On Wednesday 25 March 2015 14:13, otaksoftspamt...@gmail.com wrote:
> I have a list containing 9600 integer elements - each integer is either 0
> or 1.
>
> Starting at the front of the list, I need to combine 8 list elements into
> 1 by treating them as if they were bits of one byte with 1 and 0 denoting
> bit on/off (the 8th element would be the rightmost bit of the first byte).
>
> The end result should be a new list that is 8 x shorter than the original
> list containing integers between 0 and 255.
>
> Speed is not of utmost importance - an elegant solution is. Any
> suggestions?
Collate the list into groups of 8. Here, I pad the list with zeroes at the
end. If you prefer to drop any excess bits instead of padding them, use
itertools.izip instead of izip_longest.
import itertools
mylist = [1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1]
grouped = itertools.izip_longest(*([iter(mylist)]*8), fillvalue=0)
Now convert each group of eight into a byte:
def byte(bits):
n = 0
for b in bits:
assert b in (0, 1)
n = n*2 + b
return n
[byte(x) for x in grouped]
Or if you prefer using built-ins:
[int(''.join(str(b) for b in x), 2) for x in grouped]
I have no idea which will be faster.
Exercise for the reader: izip will drop any bits that don't make up an
octet. izip_longest will pad with zeroes on the least-significant side, e.g.
[1, 1] -> 192. How to pad on the most-significant side, or equivalently,
don't pad at all, so that [1, 1] -> 3?
--
Steve
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