How to convert a string to its Base-10 representation?

Question

Is there any python module that would help me to convert a string into a 64-bit integer? (the maximum length of this string is 8 chars, so it should fit in a long).

I would like to avoid having to write my own method.

Example:

Input String   Hex          result (Base-10 Integer)
'Y'            59           89
'YZ'           59 5a        22874
...

Roman Bodnarchuk · Accepted Answer · 2012-05-23T09:26:17.423

9

This is a job for struct:

>>> s = 'YZ'
>>> struct.unpack('>Q', '\x00' * (8 - len(s)) + s)
(22874,)

Or a bit trickier:

>>> int(s.encode('hex'), 16)
22874

edited May 23 '12 at 09:26

answered May 23 '12 at 09:17

Roman Bodnarchuk

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Great! Is there an easy way to perform the opposite conversion, i.e. get the string 'YZ' from the integer 22874? – John J. Camilleri Jun 25 '14 at 14:46
1

@JohnJ.Camilleri Sure - try `struct.pack('>H', 22874)`. – Roman Bodnarchuk Jun 25 '14 at 16:23
will this return the same number every time across versions and platforms (unlike say `hash(s)`) ? – Anentropic Feb 15 '17 at 18:50
@Anentropic yes – Roman Bodnarchuk Feb 15 '17 at 19:24

score 4 · Answer 2 · answered May 23 '12 at 09:13

4

I don't think there's a built-in method to do this, but it's easy enough to cook up:

>>> int("".join([hex(ord(x))[2:] for x in "YZ"]), 16)
22874

This goes via base 16 which can of course be optimized out. I'll leave that "as an exercise".

answered May 23 '12 at 09:13

unwind

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take a look at 'hex' encoding :) – Roman Bodnarchuk May 23 '12 at 09:26

score 3 · Answer 3 · answered May 23 '12 at 09:17

3

Another way:

sum(ord(c) << i*8 for i, c in enumerate(mystr))

answered May 23 '12 at 09:17

Tim Pietzcker

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score 1 · Answer 4 · answered May 23 '12 at 09:16

1

>>> reduce(lambda a,b: a*256+b, map(ord,'YZ'), 0)
22874

answered May 23 '12 at 09:16

Dan D.

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How to convert a string to its Base-10 representation?

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