内容简介：版权声明：本套技术专栏是作者（秦凯新）平时工作的总结和升华，通过从真实商业环境抽取案例进行总结和分享，并给出商业应用的调优建议和集群环境容量规划等内容，请持续关注本套博客。QQ邮箱地址：1120746959@qq.com，如有任何学术交流，可随时联系。
版权声明：本套技术专栏是作者（秦凯新）平时工作的总结和升华，通过从真实商业环境抽取案例进行总结和分享，并给出商业应用的调优建议和集群环境容量规划等内容，请持续关注本套博客。QQ邮箱地址：1120746959@qq.com，如有任何学术交流，可随时联系。
1 机器学习调优步骤（第一行不平衡问题处理）
2 Pandas多维特征数据预处理
-
数据初始化展示
import pandas as pd 取出第一行 loans_2007 = pd.read_csv('C:\\ML\\MLData\\filtered_loans_2007.csv', skiprows=1) print(len(loans_2007)) half_count = len(loans_2007) / 2 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) loans_2007.head(3) 复制代码
-
显示第0行数据
import pandas as pd loans_2007 = pd.read_csv("loans_2007.csv") #loans_2007.drop_duplicates() print(loans_2007.iloc[0]) print(loans_2007.shape[1]) id 1077501 member_id 1.2966e+06 loan_amnt 5000 funded_amnt 5000 funded_amnt_inv 4975 term 36 months int_rate 10.65% installment 162.87 grade B sub_grade B2 emp_title NaN emp_length 10+ years home_ownership RENT annual_inc 24000 verification_status Verified issue_d Dec-2011 loan_status Fully Paid pymnt_plan n purpose credit_card title Computer zip_code 860xx addr_state AZ dti 27.65 delinq_2yrs 0 earliest_cr_line Jan-1985 inq_last_6mths 1 open_acc 3 pub_rec 0 revol_bal 13648 revol_util 83.7% total_acc 9 initial_list_status f out_prncp 0 out_prncp_inv 0 total_pymnt 5863.16 total_pymnt_inv 5833.84 total_rec_prncp 5000 total_rec_int 863.16 total_rec_late_fee 0 recoveries 0 collection_recovery_fee 0 last_pymnt_d Jan-2015 last_pymnt_amnt 171.62 last_credit_pull_d Nov-2016 collections_12_mths_ex_med 0 policy_code 1 application_type INDIVIDUAL acc_now_delinq 0 chargeoff_within_12_mths 0 delinq_amnt 0 pub_rec_bankruptcies 0 tax_liens 0 Name: 0, dtype: object 52 复制代码 -
删除无意义列
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]) 复制代码
-
查看预测值的状态类型
print(loans_2007['loan_status'].value_counts()) Fully Paid 33902 Charged Off 5658 Does not meet the credit policy. Status:Fully Paid 1988 Does not meet the credit policy. Status:Charged Off 761 Current 201 Late (31-120 days) 10 In Grace Period 9 Late (16-30 days) 5 Default 1 Name: loan_status, dtype: int64 复制代码
-
根据贷款状态，舍弃部分不清晰结论，给出明确分类0和1，进行替换
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, } } loans_2007 = loans_2007.replace(status_replace) 复制代码 -
去除每一列值都相同的列
#let's look for any columns that contain only one unique value and remove them orig_columns = loans_2007.columns drop_columns = [] for col in orig_columns: col_series = loans_2007[col].dropna().unique() 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) ['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) 复制代码 -
空值处理
import pandas as pd loans = pd.read_csv('filtered_loans_2007.csv') null_counts = loans.isnull().sum() print(null_counts) loan_amnt 0 term 0 int_rate 0 installment 0 emp_length 0 home_ownership 0 annual_inc 0 verification_status 0 loan_status 0 pymnt_plan 0 purpose 0 title 10 addr_state 0 dti 0 delinq_2yrs 0 earliest_cr_line 0 inq_last_6mths 0 open_acc 0 pub_rec 0 revol_bal 0 revol_util 50 total_acc 0 last_credit_pull_d 2 pub_rec_bankruptcies 697 dtype: int64 loans = loans.drop("pub_rec_bankruptcies", axis=1) loans = loans.dropna(axis=0) 复制代码 -
String类型分布
print(loans.dtypes.value_counts()) object 12 float64 10 int64 1 dtype: int64 object_columns_df = loans.select_dtypes(include=["object"]) print(object_columns_df.iloc[0]) term 36 months int_rate 10.65% emp_length 10+ years home_ownership RENT verification_status Verified pymnt_plan n purpose credit_card title Computer addr_state AZ earliest_cr_line Jan-1985 revol_util 83.7% last_credit_pull_d Nov-2016 Name: 0, dtype: object cols = ['home_ownership', 'verification_status', 'emp_length', 'term', 'addr_state'] for c in cols: print(loans[c].value_counts()) RENT 18780 MORTGAGE 17574 OWN 3045 OTHER 96 NONE 3 Name: home_ownership, dtype: int64 Not Verified 16856 Verified 12705 Source Verified 9937 Name: verification_status, dtype: int64 10+ years 8821 < 1 year 4563 2 years 4371 3 years 4074 4 years 3409 5 years 3270 1 year 3227 6 years 2212 7 years 1756 8 years 1472 9 years 1254 n/a 1069 Name: emp_length, dtype: int64 36 months 29041 60 months 10457 Name: term, dtype: int64 CA 7070 NY 3788 FL 2856 TX 2714 NJ 1838 IL 1517 PA 1504 VA 1400 GA 1393 MA 1336 OH 1208 MD 1049 AZ 874 WA 834 CO 786 NC 780 CT 747 MI 722 MO 682 MN 611 NV 492 SC 470 WI 453 AL 446 OR 445 LA 435 KY 325 OK 298 KS 269 UT 256 AR 243 DC 211 RI 198 NM 188 WV 176 HI 172 NH 172 DE 113 MT 84 WY 83 AK 79 SD 63 VT 54 MS 19 TN 17 IN 9 ID 6 IA 5 NE 5 ME 3 Name: addr_state, dtype: int64 复制代码 -
String类型分布2
print(loans["purpose"].value_counts()) print(loans["title"].value_counts()) debt_consolidation 18533 credit_card 5099 other 3963 home_improvement 2965 major_purchase 2181 small_business 1815 car 1544 wedding 945 medical 692 moving 581 vacation 379 house 378 educational 320 renewable_energy 103 Name: purpose, dtype: int64 Debt Consolidation 2168 Debt Consolidation Loan 1706 Personal Loan 658 Consolidation 509 debt consolidation 502 Credit Card Consolidation 356 Home Improvement 354 Debt consolidation 333 Small Business Loan 322 Credit Card Loan 313 Personal 308 Consolidation Loan 255 Home Improvement Loan 246 personal loan 234 personal 220 Loan 212 Wedding Loan 209 consolidation 200 Car Loan 200 Other Loan 190 Credit Card Payoff 155 Wedding 152 Major Purchase Loan 144 Credit Card Refinance 143 Consolidate 127 Medical 122 Credit Card 117 home improvement 111 My Loan 94 Credit Cards 93 ... DebtConsolidationn 1 Freedom 1 Credit Card Consolidation Loan - SEG 1 SOLAR PV 1 Pay on Credit card 1 To pay off balloon payments due 1 Paying off the debt 1 Payoff ING PLOC 1 Josh CC Loan 1 House payoff 1 Taking care of Business 1 Gluten Free Bakery in ideal town for it 1 Startup Money for Small Business 1 FundToFinanceCar 1 getting ready for Baby 1 Dougs Wedding Loan 1 d rock 1 LC Loan 2 1 swimming pool repair 1 engagement 1 Cut the credit cards Loan 1 vinman 1 working hard to get out of debt 1 consolidate the rest of my debt 1 Medical/Vacation 1 2BDebtFree 1 Paying Off High Interest Credit Cards! 1 Baby on the way! 1 cart loan 1 Consolidaton 1 Name: title, dtype: int64 复制代码 -
类型转换
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) 复制代码 -
独热编码
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) 复制代码 -
查看转换类型
import pandas as pd loans = pd.read_csv("cleaned_loans2007.csv") print(loans.info()) <class 'pandas.core.frame.DataFrame'> RangeIndex: 39498 entries, 0 to 39497 Data columns (total 37 columns): loan_amnt 39498 non-null float64 int_rate 39498 non-null float64 installment 39498 non-null float64 annual_inc 39498 non-null float64 loan_status 39498 non-null int64 dti 39498 non-null float64 delinq_2yrs 39498 non-null float64 inq_last_6mths 39498 non-null float64 open_acc 39498 non-null float64 pub_rec 39498 non-null float64 revol_bal 39498 non-null float64 revol_util 39498 non-null float64 total_acc 39498 non-null float64 home_ownership_MORTGAGE 39498 non-null int64 home_ownership_NONE 39498 non-null int64 home_ownership_OTHER 39498 non-null int64 home_ownership_OWN 39498 non-null int64 home_ownership_RENT 39498 non-null int64 verification_status_Not Verified 39498 non-null int64 verification_status_Source Verified 39498 non-null int64 verification_status_Verified 39498 non-null int64 purpose_car 39498 non-null int64 purpose_credit_card 39498 non-null int64 purpose_debt_consolidation 39498 non-null int64 purpose_educational 39498 non-null int64 purpose_home_improvement 39498 non-null int64 purpose_house 39498 non-null int64 purpose_major_purchase 39498 non-null int64 purpose_medical 39498 non-null int64 purpose_moving 39498 non-null int64 purpose_other 39498 non-null int64 purpose_renewable_energy 39498 non-null int64 purpose_small_business 39498 non-null int64 purpose_vacation 39498 non-null int64 purpose_wedding 39498 non-null int64 term_ 36 months 39498 non-null int64 term_ 60 months 39498 non-null int64 dtypes: float64(12), int64(25) memory usage: 11.1 MB 复制代码
3 准确率理论及不均衡处理
-
初始定义
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]) 复制代码
-
逻辑回归不处理不均衡
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] 0.999084438406 0.998049299521 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 1 11 1 12 1 13 1 14 1 15 1 16 1 17 1 18 1 19 1 dtype: int64 复制代码 -
逻辑回归balanced处理不均衡
from sklearn.linear_model import LogisticRegression from sklearn.cross_validation import cross_val_predict 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] 0.670781771464 0.400780280192 0 1 1 0 2 0 3 1 4 1 5 0 6 0 7 0 8 0 9 0 10 1 11 0 12 1 13 1 14 0 15 0 16 1 17 1 18 1 19 0 dtype: int64 复制代码
-
逻辑回归penalty处理不均衡
from sklearn.linear_model import LogisticRegression from sklearn.cross_validation import cross_val_predict 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) 0.731799521545 0.478985635751 复制代码 -
随机森林balanced处理不均衡
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)) 复制代码
以上就是本文的全部内容，希望本文的内容对大家的学习或者工作能带来一定的帮助，也希望大家多多支持 码农网
猜你喜欢:
- 数据挖掘复习笔记---03.数据预处理
- 自然语言处理之数据预处理
- 数据挖掘的五大流程之数据预处理&特征工程
- PaddlePaddle之数据预处理
- 特征工程之数据预处理（上）
- 特征工程之数据预处理（下）
本站部分资源来源于网络，本站转载出于传递更多信息之目的，版权归原作者或者来源机构所有，如转载稿涉及版权问题，请联系我们。
需求
[美] 亚德里安•斯莱沃斯基（Adrian J. Slywotzky）、[美]卡尔•韦伯 （Karl Weber） / 魏薇、龙志勇 / 浙江人民出版社 / 2013-6 / 64.9
《财富汇•需求:缔造伟大商业传奇的根本力量》内容简介：需求，是缔造伟大商业传奇的根本力量。《财富汇•需求:缔造伟大商业传奇的根本力量》呈现了人们无法拒绝、竞争对手无法复制的需求创造的六大关键，在人们无奈接受的现状和心中真正期待的理想的这道鸿沟之上，架设起了一道桥梁。 创造需求，需要解开一个谜团，这个谜团是人类学、心理学、科技、设计、经济学、基础设施以及其他众多因素综合而成的奇特组合。《财富汇......一起来看看 《需求》 这本书的介绍吧!
