如何将包含各种ASCII码的int值转换为相应的字符串？

Question

Say I have a bunch of ASCII codes joined into one int value. For eg - My name "Rahul" yields 8297104117108( x =''.join(str(ord(c))for c in "Rahul" ). How do I convert this integer back to the word that is formed using the ASCII codes?

Answer 1

If you limit this to printable characters ( ascii >= 32 ) in an 8 bit character set (extended ASCII), there is actually no ambiguity. Each characters will use either 2 or 3 digits. Characters with two digits will be >= 32 and characters with 3 digits will be <= 255 (which happens to start with a two digit number below 32). So two consecutive digits < 32 can only be the beginning of a 3 digit character.

def decodeStr(s):
    if s == "": return ""
    code = s[:3] if s[:2] < "32" else s[:2]
    return chr(int(code)) + decodeStr(s[len(code):])

sa = decodeStr("8297104117108")
print(sa) # "Rahul"

If you have a very long string to decode, the recursive approach might hit Python's maximum recursion limit. Here is an iterative variant of the function:

def decodeStr(s):
    result = ""
    code   = ""
    for digit in s:
        code += digit
        if len(code) < 2 or code < "32" : continue
        result += chr(int(code))
        code = ""
    return result

The iterative function will run about 1.8x faster than the recursive one.

Answer 2

here is a simple, but not very efficient solution:

d = '8297104117108'
l = []
while d:
    if d[0] == '1':
        l.append(chr(int(d[:3])))
        d = d[3:]
    else:
        l.append(chr(int(d[:2])))
        d = d[2:]
print(''.join(l))

this assumes that the range of you characters is pretty limited, if you can mess with the "encryption", then pad your number to be of length 3. like so:

my_code = ''.join('{:03}'.format(ord(c))for c in "Rahul")

this will give you '082097104117108' which you can then read 3 at a time

''.join([chr(int(my_code[i:i+3])) for i in range(0,len(my_code),3)])

Answer 3

An incremental solution assuming all codes are in range(65,650) , so you can extend your text to all latin extensions :

def decode(code):
    res=''
    car=0
    for digit in code:
         car = 10*car + int(digit)
         if car >= ord('A'):
             res += chr(car)
             car=0
    return res

decode('8297104117108') is 'Rahul' , decode('8297250108') is 'Raúl' .