论文：

    题目：Personal Information in Passwords and Its Security Implications

    链接：https://ieeexplore.ieee.org/abstract/document/7931642

实验重现：

# encoding: utf-8 # author: kaiyouhu import pandas as pd import numpy as np import re class Passenger: # passenger_list = [] def __init__(self, login_email, password, name, id_number, username, phone, email): self.login_email = login_email self.password = password self.name = name self.id_number = id_number self.username = username self.phone = phone self.email = email # self.passenger_list = [] # def add(self, passenger): # self.passenger_list.append(passenger) def read_data(path): with open(path) as f: data = f.read() informations = data.splitlines() for index in range(len(informations)): informations[index] = informations[index].split('----') # print sum of information informations_sum = len(informations) print('sum: ' + str(informations_sum)) table1_header = ['RANK', 'Password', 'Amount', 'Percentage'] rank_list = list(range(1, 11)) password_list = ['123456', 'a123456', '123456a', '5201314', '111111', 'woaini1314', 'qq123456', '123123', '000000', '1qaz2wsx'] amount_list1 = [] percentage_list1 = [] for index, password in enumerate(password_list): count = 0 for information in informations: if information[1] == password: count += 1 amount_list1.append(count) for amount in amount_list1: percentage_list1.append(float(amount/informations_sum)) result = np.array(list(zip(rank_list, password_list, amount_list1, percentage_list1)), order='C') df1 = pd.DataFrame(result, columns=table1_header) print(df1) structure_list = ['D7', 'D8', 'D6', 'L2D7', 'L3D6', 'L1D7', 'L2D6', 'L3D7', 'D9', 'L2D8'] structure_list_regex = ['^\d{7}$', '^\d{8}$', '^\d{6}$', '^[a-zA-Z]{2}\d{7}$', '^[a-zA-Z]{3}\d{6}$', '^[a-zA-Z]{1}\d{7}$', '^[a-zA-Z]{2}\d{6}$', '^[a-zA-Z]{3}\d{7}$', '^\d{9}$', '^[a-zA-Z]{2}\d{8}$'] amount_list2 = [] percentage_list2 = [] for password_index, structure_regex in enumerate(structure_list_regex): count = 0 for index, information in enumerate(informations): if re.match(structure_regex, str(information[1])): count += 1 amount_list2.append(count) for amount in amount_list2: percentage_list2.append(float(amount / informations_sum)) result2 = np.array(list(zip(rank_list, structure_list, amount_list2, percentage_list2)), order='C') df2 = pd.DataFrame(result2, columns=table1_header) print(df2) rank_list = list(range(1, 7)) information_type_list = ['Birthdate', 'AccountName', 'Name', 'Email', 'IDNumber', 'CellPhone'] amount_list2 = [] percentage_list2 = [] for index, information_type in enumerate(information_type_list): count = 0 for information in informations: if information_type == 'Birthdate': if information[1].find(information[3][6:14]) != -1: count += 1 elif information_type == 'AccountName': if information[1].find(information[4]) != -1: count += 1 elif information_type == 'Name': # are you kidding? if information[1].find(information[4]) != -1: count += 1 elif information_type == 'Email': if information[1].find(information[0].split('@')[0]) != -1: count += 1 elif information_type == 'IDNumber': if information[4].find(information[1]) != -1: count += 1 elif information_type == 'CellPhone': if information[1].find(information[5]) != -1: count += 1 amount_list2.append(count) for amount in amount_list2: percentage_list2.append(float(amount / informations_sum)) result = np.array(list(zip(rank_list, information_type_list, amount_list2, percentage_list2)), order='C') df1 = pd.DataFrame(result, columns=table1_header) print(df1) pass read_data('../data/12306.txt')

输出结果：

sum: 131653 RANK Password Amount Percentage 0 1 123456 392 0.0029775242493524645 1 2 a123456 281 0.0021343987603776593 2 3 123456a 165 0.0012532946457733587 3 4 5201314 161 0.0012229117452697623 4 5 111111 157 0.0011925288447661656 5 6 woaini1314 136 0.0010330186171222835 6 7 qq123456 98 0.0007443810623381161 7 8 123123 98 0.0007443810623381161 8 9 000000 97 0.0007367853372122169 9 10 1qaz2wsx 93 0.0007064024367086204 RANK Password Amount Percentage 0 1 D7 10906 0.08283897822305607 1 2 D8 9458 0.0718403682407541 2 3 D6 9102 0.06913629009593401 3 4 L2D7 5073 0.038533113563686355 4 5 L3D6 4832 0.036702543808344666 5 6 L1D7 4778 0.03629237465154611 6 7 L2D6 4275 0.03247172491321884 7 8 L3D7 3885 0.029509392114118176 8 9 D9 3594 0.027299036102481522 9 10 L2D8 3371 0.025605189399406016 RANK Password Amount Percentage 0 1 Birthdate 5726 0.0434931220708985 1 2 AccountName 2565 0.019483034947931303 2 3 Name 2565 0.019483034947931303 3 4 Email 3979 0.030223390275952694 4 5 IDNumber 6835 0.05191678123552065 5 6 CellPhone 89 0.0006760195362050238 Process finished with exit code 0

备注：

    1实验数据自己去百度网盘找下载的，大概14M的txt，共131653条数据（有的版本可能会上下差一些条数，但基本上差不多）

    2只重现了第三个表格，前面两个基本上数据差不多，第三个，我感觉不太理解，也不知道作者具体怎么实现的（匹配细节不知道），后面的我懒得去编写输出了

substring← get_all_substring(pwd) reverse_length_sort(substring) for eachstring ∈ substring do if len(eachstring) ≥ 2 then if matchbd(eachstring,infolist) then

这里按照作者代码的理解思路是，获取密码的全部长度大于等于2的子串，然后去和身份证信息，电话号码等匹配，我就呵呵了

    3论文的内容看看就好了，个人感觉过程有点水，结论得到地太草率（本人愚钝之见，不要太在意）