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python使用正则筛选信用卡

2021-05-23 10:50水似冰 Python

这篇文章主要为大家详细介绍了python使用正则筛选信用卡,具有一定的参考价值,感兴趣的小伙伴们可以参考一下

本文实例为大家分享了python使用正则筛选信用卡的具体代码,供大家参考,具体内容如下

本文来源于两个简单的题目:

1.判断一对单词是否是" anagrams "
2.判断信用卡是否合理

判断 anagramsstrong>

anagrams 的百度翻译:由颠倒字母顺序而构成的字(短语)

而题目给出例子:

[ dog , odg ]
[ dog , dog ]
[ dog , god ]
[ dog , gdo ]

均为 anagrams 。

那思路就简单了,直接拆分字母,排序,比较就 ok 。

判断信用卡

题目给出的要求如下:

the criteria are:
• it must start with a 4,5 or 6
• it must be exactly 16 digits
• it must be numbers only
• it can have a digits in groups of 4, separated by one hyphen “-“
• it should not contain any other characters.
• it must not have any 4 repeated digits.

样例输出如下:

378282246310005 invalid
30569309025904 invalid
6011111111111117 invalid
5123-2332-3232-3213 valid

py文件

两个题目合并在一个 py 文件中。 
而入参数分别是两个文件的名字,一个是 anagram.txt ,另一个是 credit_cards.txt ,他们分别长这样:

python使用正则筛选信用卡

python使用正则筛选信用卡

最后的程序长这样:

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#! /usr/bin/python3
import re
###############################
##you need to implet the following methods:
##
##question 1
##anagram_validator()
##
##question 2
##credit_card_validator()
 
################################
 
###################################################
# question 1: check for anagrams:#
###################################################
 
def get_list(src_str):
 sub_list=[]
 word_tmp=''
 for word in src_str:
 if word != ',':
 word_tmp = word_tmp+word
 else:
 sub_list.append(word_tmp)
 word_tmp = ''
 return sub_list
 
def read_anagram(file_name):
 '''
 input: a file name
 return: a nested list of two words list
 example : [[word1,word2],[word3,word4]...etc]
 '''
 with open(file_name,encoding = 'utf-8') as f:
 return_list =[]
 file_conment = f.read()
 
 file_conment_new = re.sub(r'\n',',',file_conment)
 file_conment_new = file_conment_new+','
 subs_list = get_list(file_conment_new)
 i=0
 while i < (len(subs_list)-1):
 return_list.append([subs_list[i],subs_list[i+1]])
 i=i+2
 return return_list
 
 
 
def anagram_validator(anagram):
 '''
 input is the output from "read_anagram()".
 return: list of "anagrams" or "not anagrams" values for each two words
 example input (dog,gdo),(try,elm) then output would be ["anagrams","not anagrams"] with sequence of the input
 '''
 result_list=[]
 for i in range(len(anagram)):
 word_font = ''.join((lambda x:(x.sort(),x)[1])(list(anagram[i][0])))
 word_back = ''.join((lambda x:(x.sort(),x)[1])(list(anagram[i][1])))
 if word_font == word_back:
 result_list.append('anagrams')
 else:
 result_list.append('not anagrams')
 
 return result_list
 
############################################
# question 2: validate credit cards #
############################################
def read_credit_cards(file_name):
 '''
 input: a file name
 return tuple of numbers
 '''
 with open(file_name,encoding='utf-8') as f:
 card_conment = f.read()
 
 card_conment = re.sub(r'\[','',card_conment)
 card_conment = re.sub(r'\]','',card_conment)
 card_conment = card_conment+','
 card_list = get_list(card_conment)
 return tuple(card_list)
 
 
def credit_card_validator(numbers):
 '''
 input: tuple of numbers
 return:dictionary of credit card numbers where key is the number and value if valid or invalid
 '''
 validator_list=[]
 credit_dict={}
 for i in range(len(numbers)):
 validator_list.append(numbers[i])
 
 for i in range(len(numbers)):
 ch_num = 0
 repeat_flag = 0
 for ch in validator_list[i]:
 if ord("a")<=ord(ch)<=ord("z"):
 ch_num+=1
 break
 
 tmp_list = []
 tmp_list = (re.sub(r"[^\d]", "", validator_list[i]))
 for index in range(len(tmp_list)-4):
 if(tmp_list[index] == tmp_list[index+1] and
 tmp_list[index] == tmp_list[index+2] and
 tmp_list[index] == tmp_list[index+3] ):
 repeat_flag = 1
 break;
 
 if(validator_list[i][0] != '4' and
 validator_list[i][0] != '5' and
 validator_list[i][0] != '6' ):
 credit_dict[validator_list[i]]='invalid'
 elif( len(re.sub(r"\d", "", validator_list[i])) != 16 ):
 credit_dict[validator_list[i]]='invalid'
 elif(ch_num > 0):
 credit_dict[validator_list[i]]='invalid'
 elif(
 (re.search(r"-",validator_list[i])) and
  ( validator_list[i][4] != '-' or
  validator_list[i][9] != '-'or
  validator_list[i][14] != '-'
  )
 ):
 credit_dict[validator_list[i]]='invalid'
 elif(re.search(r"[^-\d]",validator_list[i])):
 credit_dict[validator_list[i]]='invalid'
 elif(repeat_flag == 1 ):
 credit_dict[validator_list[i]]='invalid'
 else :
 credit_dict[validator_list[i]]='valid'
 space_str = ''
 
 return credit_dict
 
 
def print_credit_card_summary(dict_o):
 '''
 input: dict
 return:
 printing summary of validation result - space between credit card and status is 40 width
 example:
 378282246310005 invalid
 30569309025904 invalid
 '''
 space_str = ''
 for key in dict_o:
 new_str = key + dict_o[key]
 if(dict_o[key] == "valid"):
 new_str_tmp = new_str[:-5]
 space_lenth = 46-len(new_str)
 for x in range(space_lenth):
 space_str +=' '
 print(new_str_tmp + space_str + 'valid')
 space_str=''
 else :
 new_str_tmp = new_str[:-7]
 space_lenth = 48-len(new_str)
 for x in range(space_lenth):
 space_str +=' '
 print(new_str_tmp + space_str + 'invalid')
 space_str=''
 
 
 
####### the code below is for testing###################
############### do not change #########################
 
 
import sys
 
if __name__ == '__main__':
 # take care of the console inputs
 if len(sys.argv) <= 1:
 sys.argv = ['', "anagram.txt", "credit_cards.txt"]
 stars = '*' * 40
 print(stars)
 print("testing question 1 --- anagrams?")
 print(stars)
 
 # testing reading_anagrams
 try:
 anagram = read_anagram(sys.argv[1])
 if not anagram:
 print("read_anagram() returns none.")
 else:
 print("anagram: ", anagram)
 print()
 except exception as e:
 print("error (readnumbers()): ", e)
 
 # testing anagram_validator
 anagrams = 0
 nanagrams = 0
 try:
 if not anagram: # question 1 has not been implemented
 print("anagram_validator() skipped....")
 else:
 result = anagram_validator(anagram)
 if result == none:
 print("anagram_validator() returns none.")
 else:
 for i in result:
  if i == "anagrams":
  anagrams += 1
  elif i == "not anagrams":
  nanagrams += 1
 print("number of valid anagrams is {} and not anagrams is {}.".format(anagrams, nanagrams))
 
 except exception as e:
 print("error (anagram_validator()):", e)
 
 # testing question 2
 
 print("\n\n" + stars)
 print("testing question 2 --- credit card validator")
 print(stars)
 
 # testing reading_credit_cards
 try:
 tup = read_credit_cards(sys.argv[2])
 if not tup:
 print("read_credit_cards() returns none.")
 else:
 print("the tuple of credit_cards: {}".format(tup))
 except exception as e:
 print("error (read_credit_cards()):", e)
 
 # testing credit_card_validator
 vcc = 0
 ivcc = 0
 try:
 if not tup: # readin_question 2 has not been implemented
 print("credit_card_validator() skipped...")
 else:
 cc_dict = credit_card_validator(tup)
 tmp_cc_dict = cc_dict
 if not cc_dict:
 print("credit_card_validator() returns none.")
 else:
 for items in cc_dict.keys():
  if cc_dict[items] == "valid":
  vcc += 1
  elif cc_dict[items] == "invalid":
  ivcc += 1
 print("number of valid credit cards is {} and invalid {}.".format(vcc, ivcc))
 except exception as e:
 print("error (credit_card_validator()):", e)
 
 # testing question 2
 print("\n\n" + stars)
 print("testing question 2b --- print credit card summary")
 print(stars)
 # testing print_credit_card_summary
 try:
 if not tmp_cc_dict: # dict credit card output has not been implemented
 print("print_credit_card_summary() skipped...")
 else:
 import io # do not delete this line
 from contextlib import redirect_stdout # do not delete this line
 
 f = io.stringio()
 with redirect_stdout(f):
 print_credit_card_summary(tmp_cc_dict)
 out = f.getvalue()
 if not out:
 print("print_credit_card_summary() returns none.")
 else:
 count44 = 0
 count46 = 0
 for line in out.splitlines():
  if len(line) - len(line.split()) == 44:
  count44 += 1
  elif len(line) - len(line.split()) == 46:
  count46 += 1
 if count44 == vcc and count46 == ivcc:
  print("your format looks good")
 else:
  print("you might have some issues in your summary format")
 
 
 except exception as e:
 print("error (print_credit_card_summary()):", e)

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持服务器之家。

原文链接:https://blog.csdn.net/qq_30650153/article/details/84110948

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