How to return the ASCII string after a given string of numbers?-CodePudding

in order to print the ASCII I need to reverse the string so that what I've done and then to take for example 41 11 etc. and convert it to char. The outcome should be a word(in this case, Hacker)

The conversion need to be according to this: 1.the value range from A to Z is from 65 to 90 2.the value range from a to z is from 97 to 122 3.the value of the space character is 32

Do you have any idea how to do that?

Thank you in advance!

that's what I get currently: I get Unicode characters

['41', '11', '01', '10', '17', '99', '99', '27'] 41 11 01 10 17 99 99 27 ) cc

s="729799107101114"

rs=s[::-1]

import re
new=re.findall('..',rs)
print(new)
n=""
for num in new:
    n =num " "
print(n)

j=''.join(chr(int(i)) for i in n.split())
print(j)

CodePudding user response：

The input presented numbers on ASCII codes. i.e 72 is the H char.

Then the outcome is look like this:

s="72 97 99 107 101 114"
j=''.join(chr(int(i)) for i in s.split())
print(j)

[Output]

Hacker

[EDIT]:

In this part, we'll tokenize the numbers as it, without space.

s="721111193211611132114101116117114110321161041013265836773733211511611410511010332971021161011143297321031051181011103211511611410511010332111102321101171099810111411563"
j=''
start=0
step=2
while start < len(s)-1:
    n=s[start:start 2]
    if 32<=int(n)<=99:
        start =2
    if int(n)<32:
        n=s[start:start 3]
        start =3        
    j=j (chr(int(n)))
print(j)

[Output2:] In this sequence numbers sample we'll get this sentence:

How to return the ASCII string after a given string of numbers?

CodePudding user response：

A slightly less verbose version:

res = ''
start = end = 0
while end < len(s):
    end = start   2   (s[start] == '1')
    res  = chr(int(s[start:end]))
    start = end