Dumb python questions
slinkp23 at yahoo.com
Sun Aug 19 03:14:17 CEST 2001
On 18 Aug 2001 12:43:29 -0700, Paul Rubin <phr-n2001 at nightsong.com> wrote:
>m.faassen at vet.uu.nl (Martijn Faassen) writes:
>> > I think I'm reacting not as much to language differences as to
>> > differences in system maturity. Whenever I write something in Perl or
>> > CL, and I need to do something like turn an array into a string,
>> Do you mean a Python list? ''.join(mylist)
>No I mean a list of character values, like (97,98,99) => "abc".
Use the chr() built-in function to convert each list item to
ascii. For this, it's handy to use either a list comprehension (new
for python 2.x), or the map() function. Either way, you can then join
the result into a string. One example:
>>> ilist = range(65, 70)
>>> alist = [chr(x) for x in ilist]
>>> # OR: alist = map(chr, ilist)
['A', 'B', 'C', 'D', 'E']
>> > somebody seems to have faced that problem before and it's been taken
>> > care of in some reasonably thought-out way. In Python, I often feel
>> > like I'm doing something that hasn't been done much before.
>> In part I suspect that's due to your inexperience with the language, but
>> please keep pointing out the things that are missing. :)
>Here's one: say I have a long int, like x = 7**77777L. How can I tell
>how many bits it takes to express that number in binary?
I think most people deal with binary data in python as strings.
fastest thing I can think of is to use oct() or hex() to convert your
number to a string representation, then compute from the length of
For example, if you can work with a count of nibbles rather than
bits, this will work:
h = hex(long(x))
return len(h) - 3 # compensate for leading '0x' and trailing 'L'
>From this you can do bytes:
n = nibbles_needed(x)
bytes, remainder = divmod(n, 2)
if remainder: return bytes + remainder
else: return bytes
A test run:
> How can I
>convert it into a packed character string (i.e. represent it in string
>form with 8 bits/char so I can write it to a file)? The answers I can
>think of involve weird kludges and/or are unacceptably slow.
Hmmm... my first thought was struct.pack but you'd first need to chop
the long up into (literally) byte-size pieces. And even when you've
done that, struct.pack() won't take a list as an argument.
Aside from struct, there's the binascii module which you might find
useful. For example, here's a pretty fast way to do what you asked
>>> from binascii import a2b_hex
>>> def charlist(x):
... x = long(x)
... # Now create a hex representation,
... # but without the leading 'Ox' or trailing 'L'.
... h = hex(x)[2:-1]
... # Split that up into actual bytes.
... bytes = [h[i:i+2] for i in range(0,len(h),2)]
... chars = [a2b_hex(b) for b in bytes]
... return chars
>>> charlist((97 * 2 ** 8) + 98)
>>> x = charlist(7**77777L)
Hope that helps!
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