Is there any "easy" way to convert a string of letters to an integer value? I know a method that would work, but it would be very tedious. Oh and in Python, I forgot to mention that. It would be like "ABC" = 123
2
-
3What would "XYZ" translate to?Robᵩ– Robᵩ2017-01-25 18:33:49 +00:00Commented Jan 25, 2017 at 18:33
-
Yes, that's quite a good question...Jacques de Hooge– Jacques de Hooge2017-01-25 18:39:01 +00:00Commented Jan 25, 2017 at 18:39
Add a comment
|
3 Answers
Pre-define a dictionary that maps letters to numbers:
d = {'A': '1', 'B': '2', 'C': '3', 'D': '4',
'E': '5', 'F': '6', 'G': '7', 'H': '8',
'I': '9', 'J': '10', 'K': '11', 'L': '12',
'M': '13', 'N': '14', 'O': '15', 'P': '16',
'Q': '17', 'R': '18', 'S': '19', 'T': '20',
'U': '21', 'V': '22', 'W': '23', 'X': '24',
'Y': '25', 'Z': '26'}
Then you can convert with:
int(''.join(d[c] for c in 'ABC'))
Comments
print ([ord (c) - ord ('A') + 1 for c in 'ABC'])
N.B. Rob's answer is better... (upto the 'i' ;=))
1 Comment
michael2103
Thanks, this is what i needed
Your problem is underspecified. There are literally infinite functions that will convert 'ABC' to 123, including this one:
def letter2number(s):
return 123
If you want to covert the strings by concatenating the decimal digits of the letter's offset into the alphabet, this one might work for you. Note that you can't round-trip the value: 24 could be X or it could be BD.
def letter2number(s):
return int(''.join('%d'%(ord(ch)-ord('A')+1) for ch in s))
assert letter2number('ABC') == 123
assert letter2number('XYZ') == 242526
3 Comments
Jacques de Hooge
So add 100 to each code (to keep it simple) and then glue together.
Robᵩ
But then
'ABC' won't yield 123, which was a requirement.Jacques de Hooge
Indeed. So it would need to be a variable lenght code, perhaps with the 0 as extension indicator. But I guess that's not what OP was looking for...