan_status"] == 1)
fn = len(predictions[fn_filter])

# True negatives
tn_filter = (predictions == 0) & (loans["loan_status"] == 0)
tn = len(predictions[tn_filter])

# Rates
tpr = tp / float((tp + fn))
fpr = fp / float((fp + tn))

print(tpr)
print(fpr)
print predictions[:20]

这也不是我们业务所需要的结果，所以需要继续爬坑，
目标是TP越大越好FP越小越好

from sklearn.linear_model import LogisticRegression
from sklearn.cross_validation import cross_val_predict
# 自定义权重项  不使用sklearn的权重项
penalty = {
    0: 5,
    1: 1
}

lr = LogisticRegression(class_weight=penalty)
kf = KFold(features.shape[0], random_state=1)
predictions = cross_val_predict(lr, features, target, cv=kf)
predictions = pd.Series(predictions)

# False positives.
fp_filter = (predictions == 1) & (loans["loan_status"] == 0)
fp = len(predictions[fp_filter])

# True positives.
tp_filter = (predictions == 1) & (loans["loan_status"] == 1)
tp = len(predictions[tp_filter])

# False negatives.
fn_filter = (predictions == 0) & (loans["loan_status"] == 1)
fn = len(predictions[fn_filter])

# True negatives
tn_filter = (predictions == 0) & (loans["loan_status"] == 0)
tn = len(predictions[tn_filter])

# Rates
tpr = tp / float((tp + fn))
fpr = fp / float((fp + tn))

print(tpr)
print(fpr)

输出 ：
0.731799521545
0.478985635751

from sklearn.ensemble import RandomForestClassifier
from sklearn.cross_validation import cross_val_predict
rf = RandomForestClassifier(n_estimators=10,class_weight="balanced", random_state=1)
#print help(RandomForestClassifier)
kf = KFold(features.shape[0], random_state=1)
predictions = cross_val_predict(rf, features, target, cv=kf)
predictions = pd.Series(predictions)

# False positives.
fp_filter = (predictions == 1) & (loans["loan_status"] == 0)
fp = len(predictions[fp_filter])

# True positives.
tp_filter = (predictions == 1) & (loans["loan_status"] == 1)
tp = len(predictions[tp_filter])

# False negatives.
fn_filter = (predictions == 0) & (loans["loan_status"] == 1)
fn = len(predictions[fn_filter])

# True negatives
tn_filter = (predictions == 0) & (loans["loan_status"] == 0)
tn = len(predictions[tn_filter])

# Rates
tpr = tp / float((tp + fn))
fpr = fp / float((fp + tn))