2009/4/29 Tim Chase <python.l...@tim.thechases.com>:
>> I want to concatenate two bits string together: say we have '1001' and
>> '111' which are represented in integer. I want to concatenate them to
>> '1001111' (also in integer form), my method is:
>> ('1001' << 3) | 111
>> which is very time consuming.
>
> You omit some key details -- namely how do you know that "1001" is 4 bits
> and not "00001001" (8-bits)? If it's a string (as your current code shows),
> you can determine the length. However, if they are actually ints, your code
> should work fine & be O(1).
Actually, what I have is a list of integer numbers [3,55,99,44], and
by using Huffman coding or fixed length coding, I will know how the
bits-length for each number. When I try to concatenate them (say
10,000 items in the list) all together, the speed is going down
quickly (because of the shifting operations of python long).
>
> This can be abstracted if you need:
>
> def combine_bits(int_a, int_b, bit_len_b):
> return (int_a << bit_len_b) | int_b
>
> a = 0x09
> b = 0x07
> print combine_bits(a, b, 3)
>
> However, if you're using gargantuan ints (as discussed before), it's a lot
> messier. You'd have to clarify the storage structure (a byte string? a
> python long?)
I am using a single python long to store all the items in the list
(say, 10,000 items), so the work does become messier...
>
> -tkc
>
> PS: You may want to CC the mailing list so that others get a crack at
> answering your questions...I've been adding it back in, but you've been
> replying just to me.
Sorry, this is the first time I am using mail-list....and always
forgot "reply to all"
Thank you very much:D
>
>
>
>
--
Li
------
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and you may find one day
you have less than you think
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