Re: How to covert ASCII to integer in Python?
- From: Mensanator <mensanator@xxxxxxx>
- Date: Fri, 30 May 2008 17:59:03 -0700 (PDT)
On May 30, 6:44 pm, Joshua Kugler <jkug...@xxxxxxxxxxx> wrote:
Skonieczny, Chris wrote:
YOU SHOULD REMOVE or CORRECT YOUR POST here:
http://mail.python.org/pipermail/python-list/2007-February/427841.html
It is not true - eg. try :
a='P' # P is ASCII , isn't it ?
b=int(a)
and what you will get ? An error !!!
Or probably you yourself should - quote :
"You probably should go through the tutorial ASAP that is located here:
http://docs.python.org/tut/"
int() converts a strings that is a valid intenter. What you're looking for
is ord().
Possibly. Perhaps the OP wants the position of 'P'
in the alphabet, in which case he wants b=64-ord(a)
or b=16.
In [1]: ord('d')
Out[1]: 100
In [2]: a='P'
In [3]: b=ord(a)
In [4]: b
Out[4]: 80
j
.
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