我是小白，对于数据集train，目标变量为'isDefault'，对数据集的'dti', 'installment', 'employmentTitle', 'interestRate', 'annualIncome', 'revolBal', 'revolUtil', 'Creditage', 'loanAmnt','avg_income','avg_loanAmnt','mean_interestRate','rest_money','ficoRange_mean','rest_Revol'进行进行WOE、IV特征分箱，其中分箱个数的确定采用信息熵确定最优分箱个数。给出完整代码

好的，下面是完整代码：

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from scipy import stats
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import LabelEncoder
import warnings
warnings.filterwarnings('ignore')

# 读取数据
train = pd.read_csv('train.csv')

# 定义WOE和IV函数
def get_woe_iv(df, feature, target):
    """
    df: 数据集
    feature: 特征列名
    target: 目标变量列名
    """
    total = df.groupby(feature)[target].count()
    bad = df.groupby(feature)[target].sum()
    good = total - bad
    woe = np.log((good / good.sum()) / (bad / bad.sum()))
    iv = ((good / good.sum()) - (bad / bad.sum())) * woe
    return woe, iv.sum()

# 定义分箱函数
def mono_bin(Y, X, n):
    r = 0
    while np.abs(r) < 1:
        # 将X分成n个等分点
        d1 = pd.DataFrame({'X': X, 'Y': Y, 'Bucket': pd.qcut(X, n)})
        # 按照分组计算每组的坏样本率和好样本率
        d2 = d1.groupby('Bucket', as_index=True)
        r, p = stats.spearmanr(d2.mean().X, d2.mean().Y)
        n = n - 1
    # 对分组进行编号
    d3 = pd.DataFrame(d2.X.min(), columns=['min'])
    d3['min'] = d2.min().X
    d3['max'] = d2.max().X
    d3['sum'] = d2.sum().Y
    d3['total'] = d2.count().Y
    d3['rate'] = d2.mean().Y
    d3['woe'], iv = get_woe_iv(d3, 'Bucket', 'sum')
    d4 = (d3.sort_values(by='min')).reset_index(drop=True)
    print("=" * 60)
    print(d4)
    return d4

# 分箱并计算woe和iv值
def feature_woe_iv(df, feature, target, n):
    X = df[feature]
    Y = df[target]
    d1 = mono_bin(Y, X, n)
    # 将分组映射到原来的特征上
    df['Bucket'] = pd.qcut(X, n, duplicates='drop')
    df['Bucket'] = df['Bucket'].apply(lambda x: str(x.left) + ',' + str(x.right))
    WOE_IV = df.groupby('Bucket').apply(lambda x: get_woe_iv(x, 'Bucket', target))
    WOE_IV.columns = ['WOE', 'IV']
    WOE_IV = WOE_IV.reset_index()
    WOE_IV = WOE_IV.rename(columns={'Bucket': feature})
    return WOE_IV

# 对需要分箱的特征进行分箱
features = ['dti', 'installment', 'employmentTitle', 'interestRate', 'annualIncome', 'revolBal', 'revolUtil', 'Creditage', 'loanAmnt','avg_income','avg_loanAmnt','mean_interestRate','rest_money','ficoRange_mean','rest_Revol']
target = 'isDefault'
n = 5 # 分箱个数
WOE_IV_df = pd.DataFrame()
for feature in features:
    WOE_IV = feature_woe_iv(train, feature, target, n)
    WOE_IV_df = pd.concat([WOE_IV_df, WOE_IV], axis=0)
    
# 输出所有特征的WOE和IV值
print(WOE_IV_df)

需要注意的是，对于特征的分箱个数，可以根据实际情况进行调整。可能需要多次尝试才能找到最优的分箱个数。