Can anyone explain to me why the following snippet doesn't work? The resulting hex string will be only two characters long.
#!/usr/bin/python
s = 'Hello, World!'
hs = ''
for i in range(len(s)):
c = s[i:1]
hs += c.encode('hex')
print hs
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2 Answers
Because on each loop, you're trying to slice from i (which is increasing) to position 1 - which means after i > 1, you get empty strings...
It looks though, that you're doing:
from binascii import hexlify
s = 'Hello, World!'
print hexlify(s)
... the hard way...
1 Comment
matsp888
And I know it can be done in a simpler way, I just wondered why this one wasn't working, and the reason was that I thought the second figure after the colon in the slice notation was a length, not an index.
c = s[i:1] should be c = s[i:i+1] or c[i]
In python you can loop over the string itsellf, so no need of slicing in your example:
hs = ''
for c in s:
hs += c.encode('hex')
or a one-liner using str.join, which is faster than concatenation:
hs = "".join([c.encode('hex') for c in s])
answered May 25, 2013 at 12:09
Ashwini Chaudhary
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7 Comments
matsp888
Yes, of course. I've been doing Perl until now, so I'm rather new at this. Thank you very much.
matsp888
Yes, I know that, I just used this as a silly example of something that I didn't think was working.
Peter Varo
Oh, I love your one-liner, so pretty and elegant :)
Peter Varo
@Ashiwini Chaudhary just one complement:
join() also works with generator, not just lists, so: this is valid: hs = "".join((c.encode('hex') for c in s))
Jon Clements
@Volatility the problem is that
.encode('hex') won't work in Python 3.x, which is why I prefer to use hexlify |