将整个 ascii 字符串转换为 int

Question

I am looking to convert a string like "ASDF" into one number like "123456789" that can easily be turned back into "ASDF", how would I go about doing this in python 3?

Answer 1

You can convert each character into its ascii value using ord and format them to be the same length using f"{1:02d}" and concatenate them together. Then when you want the string back reverse this process.

Something like

def to_num(s):
    return int("1" + "".join(map(lambda a: f"{ord(a):03d}", s)))

def to_str(n):
    num = str(n)[1:]
    chunks = [num[i:i+3] for i in range(0, len(num), 3)]
    return "".join(map(lambda a: chr(int(a)), chunks))

Answer 2

How about that:

int.from_bytes("ASDF".encode('utf-8'), byteorder='big', signed=False)
1095976006

and back

import math
(1095976006).to_bytes(math.ceil((1095976006).bit_length() / 8), byteorder = 'big', signed=False).decode('utf-8')
'ASDF'

It uses encode/decode to get utf-8 representation of Python string as byte array and then from_bytes/to_bytes to convert it to integer. SO it works with any string. Surprisingly, it is necessary to calculate number of bytes in order to use to_bytes.

Answer 3

Just figured out a possible way, since I am only using ASCII this should work:

inp = "ASDF" #Input string
bits = 8     #Amount of bits per character

num = 0      #Just some setup value, this will be our encoded number
numstr = [ord(letter) for letter in inp] #Turns the string into a list containing each of the letters ascii code
for numletter in numstr: #Loop over each letter
    num = (num<<bits)+numletter #First shift the current num stored by 8 [In binary 01010101 becomes 0101010100000000] which makes room for the next character, then adds the new character in

print("Encoded:",num) #Print the encoded number

#num = 1234, not included as we've defined it before with the encoding part
#bits = 8, same reason as above
outp = "" #Another setup value, this will be our decoded string
AND = ((2**bits)-1) #A setup constant, we'll need this later, this is a number of just 1's in binary with the length of 8.
while num>=1: #Loop over each character
     numletter = num&AND #Bitwise AND it with the constant we got earlier, this'll extract the first character in the number
     outp = chr(numletter) + outp #First convert the extracted number to it's character, then add the character to the front of the output
     num = num>>bits #Remove the extracted number and shift it back by 8

print(outp) #Print the decoded string

Answer 4

Kinda problematic by definition. Depending on the length of the string you can easily run out of ints. Unless you mean number in string representation, then it's theoretically possible, but practically you will need arithmetic that works on those int as string numbers. Basically, if you are restricted to ascii, use base 26 representation:

"F"*26^0 + "D"*26^1 + "S" * 26^ 2 and so on

Or you can just concatenate the character codes, it's also an option, although it seems rather pointless - you don't really covert string to number, you just print its representation in another way.

Or perhaps, you are talking about mapping closed set of string to integers - then just put your strings in the list and use their index as the matching int.