计算多个子字符串一次出现在字符串中的次数

Question

我正在用Python创建一个简单的脚本，该脚本可以评估分数系统上密码的强度，该分数系统根据是否包含学校的大写或小写字母，数字和符号来计算分数。

其中一项要求是，它检查UK QWERTY键盘上从左到右连续的3个字母或数字，并为每个实例取5分。例如，密码“ qwer123”将丢失“ qwe”，“ wer”和“ 123”的15点。如何做到这一点？我当前的代码如下。

def check():
  user_password_score=0
  password_capitals=False
  password_lowers=False
  password_numbers=False
  password_symbols=False
  password_explanation_check=False
  ascii_codes=[]
  password_explanation=[]
  print("The only characters allowed in the passwords are upper and lower case letters, numbers and these symbols; !, $, %, ^, &, *, (, ), _, -, = and +.\n")
  user_password=str(input("Enter the password you would like to get checked: "))
  print("")
  if len(user_password)>24 or len(user_password)<8:
    print("That password is not between 8 and 24 characters and so the Password Checker can't evaluate it.")
    menu()
  for i in user_password:
    ascii_code=ord(i)
    #print(ascii_code)
    ascii_codes.append(ascii_code)
  #print(ascii_codes)
  for i in range(len(ascii_codes)):
    if ascii_codes[i]>64 and ascii_codes[i]<90:
      password_capitals=True
    elif ascii_codes[i]>96 and ascii_codes[i]<123:
      password_lowers=True
    elif ascii_codes[i]>47 and ascii_codes[i]<58:
      password_numbers=True
    elif ascii_codes[i] in (33,36,37,94,38,42,40,41,45,95,61,43):
      password_symbols=True
    else:
      print("Your password contains characters that aren't allowed.\n")
      menu()
  if password_capitals==True:
    user_password_score+=5
  if password_lowers==True:
    user_password_score+=5
  if password_numbers==True:
    user_password_score+=5
  if password_symbols==True:
    user_password_score+=5
  if password_capitals==True and password_lowers==True and password_numbers==True and password_symbols==True:
    user_password_score+=10
  if password_numbers==False and password_symbols==False:
    user_password_score-=5
  if password_capitals==False and password_lowers==False and password_symbols==False:
    user_password_score-=5
  if password_capitals==False and password_lowers==False and password_numbers==False:
    user_password_score-=5
  #print(user_password_score)
  if user_password_score>20:
    print("Your password is strong.\n")
  else:
    print("That password is weak.\n")
  #don't forget you still need to add the thing that checks for 'qwe' and other stuff.
  menu()

Answer 1

您可以将禁止的序列存储在一组字符串中，并在每次有人使用该序列时降低分数。

password = "qwert123"
score = 42          # initial score
sequences = {       # all in lowercase because of the `lower()` in the loop
    "qwertyuiopasdfghjklzxcvbnm",
    "azertyuiopqsdfghjklmwxcvbn",
    "abcdefghijklmnopqrstuvwxyz",
    "01234567890"
}
match_length = 3                        # length threshold for the sanction
sequences.update({s[::-1] for s in sequences})      # do we allow reverse ?

for c in range(len(password)-match_length+1):
    for seq in sequences:
        if password[c:c+match_length].lower() in seq:
            score-=5
            print(f"'{password[c:c+match_length]}' => -5 !")
            break   # Don't flag the same letters more than once

print(score) # 22 (42-4*5)

Answer 2

最简单的方法是对所有可能的序列进行暴力破解。

创建4个字符串："1234567890"，"qwertyuiop"，"asdfghjkl"，"zxcvbnm"，并使用user_password中的3个字符循环遍历每个字符串。

您可以在check函数的开头初始化此列表：

sequences = ["1234567890", "qwertyuiop", "asdfghjkl", "zxcvbnm"]

，然后在for i in range(len(ascii_codes))循环中添加：

if(i<len(ascii_codes)-2):  # since we will be checking for characters up to i+2 in our loop
    flag = False  # initialize a flag to signal finding a match
    for s in sequences:  # loop through each of the 4 keyboard sequences
        if(s.find(user_password[i: i+3].lower()) != -1): 
            user_password_score -= 5
            flag = True
            break 
        if(flag): break

Answer 3

如上所述，我将创建一个相邻键序列的列表。然后，我将创建一个sliding window function来生成所有长度为3的序列，并将每个序列与密码进行匹配：

from itertools import islice

keyboard_rows = ['1234567890', 'qwertyuiop', 'asdfghjkl', 'zxcvbnm']

def window(seq, n=3):
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result
    for elem in it:
        result = result[1:] + (elem,)
        yield result

for row in keyboard_rows:

    for seq in window(row, n=3):
        if "".join(seq) in password:
            user_password_score -= 15

    # scan other direction <--
    for seq in window(row[::-1], n=3):
        if "".join(seq) in password:
            user_password_score -= 15

Answer 4

如果允许regular expressions，则可以一行完成：

import re
user_password_score = 42
pwd = 'qwer123'
user_password_score += (lambda z : -5 * len([match.group(1) for match in re.compile('(?=({0}))'.format('|'.join(["({0})".format(w) for w in [x for y in [[s[i:i+3] for i in range(0,len(s)-2)] for s in ["qwertyuiopasdfghjklzxcvbnm", "azertyuiopqsdfghjklmwxcvbn", "abcdefghijklmnopqrstuvwxyz", "01234567890"]] for x in y]]))).finditer(z) ]))(pwd)

此代码等效：

import re
user_password_score = 42
pwd = 'qwer123'
seqs = ["qwertyuiopasdfghjklzxcvbnm", "azertyuiopqsdfghjklmwxcvbn", "abcdefghijklmnopqrstuvwxyz", "01234567890"]
pattern = re.compile('(?=({0}))'.format('|'.join(["({0})".format(w) for w in [x for y in [[s[i:i+3] for i in range(0,len(s)-2)] for s in seqs] for x in y]])))
penalty = -5 * len([match.group(1) for match in pattern.finditer(pwd) ])
user_password_score += penalty

以下代码也是等效的（希望也是人类可读的）。我们将逐步打印它，以更好地了解它的作用。

import re

def build_pattern(sequences):
    all_triplets = []
    triplets = []
    for seq in sequences:
        for i in range(0, len(seq) - 2):
            triplets.append(seq[i:i+3])
        all_triplets.append(triplets)
        triplets = []
    expanded_triplets = [ x for y in all_triplets for x in y ]
    print("Plain list of triplets: " + str(expanded_triplets))
    string_pattern = '|'.join( [ "({0})".format(x) for x in expanded_triplets ] )
    lookahead_pattern = '(?=({0}))'.format(string_pattern)
    print("Regex expression: " + lookahead_pattern)
    return re.compile(lookahead_pattern)

password = 'qwer123'
user_password_score = 42
print("User password score: " + str(user_password_score))
sequences = ["qwertyuiopasdfghjklzxcvbnm", 
             "azertyuiopqsdfghjklmwxcvbn", 
             "abcdefghijklmnopqrstuvwxyz", 
             "01234567890"]
pattern = build_pattern(sequences)
matches = [ match.group(1) for match in pattern.finditer(password) ]
print("Matches : " + str(matches))
matches_count = len(matches)
penalty = -5 * matches_count
print("Penalty: " + str(penalty))
user_password_score += penalty
print("Final score: " + str(user_password_score))

这是输出：

User password score: 42
Plain list of triplets: ['qwe', 'wer', 'ert', 'rty', 'tyu', 'yui', 'uio', 'iop', 'opa', 'pas', 'asd', 'sdf', 'dfg', 'fgh', 'ghj', 'hjk', 'jkl', 'klz', 'lzx', 'zxc', 'xcv', 'cvb', 'vbn', 'bnm', 'aze', 'zer', 'ert', 'rty', 'tyu', 'yui', 'uio', 'iop', 'opq', 'pqs', 'qsd', 'sdf', 'dfg', 'fgh', 'ghj', 'hjk', 'jkl', 'klm', 'lmw', 'mwx', 'wxc', 'xcv', 'cvb', 'vbn', 'abc', 'bcd', 'cde', 'def', 'efg', 'fgh', 'ghi', 'hij', 'ijk', 'jkl', 'klm', 'lmn', 'mno', 'nop', 'opq', 'pqr', 'qrs', 'rst', 'stu', 'tuv', 'uvw', 'vwx', 'wxy', 'xyz', '012', '123', '234', '345', '456', '567', '678', '789', '890']
Regex expression: (?=((qwe)|(wer)|(ert)|(rty)|(tyu)|(yui)|(uio)|(iop)|(opa)|(pas)|(asd)|(sdf)|(dfg)|(fgh)|(ghj)|(hjk)|(jkl)|(klz)|(lzx)|(zxc)|(xcv)|(cvb)|(vbn)|(bnm)|(aze)|(zer)|(ert)|(rty)|(tyu)|(yui)|(uio)|(iop)|(opq)|(pqs)|(qsd)|(sdf)|(dfg)|(fgh)|(ghj)|(hjk)|(jkl)|(klm)|(lmw)|(mwx)|(wxc)|(xcv)|(cvb)|(vbn)|(abc)|(bcd)|(cde)|(def)|(efg)|(fgh)|(ghi)|(hij)|(ijk)|(jkl)|(klm)|(lmn)|(mno)|(nop)|(opq)|(pqr)|(qrs)|(rst)|(stu)|(tuv)|(uvw)|(vwx)|(wxy)|(xyz)|(012)|(123)|(234)|(345)|(456)|(567)|(678)|(789)|(890)))
Matches : ['qwe', 'wer', '123']
Penalty: -15
Final score: 27

在build_pattern函数中，[ x for y in all_triplets for x in y ]是将列表列表扩展为纯列表的一种技巧。在(lmw)|(mwx)|(wxc)中使用的类似finditer()的正则表达式模式告诉我们要查找lmw，mwx和wxc的所有匹配项。当我们将这种模式包装在前瞻（(?=())）中时，我们在告诉re它也应该在结果中包含重叠的匹配项。