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Changing version number to single digits python

I have a version number in a file like this:
Testing x.x.x.x
So I am grabbing it off like this:
import re
def increment(match):
# convert the four matches to integers
a,b,c,d = [int(x) for x in match.groups()]
# return the replacement string
return f'{a}.{b}.{c}.{d}'
lines = open('file.txt', 'r').readlines()
lines[3] = re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, lines[3])
I want to make it so if the last digit is a 9... then change it to 0 and then change the previous digit to a 1. So 1.1.1.9 changes to 1.1.2.0.
I did that by doing:
def increment(match):
# convert the four matches to integers
a,b,c,d = [int(x) for x in match.groups()]
# return the replacement string
if (d == 9):
return f'{a}.{b}.{c+1}.{0}'
elif (c == 9):
return f'{a}.{b+1}.{0}.{0}'
elif (b == 9):
return f'{a+1}.{0}.{0}.{0}'
Issue occurs when its 1.1.9.9 or 1.9.9.9. Where multiple digits need to rounded. How can I handle this issue?

Use integer addition?
def increment(match):
# convert the four matches to integers
a,b,c,d = [int(x) for x in match.groups()]
*a,b,c,d = [int(x) for x in str(a*1000 + b*100 + c*10 + d + 1)]
a = ''.join(map(str,a)) # fix for 2 digit 'a'
# return the replacement string
return f'{a}.{b}.{c}.{d}'

If your versions are never going to go beyond 10, it is better to just convert it to an integer, increment it and then convert back to a string.
This allows you to go up to as many version numbers as you require and you are not limited to thousands.
def increment(match):
match = match.replace('.', '')
match = int(match)
match += 1
match = str(match)
output = '.'.join(match)
return output

Add 1 to the last element. If it's more than 9, set it to 0 and do the same for the previous element. Repeat as necessary:
import re
def increment(match):
# convert the four matches to integers
g = [int(x) for x in match.groups()]
# increment, last one first
pos = len(g)-1
g[pos] += 1
while pos > 0:
if g[pos] > 9:
g[pos] = 0
pos -= 1
g[pos] += 1
else:
break
# return the replacement string
return '.'.join(str(x) for x in g)
print (re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, '1.8.9.9'))
print (re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, '1.9.9.9'))
print (re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, '9.9.9.9'))
Result:
1.9.0.0
2.0.0.0
10.0.0.0

python intelligent hexadecimal numbers generator

I want to be able to generate 12 character long chain, of hexadecimal, BUT with no more than 2 identical numbers duplicate in the chain: 00 and not 000
Because, I know how to generate ALL possibilites, including 00000000000 to FFFFFFFFFFF, but I know that I won't use all those values, and because the size of the file generated with ALL possibilities is many GB long, I want to reduce the size by avoiding the not useful generated chains.
So my goal is to have results like 00A300BF8911 and not like 000300BF8911
Could you please help me to do so?
Many thanks in advance!

if you picked the same one twice, remove it from the choices for a round:
import random
hex_digits = set('0123456789ABCDEF')
result = ""
pick_from = hex_digits
for digit in range(12):
cur_digit = random.sample(hex_digits, 1)[0]
result += cur_digit
if result[-1] == cur_digit:
pick_from = hex_digits - set(cur_digit)
else:
pick_from = hex_digits
print(result)
Since the title mentions generators. Here's the above as a generator:
import random
hex_digits = set('0123456789ABCDEF')
def hexGen():
while True:
result = ""
pick_from = hex_digits
for digit in range(12):
cur_digit = random.sample(hex_digits, 1)[0]
result += cur_digit
if result[-1] == cur_digit:
pick_from = hex_digits - set(cur_digit)
else:
pick_from = hex_digits
yield result
my_hex_gen = hexGen()
counter = 0
for result in my_hex_gen:
print(result)
counter += 1
if counter > 10:
break
Results:
1ECC6A83EB14
D0897DE15E81
9C3E9028B0DE
CE74A2674AF0
9ECBD32C003D
0DF2E5DAC0FB
31C48E691C96
F33AAC2C2052
CD4CEDADD54D
40A329FF6E25
5F5D71F823A4
You could also change the while true loop to only produce a certain number of these based on a number passed into the function.

I interpret this question as, "I want to construct a rainbow table by iterating through all strings that have the following qualities. The string has a length of 12, contains only the characters 0-9 and A-F, and it never has the same character appearing three times in a row."
def iter_all_strings_without_triplicates(size, last_two_digits = (None, None)):
a,b = last_two_digits
if size == 0:
yield ""
else:
for c in "0123456789ABCDEF":
if a == b == c:
continue
else:
for rest in iter_all_strings_without_triplicates(size-1, (b,c)):
yield c + rest
for s in iter_all_strings_without_triplicates(12):
print(s)
Result:
001001001001
001001001002
001001001003
001001001004
001001001005
001001001006
001001001007
001001001008
001001001009
00100100100A
00100100100B
00100100100C
00100100100D
00100100100E
00100100100F
001001001010
001001001011
...
Note that there will be several hundred terabytes' worth of values outputted, so you aren't saving much room compared to just saving every single string, triplicates or not.

import string, random
source = string.hexdigits[:16]
result = ''
while len(result) < 12 :
idx = random.randint(0,len(source))
if len(result) < 3 or result[-1] != result[-2] or result[-1] != source[idx] :
result += source[idx]

You could extract a random sequence from a list of twice each hexadecimal digits:
digits = list('1234567890ABCDEF') * 2
random.shuffle(digits)
hex_number = ''.join(digits[:12])
If you wanted to allow shorter sequences, you could randomize that too, and left fill the blanks with zeros.
import random
digits = list('1234567890ABCDEF') * 2
random.shuffle(digits)
num_digits = random.randrange(3, 13)
hex_number = ''.join(['0'] * (12-num_digits)) + ''.join(digits[:num_digits])
print(hex_number)

You could use a generator iterating a window over the strings your current implementation yields. Sth. like (hex_str[i:i + 3] for i in range(len(hex_str) - window_size + 1)) Using len and set you could count the number of different characters in the slice. Although in your example it might be easier to just compare all 3 characters.

You can create an array from 0 to 255, and use random.sample with your list to get your list

Caesar Cipher shift by two letters

def main():
cc = (input("Enter Message to Encrypt\n"))#user input
shift = int(2) #shift length
a=["a","b","c","d","e","f","g","h","i","j","k","l",
"m","n","o","p","q","r","s","t","u","v","w","x","y","z"] #reference list
newa={} #new shifted reference list
for i in range (0,len(a)):
newa [a[i]]=a[(i+shift)%len(a)]
#adds shifted 2 alaphabet into newalaphabet
#% moodulus used to wrap
for i in cc: #iterates through cc
if i in a:
a[i]=cc[i]
a[i]=newa[i]
main()
So I need input from the user #cc
the shift needs to be two
I used an alphabet list
then shift the alphabet by two to create newa
but I do not know how to apply the new alphabet to my user's input

Use a dictionary to map inputs to outputs
shifted_a = a[-shift:] + a[:-shift]
cipher = {a[i]: shifted_a[i] for i in range(len(a))}
output = ''.join(cipher[char] for char in cc)

Iterate through the string cc and replace all the alphabets using the get method of newa. Characters that are not in the dictionary are left as is, by passing them as the default to newa.get when the key is missing:
newa = {}
for i, x in enumerate(a):
newa[x] = a[(i+shift) % len(a)]
encrypted_text = ''.join(newa.get(i, i) for i in cc)
Python's builtin enumerate can be used in place of range(len(a)) in this case where you need the items in a and their respective indices.

Use mapping for every char, then join them back to create the encrypted message:
''.join(map(lambda x: chr((ord(x) - 97 + shift) % 26 + 97) if x in alphabet else x, cc.lower()))
Integrate it like that:
import string
alphabet = string.ascii_lowercase
cc = input('Enter string to encode: ')
shift = 2 # could be any number
encrypted = ''.join(map(lambda x: chr((ord(x) - 97 + shift) % 26 + 97) if x in alphabet else x, cc.lower()))
cc.lower() for the letters to be all same case (to map using constant ord)
chr((ord(x) - 97 + shift) % 26 + 97) :
get the value of the number minus 97 (0 for a, 1 for b, etc.).
apply the shift (a turns to c, etc.).
modulate by 26 to prevent letters like z from exceeding (25 + 2 = 27, 27 % 26 = 1 = b).
add 97 to bring the letter back to ascii standard (97 for a, 98 for b, etc.)
if x in alphabet else x cover for signs that are not letter (if you want to ignore spaces and punctuation use if x in alphabet else '' instead).

I would just build transition table and use it to decode string.
import string
shift = 2
letters = string.ascii_lowercase + string.ascii_uppercase
transtable = str.maketrans({letters[i]: letters[(i + shift) % len(letters)]
for i in range(len(letters))})
cc = input('Enter string to encode: ')
print(cc.translate(transtable))

I'll throw my solution in there. It should be pretty clear how it works...
import string
index_lookup = {letter: index for index, letter in enumerate(string.ascii_lowercase)}
def caesar_letter(l, shift=2):
new_index = index_lookup[l] + shift
return string.ascii_lowercase[new_index % len(index_lookup)]
def caesar_word(s):
return ''.join([caesar_letter(letter) for letter in s])
I think the above is better for readability but if you're opposed to imports...
index_lookup = {chr(idx): idx - ord('a') for idx in range(ord('a'), ord('z')+1)}
...
In [5]: caesar_word('abcdefghijklmnopqrstuvwxyz')
Out[5]: 'cdefghijklmnopqrstuvwxyzab'

Python Simple Encryption program: How to loop Z back to A

I'm trying to make an encryption function that encrypts plaintext messages but the problem is that if i input a key too large that goes past 'Z' then it goes onto greater unicode values.
My code:
def encrypt(var1,var2):
var3 = ""
for i in range(0, len(var1)):
if ord(var1[i])>64 and ord(var1[i])<90:
var3=var3+chr((ord(var1[i])+var2))
elif ord(var1[i])+var2>90:
???
else:
continue
return(var3)
How do I get it to loop 'Z' back to 'A'. I think I have to make an if statement like this but I'm not sure what to put into it.
elif ord(var1[i])+var2>90:
???

Here is my one! Im using the modulus operator to wrap around every 26 numbers (the number of letter between a-z). I also handle upper on lowercase separately.
def encrypt(data, shift):
result = ''
for c in data:
c_num = ord(c)
# is the letter lower case a - z?
if (c_num >= ord('a')) and (c_num <= ord('z')):
# get the letter number from 0 - 26
c_num = c_num - ord('a')
# shift the number
c_num += shift
# wrap the number every 26 numbers
c_num = c_num % 26
# now increase a by the new amount
c_num += ord('a')
result += chr(c_num)
# is the letter upper case A - Z?
elif (c_num >= ord('A')) and (c_num <= ord('Z')):
# get the letter number from 0 - 26
c_num = c_num - ord('A')
# shift the number
c_num += shift
# wrap the number every 26 numbers
c_num = c_num % 26
# now increase a by the new amount
c_num += ord('A')
result += chr(c_num)
return result
encrypt('aAbB', 2)
'cCdD'
encrypt('afZz', 2)
'chBb'
Here is the code golf version using list comprehension just for fun!
def encrypt(data, shift):
return ''.join([chr(((ord(c) - ord('a') + shift) % 26) + ord('a')) if ord(c) in range(ord('a'), ord('z')+1) else chr(((ord(c) - ord('A') + shift) % 26) + ord('A')) for c in data])

A straight-forward way would be to check if you have passed beyond Z, and modify the character in that case:
...
if var1[i] >= 'A' and var1[i] <= 'Z':
translated_char = chr(ord(var1[i])+var2)
if translated_char > 'Z':
# If the resulting character is beyond Z,
# we go 26 characters back
translated_char = chr(ord(translated_char)-26)
# Append the translated character to the output string
var3 += translated_char
...
You may want to consider more descriptive variable names -- you'll thank yourself if you revisit your code after two months :-)

I would recommend using the modulus operator to do what you are wanting. In python that is the % character. In modulus math. X % Y tells us what the remainder of X / Y is. For example. 27 % 26 is 1. Using this you can get your wrap around that you want. Here is a sample bit of code to encrypt a single character
def encrypt_character( valToEncrypt, keyVal ):
# Update the character to be our standard Alphabet mapping
# A -> 0; B->1 ... Z -> 25
x = ord(valToEncrypt) - ord('A')
# Perform the Encryption
retVal = ( x + keyVal ) % 26
# Translate back to the standard ASCII mapping of the character
# for display in python and translate it back into a string
retVal = chr(retVal + ord('A'))
return retVal
# end encrypt_character
Now if we feed the character "A" Into our encryption algorithm with a key of 13 we get "N" as shown:
>>> encrypt_character("A", 13)
'N'
The decrypt algorithm is very similar, except you do subtraction instead of addtion
def decrypt_character( valToDecrypt, keyVal ):
# Update the character to be our standard Alphabet mapping
# A -> 0; B->1 ... Z -> 25
x = ord(valToDecrypt) - ord('A')
retVal = ( x - keyVal ) % 26
# Translate back to the standard ASCII mapping of the character
# for display in python and translate it back into a string
retVal = chr(retVal + ord('A'))
return retVal
To encrypt a string you can use the following function:
from re import sub
def encrypt_message( message, key ):
# Convert the message text into a plain text with all spaces and
# punctuation removed.
plainText = sub(r'[^A-Z]', '', message.upper())
cipherText = ""
charIndex = 0
# Encrypt the message 1 character at a time
while charIndex < len(plainText):
cipherText += \
encrypt_character( plainText[charIndex], key)
charIndex += 1
return cipherText
This function can be called:
>>> encrypt_message("HELLO World!", key=23)
'EBIILTLOIA'
The decryption function is very similar to the encryption function, except it calls the decrypt utility instead of the encrypt utility.

converting alphanumeric string to int and vice versa in python

I am trying to convert alphanumeric string with maximum length of 40 characters to an integer as small as possible so that we can easily save and retrieve from database. I am not aware if there is any python method existing for it or any simple algorithms we can use. To be specific my string will have only characters 0-9 and a-g. So kindly help with any suggestions on how we can uniquely convert from string to int and vice versa. I am using Python 2.7 on Cent os 6.5

This is not that difficult:
def str2int(s, chars):
i = 0
for c in reversed(s):
i *= len(chars)
i += chars.index(c)
return i
def int2str(i, chars):
s = ""
while i:
s += chars[i % len(chars)]
i //= len(chars)
return s
Example:
>>> chars = "".join(str(n) for n in range(10)) + "abcdefg"
>>> str2int("0235abg02", chars)
14354195089
>>> int2str(_, chars)
'0235abg02'
Basically if you want to encode n characters into an integer you interpret it as base-n.

There are 17 symbols in your input, so you can treat is as a base-17 number:
>>> int('aga0',17)
53924
For the reverse conversion, there are lots of solutions over here.

Improving on the above answers:
# The location of a character in the string matters.
chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
charsLen = len(chars)
def numberToStr(num):
s = ""
while num:
s = self.chars[num % charsLen] + s
num //= charsLen
return s # Or e.g. "s.zfill(10)"
Can handle strings with leading 0s:
def strToNumber(numStr):
num = 0
for i, c in enumerate(reversed(numStr)):
num += chars.index(c) * (charsLen ** i)
return num